SlideShare a Scribd company logo
1 of 28
PRACTICE EXERCISES
                                    LIS 397C


                   Introduction to Research in Information Studies
                                School of Information
                            University of Texas at Austin

                                      Dr. Philip Doty



                                         Version 3.5




Copyright Philip Doty, University of Texas at Austin, August 2004    1
LIS 397C
                                           Doty
                                   Practice Exercises 3.5


1.     Define the following terms and symbols:
       n
        ∑x
       x
        x
       s
       s2
       N
       µ
        σ
        σ   2

       coefficient of variation (CV)
       mode
       median
       arithmetic mean
       range
       variance
       interquartile range (IQR)
       standard deviation
       sample
       statistic
       parameter
       frequency distribution

2.     A sample of the variable x assumes the following values:

       9        11     13      3       7       2       8       9    6   10

       Compute:
       (a) n
       (b) ∑   x
       (c) x
       (d) s
       (e) s 2
       (f) median
       (g) mode
       (h) range


Copyright Philip Doty, University of Texas at Austin, August 2004            2
(i) CV




Copyright Philip Doty, University of Texas at Austin, August 2004   3
3.     A sample of the variable x assumes the following values:

       57      51      58      52      50      59      57      51   59   56
       50      53      54      50      57      51      53      55   52   54

       Generate a frequency distribution indicating x, frequency of x, cumulative
       frequency of x, relative frequency of x, and cumulative relative frequency
       of x.

4.     For the frequency distribution in problem 3, compute:

       (a) n
       (b) ∑   x
       (c) x
       (d) s
       (e) s 2
       (f) median
       (g) mode
       (h) range
       (i) CV

5.     Generate a histogram for the data in problem 3.

6.     Generate a frequency polygon for the data in problem 3.

7.     Define the following terms:

       skewness
       ordinate
       abscissa
       central tendency
       bimodal
       ordered pair
       Cartesian plane
       stem-and-leaf plot
       outlier
       reliability
       dispersion or variability
       negatively skewed
       positively skewed
       validity
       box plot
       whiskers


Copyright Philip Doty, University of Texas at Austin, August 2004               4
8.     What is the relationship between or among the terms?

       (a) sample/population
       (b) x /µ
       (c) s/ σ
       (d) variance/standard deviation
       (e) n/N
       (f) statistic/parameter
       (g) mean, median, mode of the normal curve
       (h) coefficient of variation/IQR

9.     Graph the following sample distributions (histogram and frequency
       polygon) using three different pairs of axes.

       Distribution 1          Distribution 2          Distribution 3

       Score | Frequency Score | Frequency Score | Frequency

       25               4       1              25       2            8
       26              10       2              31       4           20
       27               6       3              40       6           25
       28               3       4              44       8           35
       29               3       5              51      10           40
       30               1       6              19      12           45
       53               1       7              10      13           24
                               85               1      14           20

10.    For each of the distributions in problem 9, answer the following questions.

       (a)     Is the curve of the distribution positively or negatively skewed?
       (b)     What is n?
       (c)     Is the mode > median? Compute the answer and also answer it
               graphically, i.e., label the position of the mode and the median on
               the curve.
       (d)     Is the mean > mode? Compute the answer and also answer it
               graphically as in part (c) of this question.
       (e)     What is the variance of the distribution?
       (f)     What is the standard deviation of the distribution?
       (g)     What is the range of the distribution?
       (h)     What is the coefficient of variation of the distribution?
       (i)     Which measure of central tendency, mode, median, or (arithmetic)
               mean, is the fairest and clearest description of the distribution?


Copyright Philip Doty, University of Texas at Austin, August 2004                    5
Why?




Copyright Philip Doty, University of Texas at Austin, August 2004   6
11.    Define:
       error model
       quartile
       percentile
       Q1
       Q2
       Q3
       freq (x)
       cf (x)
       rel freq (x)
       cum rel freq (x)
       PR
       z-scores
        χ
       deviation score
       s of z-scores
        x of z-scores
        ∑x
       centile
       Interquartile Range (IQR)
       non-response bias
       self-selection

12.    Define the relationship(s) between or among the terms:
       Q 2 /median
       median/N or n
       z-score/raw score
       z-score/deviation score/standard deviation
       Q 1/ Q 2/ Q 3
       z-score/ χ /s or    σ
       median/fifth centile
       cf (x)/freq (x)/PR
       Literary Digest poll/bias

13.    The observations of the values of variable x can be summarized in the
       population frequency distribution below.

       x       freq (x)

       9         3
       8         9
       6         5


Copyright Philip Doty, University of Texas at Austin, August 2004              7
3         2
       2         6




Copyright Philip Doty, University of Texas at Austin, August 2004   8
For this distribution of x, calculate:

       (a) Cumulative frequency, relative frequency, and cumulative relative
            frequency for each value
       (b) N
       (c) the range
       (d) median
       (e) mode
       (f) µ
       (g)  σ
       (h) Q 1 , Q 2 , and Q 3
       (i) CV (coefficient of variation)
       (j) IQR
       (k) the percentile rank of x = 8, x = 2, x = 3
       (l) z-scores for x = 6, x = 8, x = 2, x = 3, x = 9

14.    Generate a box plot for the data in problem 13.

15.    Generate a box plot for the data in problem 3.

16.    For a normally distributed distribution of variable x, where µ = 50 and   σ
       = 2.5 [ND (50, 2.5)], calculate:

       (a) the percentile rank of x = 45
       (b) the z-score of x = 52.6
       (c) the percentile rank of x = 58
       (d) the 29.12th percentile
       (e) the 89.74th percentile
       (f) the z-score of x = 45
       (g) the percentile rank of x = 49

17.    Define:
        α
       sampling distribution
       Central Limit Theorem
       Standard Error (SE)
       ND(µ, )  σ
       decile
       descriptive statistics
       inferential statistics
       effect size
       confidence interval (C.I.) on µ


Copyright Philip Doty, University of Texas at Austin, August 2004                    9
Student's t
       degrees of freedom (df)
       random sampling
       stratified random sample




Copyright Philip Doty, University of Texas at Austin, August 2004   10
18.    Define the relationship(s) between or among the terms:

       E( x )/µ
        α /df/t
       Central Limit Theorem/C.I. on µ
        α /C.I. on µ when     σ
                             is known
        α /C.I. on µ when     σ
                             is not known
       z/confidence interval on µ
       t/C.I. on µ
       t/z

19.    The following values indicate the number of microcomputer applications
       available to a sample of 10 computer users.

       2, 5, 9, 5, 3, 6, 6, 3, 1, 13

       For the population from which the sample was drawn, µ = 4.1 and
        σ= 2.93.

       Calculate:

       a. The expected value of the mean of the sampling distribution of means
       b. The standard deviation of the sampling distribution of means.

20.    From previous research, we know that the standard deviation of the ages
       of public library users is 3.9 years. If the "average" age of a sample of 90
       public library users is 20.3 years, construct:

       a. a 95% C.I. around µ
       b. a 90% C.I. around µ
       c. a 99% C.I. around µ.

       What does a 95% interval around µ mean?

       What is our best estimate of µ?

21.    The sample size in problem 20 was increased to 145, while the "average"
       age of the sample remained at 20.3 years. Construct three confidence
       intervals around µ with the same levels of confidence as in Question 20.

22.    Determine t for a C.I. of 95% around µ when n equals 20, 9, ∞ , and 1.



Copyright Philip Doty, University of Texas at Austin, August 2004                 11
23.    Determine t on the same values of n (20, 9, ∞ , and 1) as in Question 22,
       but for a C.I. of 99% around µ. Should this new interval be narrower or
       wider than a 95% confidence interval on µ? Why? Answer the question
       both conceptually and algebraically.

24.    The following frequency distribution gives the values for variable x in a
       sample drawn from a larger population.

               x               freq(x)

               43               6
               42              11
               39               3
               36               7
               35               7
               34              15

       Calculate:

       (a) x
       (b) s
       (c) SEµ = σx
       (d) E( x )
       (e) Q 1 , Q 2 , and Q 3
       (f) mode
       (g) CV
       (h) IQR
       (i) a 95% C.I. on µ
       (j) the width of the confidence interval in part (i)
       (k) a 99% C.I. on µ
       (l) the width of the confidence interval in part (k)
       (m) PR (percentile rank) of x = 42
       (n) our best estimates of µ and       σ
                                          from the data.

25.    Generate a box plot for the data in problem 24.

26.    Construct a stem-and-leaf plot for the following data set; indicate Q 1 , Q 2
       , and Q 3 on the plot; and generate the six-figure summary. Be sure that
       you are able to identify the stems and the leaves and to identify their units
       of measurement.

       The heights of members of an extended family were measured in inches.
       The observations were: 62, 48, 56, 37, 37, 26, 74, 66, 29, 49, 72, 77, 69, 62,


Copyright Philip Doty, University of Texas at Austin, August 2004                       12
and 64.




Copyright Philip Doty, University of Texas at Austin, August 2004   13
27.    H 0 : There is no relationship between computer expertise and minutes
       spent doing known-item searches in an OPAC at      = 0.10.   α
       Should we reject the H0 given the following data? Remember that the
       acceptable error rate is 0.10.

                                       TIME (MINS)

       EXPERTISE               ≤5              > 5, ≤ 10                > 10

       Novice                  14              20                       19
       Intermediate            15              16                        9
       Expert                  22              11                        2


28.    Answer Question 27 at an acceptable error rate of 0.05.

29.    Define:

       statistical hypothesis
        Ho
        H1
       p
       Type I error
       Type II error
       χ2
       nonparametric
       contingency table
       statistically significant
       E (expected value) in χ 2
       O (observed value) in χ 2

30.    Discuss the relationship(s) between or among the terms:

        α/p
       df/R/C [in χ 2 situation]
       H o / H1
             α
       χ 2 / /df
        α/Type I error
       E/O/ χ 2
       Type I/Type II error



Copyright Philip Doty, University of Texas at Austin, August 2004              14
SELECTED ANSWERS 3.4

2.     (a) n = 10

       (b)   ∑ = 78
              x


              ∑78
                x
       (c) x = = =   7.8
              n   10


       (d) s =
                    ∑ 714 −
                    x −
                      nx2

                         =
                           608.4
                                2

                                                  = =
                                                   11.73 3.43
                        n−
                         1                 9

               2
       (e) s       = 11.73

                      n +1            10 +1
       (f) P(med) =        th score =        th score = 5.5th score
                        2               2
                                 5thscore + 6thscore    8+9
             med = 5.5th score =                      =         = 8.5
                                          2               2

       (g) mode = 9

       (h) range = highest value - lowest value = 13 - 2 = 11

                                                 s 3.43
       (i) coefficient of variation (CV) =         = =  0.44
                                                 x  7.8

4.     (a) n = 20

       (b)   ∑ = 1079
              x


              ∑ x 1079
       (c) x = = =     53.95
              n    20


       (d) s =
                     ∑ 58395− = =
                     x −
                       nx
                          =
                            2
                            58212
                                 2

                                  9.63 3.1
                        n−
                         1                  19

               2
       (e) s       = 9.63

                            n +1            20 +1
       (f) P(med) =              th score =       th score = 10.5th score
                              2               2




Copyright Philip Doty, University of Texas at Austin, August 2004           15
53 + 54
           median =               =53.5
                             2

       (g) mode = 50, 51, 57 (trimodal)

       (h) range = 59 - 50 = 9

              s  3.1
       (i) CV = =     =
                      0.06
              x 53.95


10.    (a) Pos, Pos, Neg

       (b) n = 28, 221, 217

       (c) mode = 26, 5, 12

           P(median) = 14.5th, 111th, 109th observations
           median = 26.5, 4, 10

           mode1 < median1 ; mode2 > median2; mode3 > median3

                    ∑ ; 2058 =
       (d) mean = x = =
                       x 776 907
                            ;       27.7,4.1,9.5
                     n   28 221 217

           mean1 > mode1; mode2 > mean2; mode3 > mean3


       (e) s 2   ∑
                 =
                  x − 22218 −
                    nx2

                       =
                              2
                            (28)(27.7)         733.88
                                               = =
                                                      2
                                                      27.18
            1
                     n−
                      1                   28 −
                                             1   27

             2 10887 −
                     (221)(4.1)2 7171.99
           s2 =                 =        =
                                         32.60
                     220           220

             2 21948 −
                     (217)(9.5) 2 2363.75
           s3 =                  =        =
                                          10.94
                     216            216

       (f) s1,s2 ,s3 =
                     s2 =
                        27.18, 32.60 , 10.94 =
                                             5.21,5.71,3.31

       (g) Range = highest observation - lowest observation = H - L = 53 -25 = 28;
                 85 - 1 = 84; 14 - 2 = 12

              s 5.21 5.71 3.31
       (h) CV = ,
               =         ,     =
                               0.19,1.39,0.35
              x 27.7 4.1 9.5


Copyright Philip Doty, University of Texas at Austin, August 2004               16
13. (a) x      freq x          cum freq x      rel freq x       cum rel freq x

       9        3                   25            0.12                1.00
       8        9                   22            0.36                0.88
       6        5                  13             0.20                0.52
       3        2                   8             0.08                0.32
       2        6                   6             0.24                0.24
               25                                 1.00

       (b) N = 25

       (c) Range = H - L = 9 - 2 = 7

       (d) P(median) = N + 1 th position = 13th observation; median = 6
                         2

       (e) mode = 8


       (f)   µ
             =
                ∑ =
                x 147
                 =    5.88
                 N      25

                 Nµ
              ∑ 1041−
             σ x −= (25)(5.88)
                        2      2                         2
       (g)                                                   174.64
             =                                               =      ==
                                                                    7.07 2.66
                        N                    25                25

       (h) P(Q2) = 13th observation; Q2 = 6

                     ↓
                    1+ 2 ) 1+
                      P(Q    13
             P(Q1 ) =     = = 7th observation from the beginning; Q1=
                      2     2
3

                      ↓
                     1+ 2 ) 1+
                       P(Q    13
             P(Q 3 ) =     = = 7th observation from the end; Q3= 8
                       2     2

                  σ =
                   2.66
       (i) CV ==        0.45
                  µ5.88

       (j) IQR = Q3 - Q1 = 8 - 3 = 5

       (k) PR = % lower + 1/2 % of that score

               PR(8) = 0.52 + (0.36) = 0.70
                                2



Copyright Philip Doty, University of Texas at Austin, August 2004                17
PR(2) = 0 + (0.24) = 0.12
                             2
               PR(3) = 0.24 + (0.08) = 0.28
                                 2




Copyright Philip Doty, University of Texas at Austin, August 2004   18
x−µ
       (l) z =   (for populations)
                  σ

            z6 =
                  µ 5.88 =
                x − 6−
                   =
                                  8−
                         0.05;z 8 =
                                     5.88
                                          =
                                                   2−
                                          0.80;z 2 =
                                                      5.88
                                                           =
                                                           −1.46
                  σ 2.66           2.66             2.66

                3− 5.88         9− 5.88
            z3 =        =
                        − 9=
                         1.08;z         =
                                        1.17
                 2.66            2.66


16. ND(50, 2.5)

                       x− − µ
                         45 50
       (a) PR(45); z45 = = = ⇒ − −
                            σ
                           2.5
                               2.00 0.4772


           0.5000 - 0.4772 = 0.0228, 2.28%, 2.28th percentile

                 52.6 − 50
       (b) z52.6 =         =
                           1.04
                    2.5

                       58 − 50
       (c) PR(58); z58 =       =⇒
                               3.2 0.4993
                         2.5

           0.5000 + 0.4993 = 0.9993, 99.93%, 99.93rd percentile

       (d) find the 29.12th percentile

           0.5000 - 0.2912 = 0.2088    ⇒ z = - 0.55
                 x− 50
            − =
            0.55
                  2.5
              x − =1.375
                 50 −
              x= 48.63

       (e) find the 89.74th percentile

           0.8974 - 0.5000 = 0.3974    ⇒ z = 1.27
                      x− 50
             z=1.27 =
                       2.5
             x= 53.175

               45 − 50
       (f) z45 =       =
                       −2.00
                 2.5


Copyright Philip Doty, University of Texas at Austin, August 2004   19
49 − −
                            50   1
       (g) PR(49); z49 = = − −     =⇒
                                   0.4 0.1554
                         2.5   2.5

              0.5000 – 0.1554 =0.3446, 34.46%, 34.46th percentile

19.    2, 5, 9, 5, 3, 6, 6, 3, 1, 13

       (a) E(x ) =
                 µ=
                  4.1

                 σ2.93 0.93
       (b) SE µσ
              == = =
               x
                 n 10

20.     σ=3.9 years, n = 90, x =20.3 years
       (a) 95% C.I. on µ     x− µ µ
                                zSE µ ≤
                                    ≤ zSE  x+
            SE µ σ
                 == = =
                        σ 3.9 0.41
                    x
                         n   90
                                       0.95
            z95% = 1.96 [Remember that      =
                                            0.4750 ⇒ z ]
                                                   1.96 =
                                         2

            x− µ µ
             zSE µ ≤
                 ≤ zSE
                   x+

           20.3 - (1.96)(0.41) ≤ µ ≤ 20.3 + (1.96)(0.41)

           19.5 ≤ µ ≤ 21.1                                     interval width = 1.6

       (b) 90% C.I. on µ        z90%= 1.65

            x− µ µ
             zSE µ ≤
                 ≤ zSE
                   x+

           20.3 - (1.65)(0.41) ≤ µ ≤ 20.3 + (1.65)(0.41)

           19.62 ≤ µ ≤ 20.98                                   interval width = 1.36

       (c) 99% C.I. on µ        z99%= 2.58

           x− µ µ
            zSE µ ≤
                ≤ zSE
                  x+

           20.3 - (2.58)(0.41) ≤ µ ≤ 20.3 + (2.58)(0.41)

           19.25 ≤ µ ≤ 21.35                                   interval width = 2.1



Copyright Philip Doty, University of Texas at Austin, August 2004                      20
Our best estimate of µ is     x , 20.3 years




Copyright Philip Doty, University of Texas at Austin, August 2004   21
21. N = 145,   x   = 20.3 years,   σ = 3.9 years
   (a) 95% C.I. on µ             19.67 ≤ µ ≤ 20.93     interval width = 1.26



22. 95% C.I. on µ,     ∴α=0.05
   df = n - 1
   n = 20, 9,  ∞ ,1
   t = 2.093, 2.306, 1.960, ?



23. 99% C.I. on µ,     ∴α=0.01
   df = n - 1
   t = 2.861, 3.355, 2.576, ?


          ∑  x 1844
24. (a) x = = =     37.63
           n    49


   (b) s =
              ∑= − =
              x −
                nx 2      2
                   70048 69385
                               3.72
                   n−
                    1                 48


   (c) SE µσ
                s   3.72
           =x =     = =  0.537
               n −6.93
                  1

   (d) E(x ) =
             µ

               n+ 1
   (e) P(Q2 ) =     th position = 25th observation; Q2 = 36
                2

                ↓
               1+ 2 ) 1+
                 P(Q    25
        P(Q1 ) =     = = 13th observation from the beginning; Q1= 34
                 2     2

                 ↓
                1+ 2 ) 1+
                  P(Q    25
        P(Q 3 ) =     = = 13th observation from the end; Q3= 42
                  2     2

   (f) mode = 34




Copyright Philip Doty, University of Texas at Austin, August 2004              22
s  3.72
   (g) CV = =     =
                  0.10
          x 37.63

   (h) IQR = Q3 – Q1 = 42 - 34 = 8




Copyright Philip Doty, University of Texas at Austin, August 2004   23
(i) 95% C.I. on µ, µ,   ∴α=0.05
       SEµ = 0.537 (see (c) above)
       df = 48, t = 2.021

        x− µ µ
         zSE µ ≤
             ≤ zSE
               x+

       37.63 - (2.021)(0.537) ≤ µ ≤ 37.63 + (2.021)(0.537)

       37.63 - 1.085 ≤ µ ≤ 37.63 + 1.085

       36.5 ≤ µ ≤ 38.7

   (m) PR42 = % below + 1/2 % of that score = 0.65 + (0.22) = 0.76
                                                        2




Copyright Philip Doty, University of Texas at Austin, August 2004    24
27.

                                        TIME (MINS)

                                ≤5             >5, ≤10              >10
            EXPERTISE

            Novice              14 (21.12)     20 (19.46)           19 (12.42)      53

            Intermediate        15 (15.94)     16 (14.69)            9 (9.38)       40

            Expert              22 (13.95)     11 (12.85)            2 (8.20)       35
                                    51             47                   30         128


        (O − )2
            E                  RxC
      χ=2
                             E=
           E                    n


      χ −                   (15 − 2 (22 − 2 (20 − 2 (16 − 2
                        2
       (14 21.12)                15.94)    13.95)   19.46)   14.69)
      = 2
                            +           +         +        +
          21.12                 15.94    13.95     19.46    14.69

            (11 −2 (19 −2 (9 − (2 −
                 12.85)   12.42)   9.38) 2   8.20)2
            +           +        +         +
                12.85    12.42    9.38      8.20


      χ 2 = 2.40 + 0.06 + 4.65 + 0.01 + 0.12 + 0.27 + 3.49 + 0.02 + 4.69 = 15.71


      df = (R - 1)(C - 1) = 2 x 2 = 4

   χ 2 (4, 0.10) = 7.78

      15.71 > 7.78, reject H0


28. χ 2 (4, 0.05) = 9.49

      15.71 > 9.49, reject H0




Copyright Philip Doty, University of Texas at Austin, August 2004                        25
27.

                                        TIME (MINS)

                                ≤5             >5, ≤10              >10
            EXPERTISE

            Novice              14 (21.12)     20 (19.46)           19 (12.42)      53

            Intermediate        15 (15.94)     16 (14.69)            9 (9.38)       40

            Expert              22 (13.95)     11 (12.85)            2 (8.20)       35
                                    51             47                   30         128


        (O − )2
            E                  RxC
      χ=2
                             E=
           E                    n


      χ −                   (15 − 2 (22 − 2 (20 − 2 (16 − 2
                        2
       (14 21.12)                15.94)    13.95)   19.46)   14.69)
      = 2
                            +           +         +        +
          21.12                 15.94    13.95     19.46    14.69

            (11 −2 (19 −2 (9 − (2 −
                 12.85)   12.42)   9.38) 2   8.20)2
            +           +        +         +
                12.85    12.42    9.38      8.20


      χ 2 = 2.40 + 0.06 + 4.65 + 0.01 + 0.12 + 0.27 + 3.49 + 0.02 + 4.69 = 15.71


      df = (R - 1)(C - 1) = 2 x 2 = 4

   χ 2 (4, 0.10) = 7.78

      15.71 > 7.78, reject H0


28. χ 2 (4, 0.05) = 9.49

      15.71 > 9.49, reject H0




Copyright Philip Doty, University of Texas at Austin, August 2004                        25
27.

                                        TIME (MINS)

                                ≤5             >5, ≤10              >10
            EXPERTISE

            Novice              14 (21.12)     20 (19.46)           19 (12.42)      53

            Intermediate        15 (15.94)     16 (14.69)            9 (9.38)       40

            Expert              22 (13.95)     11 (12.85)            2 (8.20)       35
                                    51             47                   30         128


        (O − )2
            E                  RxC
      χ=2
                             E=
           E                    n


      χ −                   (15 − 2 (22 − 2 (20 − 2 (16 − 2
                        2
       (14 21.12)                15.94)    13.95)   19.46)   14.69)
      = 2
                            +           +         +        +
          21.12                 15.94    13.95     19.46    14.69

            (11 −2 (19 −2 (9 − (2 −
                 12.85)   12.42)   9.38) 2   8.20)2
            +           +        +         +
                12.85    12.42    9.38      8.20


      χ 2 = 2.40 + 0.06 + 4.65 + 0.01 + 0.12 + 0.27 + 3.49 + 0.02 + 4.69 = 15.71


      df = (R - 1)(C - 1) = 2 x 2 = 4

   χ 2 (4, 0.10) = 7.78

      15.71 > 7.78, reject H0


28. χ 2 (4, 0.05) = 9.49

      15.71 > 9.49, reject H0




Copyright Philip Doty, University of Texas at Austin, August 2004                        25
27.

                                        TIME (MINS)

                                ≤5             >5, ≤10              >10
            EXPERTISE

            Novice              14 (21.12)     20 (19.46)           19 (12.42)      53

            Intermediate        15 (15.94)     16 (14.69)            9 (9.38)       40

            Expert              22 (13.95)     11 (12.85)            2 (8.20)       35
                                    51             47                   30         128


        (O − )2
            E                  RxC
      χ=2
                             E=
           E                    n


      χ −                   (15 − 2 (22 − 2 (20 − 2 (16 − 2
                        2
       (14 21.12)                15.94)    13.95)   19.46)   14.69)
      = 2
                            +           +         +        +
          21.12                 15.94    13.95     19.46    14.69

            (11 −2 (19 −2 (9 − (2 −
                 12.85)   12.42)   9.38) 2   8.20)2
            +           +        +         +
                12.85    12.42    9.38      8.20


      χ 2 = 2.40 + 0.06 + 4.65 + 0.01 + 0.12 + 0.27 + 3.49 + 0.02 + 4.69 = 15.71


      df = (R - 1)(C - 1) = 2 x 2 = 4

   χ 2 (4, 0.10) = 7.78

      15.71 > 7.78, reject H0


28. χ 2 (4, 0.05) = 9.49

      15.71 > 9.49, reject H0




Copyright Philip Doty, University of Texas at Austin, August 2004                        25

More Related Content

What's hot

Why recursion is impotant_new_my.docx
Why recursion is impotant_new_my.docxWhy recursion is impotant_new_my.docx
Why recursion is impotant_new_my.docxMiracule D Gavor
 
0580 w13 ms_41
0580 w13 ms_410580 w13 ms_41
0580 w13 ms_41King Ali
 
Reliable ABC model choice via random forests
Reliable ABC model choice via random forestsReliable ABC model choice via random forests
Reliable ABC model choice via random forestsChristian Robert
 
Uncertainty quantification of groundwater contamination
Uncertainty quantification of groundwater contaminationUncertainty quantification of groundwater contamination
Uncertainty quantification of groundwater contaminationAlexander Litvinenko
 
ISI MSQE Entrance Question Paper (2012)
ISI MSQE Entrance Question Paper (2012)ISI MSQE Entrance Question Paper (2012)
ISI MSQE Entrance Question Paper (2012)CrackDSE
 
ISI MSQE Entrance Question Paper (2011)
ISI MSQE Entrance Question Paper (2011)ISI MSQE Entrance Question Paper (2011)
ISI MSQE Entrance Question Paper (2011)CrackDSE
 
ISI MSQE Entrance Question Paper (2010)
ISI MSQE Entrance Question Paper (2010)ISI MSQE Entrance Question Paper (2010)
ISI MSQE Entrance Question Paper (2010)CrackDSE
 
ISI MSQE Entrance Question Paper (2006)
ISI MSQE Entrance Question Paper (2006)ISI MSQE Entrance Question Paper (2006)
ISI MSQE Entrance Question Paper (2006)CrackDSE
 
0580_w13_qp_41
0580_w13_qp_410580_w13_qp_41
0580_w13_qp_41King Ali
 
ISI MSQE Entrance Question Paper (2008)
ISI MSQE Entrance Question Paper (2008)ISI MSQE Entrance Question Paper (2008)
ISI MSQE Entrance Question Paper (2008)CrackDSE
 
352735326 rsh-qam11-tif-05-doc
352735326 rsh-qam11-tif-05-doc352735326 rsh-qam11-tif-05-doc
352735326 rsh-qam11-tif-05-docFiras Husseini
 
ISI MSQE Entrance Question Paper (2005)
ISI MSQE Entrance Question Paper (2005)ISI MSQE Entrance Question Paper (2005)
ISI MSQE Entrance Question Paper (2005)CrackDSE
 
ISI MSQE Entrance Question Paper (2009)
ISI MSQE Entrance Question Paper (2009)ISI MSQE Entrance Question Paper (2009)
ISI MSQE Entrance Question Paper (2009)CrackDSE
 
Machine Learning and Data Mining - Decision Trees
Machine Learning and Data Mining - Decision TreesMachine Learning and Data Mining - Decision Trees
Machine Learning and Data Mining - Decision Treeswebisslides
 
Support Vector Machines (SVM)
Support Vector Machines (SVM)Support Vector Machines (SVM)
Support Vector Machines (SVM)amitpraseed
 
0580 s10 qp_41
0580 s10 qp_410580 s10 qp_41
0580 s10 qp_41King Ali
 
ISI MSQE Entrance Question Paper (2013)
ISI MSQE Entrance Question Paper (2013)ISI MSQE Entrance Question Paper (2013)
ISI MSQE Entrance Question Paper (2013)CrackDSE
 

What's hot (20)

Why recursion is impotant_new_my.docx
Why recursion is impotant_new_my.docxWhy recursion is impotant_new_my.docx
Why recursion is impotant_new_my.docx
 
ICPR 2016
ICPR 2016ICPR 2016
ICPR 2016
 
0580 w13 ms_41
0580 w13 ms_410580 w13 ms_41
0580 w13 ms_41
 
Reliable ABC model choice via random forests
Reliable ABC model choice via random forestsReliable ABC model choice via random forests
Reliable ABC model choice via random forests
 
Uncertainty quantification of groundwater contamination
Uncertainty quantification of groundwater contaminationUncertainty quantification of groundwater contamination
Uncertainty quantification of groundwater contamination
 
ISI MSQE Entrance Question Paper (2012)
ISI MSQE Entrance Question Paper (2012)ISI MSQE Entrance Question Paper (2012)
ISI MSQE Entrance Question Paper (2012)
 
ISI MSQE Entrance Question Paper (2011)
ISI MSQE Entrance Question Paper (2011)ISI MSQE Entrance Question Paper (2011)
ISI MSQE Entrance Question Paper (2011)
 
ISI MSQE Entrance Question Paper (2010)
ISI MSQE Entrance Question Paper (2010)ISI MSQE Entrance Question Paper (2010)
ISI MSQE Entrance Question Paper (2010)
 
ISI MSQE Entrance Question Paper (2006)
ISI MSQE Entrance Question Paper (2006)ISI MSQE Entrance Question Paper (2006)
ISI MSQE Entrance Question Paper (2006)
 
0580_w13_qp_41
0580_w13_qp_410580_w13_qp_41
0580_w13_qp_41
 
Assignment4
Assignment4Assignment4
Assignment4
 
ISI MSQE Entrance Question Paper (2008)
ISI MSQE Entrance Question Paper (2008)ISI MSQE Entrance Question Paper (2008)
ISI MSQE Entrance Question Paper (2008)
 
352735326 rsh-qam11-tif-05-doc
352735326 rsh-qam11-tif-05-doc352735326 rsh-qam11-tif-05-doc
352735326 rsh-qam11-tif-05-doc
 
ISI MSQE Entrance Question Paper (2005)
ISI MSQE Entrance Question Paper (2005)ISI MSQE Entrance Question Paper (2005)
ISI MSQE Entrance Question Paper (2005)
 
ISI MSQE Entrance Question Paper (2009)
ISI MSQE Entrance Question Paper (2009)ISI MSQE Entrance Question Paper (2009)
ISI MSQE Entrance Question Paper (2009)
 
Machine Learning and Data Mining - Decision Trees
Machine Learning and Data Mining - Decision TreesMachine Learning and Data Mining - Decision Trees
Machine Learning and Data Mining - Decision Trees
 
Support Vector Machines (SVM)
Support Vector Machines (SVM)Support Vector Machines (SVM)
Support Vector Machines (SVM)
 
0580 s10 qp_41
0580 s10 qp_410580 s10 qp_41
0580 s10 qp_41
 
ISI MSQE Entrance Question Paper (2013)
ISI MSQE Entrance Question Paper (2013)ISI MSQE Entrance Question Paper (2013)
ISI MSQE Entrance Question Paper (2013)
 
Visual Techniques
Visual TechniquesVisual Techniques
Visual Techniques
 

Viewers also liked

Writing a research protocol 2011
Writing a research protocol 2011Writing a research protocol 2011
Writing a research protocol 2011Forensic Pathology
 
Writing a research protocol 2011
Writing a research protocol 2011Writing a research protocol 2011
Writing a research protocol 2011Forensic Pathology
 
Stat3 central tendency & dispersion
Stat3 central tendency & dispersionStat3 central tendency & dispersion
Stat3 central tendency & dispersionForensic Pathology
 
Haemodynamic disorders , thromboembolism and shock by Dr Nadeem (RMC)
Haemodynamic disorders , thromboembolism and shock by Dr Nadeem (RMC)Haemodynamic disorders , thromboembolism and shock by Dr Nadeem (RMC)
Haemodynamic disorders , thromboembolism and shock by Dr Nadeem (RMC)Hassan Ahmad
 
Hemodynamic disorders, thrombosis and shock (practical pathology)
Hemodynamic disorders, thrombosis and shock (practical pathology)Hemodynamic disorders, thrombosis and shock (practical pathology)
Hemodynamic disorders, thrombosis and shock (practical pathology)Mohaned Lehya
 

Viewers also liked (15)

Exercise 1 12 no solution
Exercise 1 12 no solutionExercise 1 12 no solution
Exercise 1 12 no solution
 
Writing a research protocol 2011
Writing a research protocol 2011Writing a research protocol 2011
Writing a research protocol 2011
 
Stat5 the t test
Stat5 the t testStat5 the t test
Stat5 the t test
 
Stat6 chi square test
Stat6 chi square testStat6 chi square test
Stat6 chi square test
 
Prac excises 3[1].5
Prac excises 3[1].5Prac excises 3[1].5
Prac excises 3[1].5
 
Some statistical defenition
Some statistical defenitionSome statistical defenition
Some statistical defenition
 
Writing a research protocol 2011
Writing a research protocol 2011Writing a research protocol 2011
Writing a research protocol 2011
 
Exercise 1 12 no solution
Exercise 1 12 no solutionExercise 1 12 no solution
Exercise 1 12 no solution
 
Hemodynamics disorders
Hemodynamics disorders Hemodynamics disorders
Hemodynamics disorders
 
Stat3 central tendency & dispersion
Stat3 central tendency & dispersionStat3 central tendency & dispersion
Stat3 central tendency & dispersion
 
Haemodynamic disorders , thromboembolism and shock by Dr Nadeem (RMC)
Haemodynamic disorders , thromboembolism and shock by Dr Nadeem (RMC)Haemodynamic disorders , thromboembolism and shock by Dr Nadeem (RMC)
Haemodynamic disorders , thromboembolism and shock by Dr Nadeem (RMC)
 
Hemodynamic disorders
Hemodynamic disordersHemodynamic disorders
Hemodynamic disorders
 
IMMUNOPATHOLOGY
IMMUNOPATHOLOGYIMMUNOPATHOLOGY
IMMUNOPATHOLOGY
 
Hemodynamic disorders, thrombosis and shock (practical pathology)
Hemodynamic disorders, thrombosis and shock (practical pathology)Hemodynamic disorders, thrombosis and shock (practical pathology)
Hemodynamic disorders, thrombosis and shock (practical pathology)
 
Hemodynamic Disorders
Hemodynamic DisordersHemodynamic Disorders
Hemodynamic Disorders
 

Similar to Prac ex'cises 3[1].5

ISM_Session_5 _ 23rd and 24th December.pptx
ISM_Session_5 _ 23rd and 24th December.pptxISM_Session_5 _ 23rd and 24th December.pptx
ISM_Session_5 _ 23rd and 24th December.pptxssuser1eba67
 
Paper Summary of Disentangling by Factorising (Factor-VAE)
Paper Summary of Disentangling by Factorising (Factor-VAE)Paper Summary of Disentangling by Factorising (Factor-VAE)
Paper Summary of Disentangling by Factorising (Factor-VAE)준식 최
 
Chapter 8 Of Rock Engineering
Chapter 8 Of  Rock  EngineeringChapter 8 Of  Rock  Engineering
Chapter 8 Of Rock EngineeringNgo Hung Long
 
Fin500J_topic10_Probability_2010_0000000
Fin500J_topic10_Probability_2010_0000000Fin500J_topic10_Probability_2010_0000000
Fin500J_topic10_Probability_2010_0000000Tushar Chaudhari
 
Bayesian inference for mixed-effects models driven by SDEs and other stochast...
Bayesian inference for mixed-effects models driven by SDEs and other stochast...Bayesian inference for mixed-effects models driven by SDEs and other stochast...
Bayesian inference for mixed-effects models driven by SDEs and other stochast...Umberto Picchini
 
Pattern learning and recognition on statistical manifolds: An information-geo...
Pattern learning and recognition on statistical manifolds: An information-geo...Pattern learning and recognition on statistical manifolds: An information-geo...
Pattern learning and recognition on statistical manifolds: An information-geo...Frank Nielsen
 
HW1_STAT206.pdfStatistical Inference II J. Lee Assignment.docx
HW1_STAT206.pdfStatistical Inference II J. Lee Assignment.docxHW1_STAT206.pdfStatistical Inference II J. Lee Assignment.docx
HW1_STAT206.pdfStatistical Inference II J. Lee Assignment.docxwilcockiris
 
Possible applications of low-rank tensors in statistics and UQ (my talk in Bo...
Possible applications of low-rank tensors in statistics and UQ (my talk in Bo...Possible applications of low-rank tensors in statistics and UQ (my talk in Bo...
Possible applications of low-rank tensors in statistics and UQ (my talk in Bo...Alexander Litvinenko
 
Multiple estimators for Monte Carlo approximations
Multiple estimators for Monte Carlo approximationsMultiple estimators for Monte Carlo approximations
Multiple estimators for Monte Carlo approximationsChristian Robert
 
An investigation of inference of the generalized extreme value distribution b...
An investigation of inference of the generalized extreme value distribution b...An investigation of inference of the generalized extreme value distribution b...
An investigation of inference of the generalized extreme value distribution b...Alexander Decker
 
Bayesian phylogenetic inference_big4_ws_2016-10-10
Bayesian phylogenetic inference_big4_ws_2016-10-10Bayesian phylogenetic inference_big4_ws_2016-10-10
Bayesian phylogenetic inference_big4_ws_2016-10-10FredrikRonquist
 
BUS173 Lecture 4.pptx
BUS173 Lecture 4.pptxBUS173 Lecture 4.pptx
BUS173 Lecture 4.pptxSusantoSaha1
 
Statistics and Probability Reviewer.pptx
Statistics and Probability Reviewer.pptxStatistics and Probability Reviewer.pptx
Statistics and Probability Reviewer.pptxJERWINTELACAS
 

Similar to Prac ex'cises 3[1].5 (20)

Data Analysis Assignment Help
Data Analysis Assignment Help Data Analysis Assignment Help
Data Analysis Assignment Help
 
ISM_Session_5 _ 23rd and 24th December.pptx
ISM_Session_5 _ 23rd and 24th December.pptxISM_Session_5 _ 23rd and 24th December.pptx
ISM_Session_5 _ 23rd and 24th December.pptx
 
Paper Summary of Disentangling by Factorising (Factor-VAE)
Paper Summary of Disentangling by Factorising (Factor-VAE)Paper Summary of Disentangling by Factorising (Factor-VAE)
Paper Summary of Disentangling by Factorising (Factor-VAE)
 
Chapter 8 Of Rock Engineering
Chapter 8 Of  Rock  EngineeringChapter 8 Of  Rock  Engineering
Chapter 8 Of Rock Engineering
 
Fin500J_topic10_Probability_2010_0000000
Fin500J_topic10_Probability_2010_0000000Fin500J_topic10_Probability_2010_0000000
Fin500J_topic10_Probability_2010_0000000
 
Bayesian inference for mixed-effects models driven by SDEs and other stochast...
Bayesian inference for mixed-effects models driven by SDEs and other stochast...Bayesian inference for mixed-effects models driven by SDEs and other stochast...
Bayesian inference for mixed-effects models driven by SDEs and other stochast...
 
Sampling distribution
Sampling distributionSampling distribution
Sampling distribution
 
Pattern learning and recognition on statistical manifolds: An information-geo...
Pattern learning and recognition on statistical manifolds: An information-geo...Pattern learning and recognition on statistical manifolds: An information-geo...
Pattern learning and recognition on statistical manifolds: An information-geo...
 
HW1_STAT206.pdfStatistical Inference II J. Lee Assignment.docx
HW1_STAT206.pdfStatistical Inference II J. Lee Assignment.docxHW1_STAT206.pdfStatistical Inference II J. Lee Assignment.docx
HW1_STAT206.pdfStatistical Inference II J. Lee Assignment.docx
 
Possible applications of low-rank tensors in statistics and UQ (my talk in Bo...
Possible applications of low-rank tensors in statistics and UQ (my talk in Bo...Possible applications of low-rank tensors in statistics and UQ (my talk in Bo...
Possible applications of low-rank tensors in statistics and UQ (my talk in Bo...
 
Multivariate Methods Assignment Help
Multivariate Methods Assignment HelpMultivariate Methods Assignment Help
Multivariate Methods Assignment Help
 
Multiple estimators for Monte Carlo approximations
Multiple estimators for Monte Carlo approximationsMultiple estimators for Monte Carlo approximations
Multiple estimators for Monte Carlo approximations
 
main
mainmain
main
 
An investigation of inference of the generalized extreme value distribution b...
An investigation of inference of the generalized extreme value distribution b...An investigation of inference of the generalized extreme value distribution b...
An investigation of inference of the generalized extreme value distribution b...
 
Bayesian phylogenetic inference_big4_ws_2016-10-10
Bayesian phylogenetic inference_big4_ws_2016-10-10Bayesian phylogenetic inference_big4_ws_2016-10-10
Bayesian phylogenetic inference_big4_ws_2016-10-10
 
BUS173 Lecture 4.pptx
BUS173 Lecture 4.pptxBUS173 Lecture 4.pptx
BUS173 Lecture 4.pptx
 
STATISTICS 1
STATISTICS 1STATISTICS 1
STATISTICS 1
 
Talk litvinenko gamm19
Talk litvinenko gamm19Talk litvinenko gamm19
Talk litvinenko gamm19
 
Statistics and Probability Reviewer.pptx
Statistics and Probability Reviewer.pptxStatistics and Probability Reviewer.pptx
Statistics and Probability Reviewer.pptx
 
Statistical analysis by iswar
Statistical analysis by iswarStatistical analysis by iswar
Statistical analysis by iswar
 

More from Forensic Pathology (20)

Exercise 1 12 no solution
Exercise 1 12 no solutionExercise 1 12 no solution
Exercise 1 12 no solution
 
Analysis of variance
Analysis of varianceAnalysis of variance
Analysis of variance
 
Stat3 central tendency & dispersion
Stat3 central tendency & dispersionStat3 central tendency & dispersion
Stat3 central tendency & dispersion
 
Stat 4 the normal distribution & steps of testing hypothesis
Stat 4 the normal distribution & steps of testing hypothesisStat 4 the normal distribution & steps of testing hypothesis
Stat 4 the normal distribution & steps of testing hypothesis
 
Stat 2 data presentation2
Stat 2 data presentation2Stat 2 data presentation2
Stat 2 data presentation2
 
Correlation 3rd
Correlation 3rdCorrelation 3rd
Correlation 3rd
 
Stat 1 variables & sampling
Stat 1 variables & samplingStat 1 variables & sampling
Stat 1 variables & sampling
 
Pancreas 2
Pancreas 2Pancreas 2
Pancreas 2
 
Pancreas 1
Pancreas 1Pancreas 1
Pancreas 1
 
Liver 3
Liver 3Liver 3
Liver 3
 
Liver 2
Liver 2Liver 2
Liver 2
 
Liver 1
Liver 1Liver 1
Liver 1
 
The biliary tract
The biliary tractThe biliary tract
The biliary tract
 
Valvular heart diseases 4
Valvular heart diseases 4Valvular heart diseases 4
Valvular heart diseases 4
 
Pericardial dis.&cardiactumors 5
Pericardial dis.&cardiactumors 5Pericardial dis.&cardiactumors 5
Pericardial dis.&cardiactumors 5
 
Ischemic heart diseases 2
Ischemic heart diseases 2Ischemic heart diseases 2
Ischemic heart diseases 2
 
Hypertensive hd, and cardiomyopathy 3
Hypertensive hd, and cardiomyopathy 3Hypertensive hd, and cardiomyopathy 3
Hypertensive hd, and cardiomyopathy 3
 
Congenital heart diseases 1
Congenital heart diseases 1Congenital heart diseases 1
Congenital heart diseases 1
 
Arteriosclerosis 6
Arteriosclerosis 6Arteriosclerosis 6
Arteriosclerosis 6
 
Aneurysms & dissection 7
Aneurysms & dissection 7Aneurysms & dissection 7
Aneurysms & dissection 7
 

Prac ex'cises 3[1].5

  • 1. PRACTICE EXERCISES LIS 397C Introduction to Research in Information Studies School of Information University of Texas at Austin Dr. Philip Doty Version 3.5 Copyright Philip Doty, University of Texas at Austin, August 2004 1
  • 2. LIS 397C Doty Practice Exercises 3.5 1. Define the following terms and symbols: n ∑x x x s s2 N µ σ σ 2 coefficient of variation (CV) mode median arithmetic mean range variance interquartile range (IQR) standard deviation sample statistic parameter frequency distribution 2. A sample of the variable x assumes the following values: 9 11 13 3 7 2 8 9 6 10 Compute: (a) n (b) ∑ x (c) x (d) s (e) s 2 (f) median (g) mode (h) range Copyright Philip Doty, University of Texas at Austin, August 2004 2
  • 3. (i) CV Copyright Philip Doty, University of Texas at Austin, August 2004 3
  • 4. 3. A sample of the variable x assumes the following values: 57 51 58 52 50 59 57 51 59 56 50 53 54 50 57 51 53 55 52 54 Generate a frequency distribution indicating x, frequency of x, cumulative frequency of x, relative frequency of x, and cumulative relative frequency of x. 4. For the frequency distribution in problem 3, compute: (a) n (b) ∑ x (c) x (d) s (e) s 2 (f) median (g) mode (h) range (i) CV 5. Generate a histogram for the data in problem 3. 6. Generate a frequency polygon for the data in problem 3. 7. Define the following terms: skewness ordinate abscissa central tendency bimodal ordered pair Cartesian plane stem-and-leaf plot outlier reliability dispersion or variability negatively skewed positively skewed validity box plot whiskers Copyright Philip Doty, University of Texas at Austin, August 2004 4
  • 5. 8. What is the relationship between or among the terms? (a) sample/population (b) x /µ (c) s/ σ (d) variance/standard deviation (e) n/N (f) statistic/parameter (g) mean, median, mode of the normal curve (h) coefficient of variation/IQR 9. Graph the following sample distributions (histogram and frequency polygon) using three different pairs of axes. Distribution 1 Distribution 2 Distribution 3 Score | Frequency Score | Frequency Score | Frequency 25 4 1 25 2 8 26 10 2 31 4 20 27 6 3 40 6 25 28 3 4 44 8 35 29 3 5 51 10 40 30 1 6 19 12 45 53 1 7 10 13 24 85 1 14 20 10. For each of the distributions in problem 9, answer the following questions. (a) Is the curve of the distribution positively or negatively skewed? (b) What is n? (c) Is the mode > median? Compute the answer and also answer it graphically, i.e., label the position of the mode and the median on the curve. (d) Is the mean > mode? Compute the answer and also answer it graphically as in part (c) of this question. (e) What is the variance of the distribution? (f) What is the standard deviation of the distribution? (g) What is the range of the distribution? (h) What is the coefficient of variation of the distribution? (i) Which measure of central tendency, mode, median, or (arithmetic) mean, is the fairest and clearest description of the distribution? Copyright Philip Doty, University of Texas at Austin, August 2004 5
  • 6. Why? Copyright Philip Doty, University of Texas at Austin, August 2004 6
  • 7. 11. Define: error model quartile percentile Q1 Q2 Q3 freq (x) cf (x) rel freq (x) cum rel freq (x) PR z-scores χ deviation score s of z-scores x of z-scores ∑x centile Interquartile Range (IQR) non-response bias self-selection 12. Define the relationship(s) between or among the terms: Q 2 /median median/N or n z-score/raw score z-score/deviation score/standard deviation Q 1/ Q 2/ Q 3 z-score/ χ /s or σ median/fifth centile cf (x)/freq (x)/PR Literary Digest poll/bias 13. The observations of the values of variable x can be summarized in the population frequency distribution below. x freq (x) 9 3 8 9 6 5 Copyright Philip Doty, University of Texas at Austin, August 2004 7
  • 8. 3 2 2 6 Copyright Philip Doty, University of Texas at Austin, August 2004 8
  • 9. For this distribution of x, calculate: (a) Cumulative frequency, relative frequency, and cumulative relative frequency for each value (b) N (c) the range (d) median (e) mode (f) µ (g) σ (h) Q 1 , Q 2 , and Q 3 (i) CV (coefficient of variation) (j) IQR (k) the percentile rank of x = 8, x = 2, x = 3 (l) z-scores for x = 6, x = 8, x = 2, x = 3, x = 9 14. Generate a box plot for the data in problem 13. 15. Generate a box plot for the data in problem 3. 16. For a normally distributed distribution of variable x, where µ = 50 and σ = 2.5 [ND (50, 2.5)], calculate: (a) the percentile rank of x = 45 (b) the z-score of x = 52.6 (c) the percentile rank of x = 58 (d) the 29.12th percentile (e) the 89.74th percentile (f) the z-score of x = 45 (g) the percentile rank of x = 49 17. Define: α sampling distribution Central Limit Theorem Standard Error (SE) ND(µ, ) σ decile descriptive statistics inferential statistics effect size confidence interval (C.I.) on µ Copyright Philip Doty, University of Texas at Austin, August 2004 9
  • 10. Student's t degrees of freedom (df) random sampling stratified random sample Copyright Philip Doty, University of Texas at Austin, August 2004 10
  • 11. 18. Define the relationship(s) between or among the terms: E( x )/µ α /df/t Central Limit Theorem/C.I. on µ α /C.I. on µ when σ is known α /C.I. on µ when σ is not known z/confidence interval on µ t/C.I. on µ t/z 19. The following values indicate the number of microcomputer applications available to a sample of 10 computer users. 2, 5, 9, 5, 3, 6, 6, 3, 1, 13 For the population from which the sample was drawn, µ = 4.1 and σ= 2.93. Calculate: a. The expected value of the mean of the sampling distribution of means b. The standard deviation of the sampling distribution of means. 20. From previous research, we know that the standard deviation of the ages of public library users is 3.9 years. If the "average" age of a sample of 90 public library users is 20.3 years, construct: a. a 95% C.I. around µ b. a 90% C.I. around µ c. a 99% C.I. around µ. What does a 95% interval around µ mean? What is our best estimate of µ? 21. The sample size in problem 20 was increased to 145, while the "average" age of the sample remained at 20.3 years. Construct three confidence intervals around µ with the same levels of confidence as in Question 20. 22. Determine t for a C.I. of 95% around µ when n equals 20, 9, ∞ , and 1. Copyright Philip Doty, University of Texas at Austin, August 2004 11
  • 12. 23. Determine t on the same values of n (20, 9, ∞ , and 1) as in Question 22, but for a C.I. of 99% around µ. Should this new interval be narrower or wider than a 95% confidence interval on µ? Why? Answer the question both conceptually and algebraically. 24. The following frequency distribution gives the values for variable x in a sample drawn from a larger population. x freq(x) 43 6 42 11 39 3 36 7 35 7 34 15 Calculate: (a) x (b) s (c) SEµ = σx (d) E( x ) (e) Q 1 , Q 2 , and Q 3 (f) mode (g) CV (h) IQR (i) a 95% C.I. on µ (j) the width of the confidence interval in part (i) (k) a 99% C.I. on µ (l) the width of the confidence interval in part (k) (m) PR (percentile rank) of x = 42 (n) our best estimates of µ and σ from the data. 25. Generate a box plot for the data in problem 24. 26. Construct a stem-and-leaf plot for the following data set; indicate Q 1 , Q 2 , and Q 3 on the plot; and generate the six-figure summary. Be sure that you are able to identify the stems and the leaves and to identify their units of measurement. The heights of members of an extended family were measured in inches. The observations were: 62, 48, 56, 37, 37, 26, 74, 66, 29, 49, 72, 77, 69, 62, Copyright Philip Doty, University of Texas at Austin, August 2004 12
  • 13. and 64. Copyright Philip Doty, University of Texas at Austin, August 2004 13
  • 14. 27. H 0 : There is no relationship between computer expertise and minutes spent doing known-item searches in an OPAC at = 0.10. α Should we reject the H0 given the following data? Remember that the acceptable error rate is 0.10. TIME (MINS) EXPERTISE ≤5 > 5, ≤ 10 > 10 Novice 14 20 19 Intermediate 15 16 9 Expert 22 11 2 28. Answer Question 27 at an acceptable error rate of 0.05. 29. Define: statistical hypothesis Ho H1 p Type I error Type II error χ2 nonparametric contingency table statistically significant E (expected value) in χ 2 O (observed value) in χ 2 30. Discuss the relationship(s) between or among the terms: α/p df/R/C [in χ 2 situation] H o / H1 α χ 2 / /df α/Type I error E/O/ χ 2 Type I/Type II error Copyright Philip Doty, University of Texas at Austin, August 2004 14
  • 15. SELECTED ANSWERS 3.4 2. (a) n = 10 (b) ∑ = 78 x ∑78 x (c) x = = = 7.8 n 10 (d) s = ∑ 714 − x − nx2 = 608.4 2 = = 11.73 3.43 n− 1 9 2 (e) s = 11.73 n +1 10 +1 (f) P(med) = th score = th score = 5.5th score 2 2 5thscore + 6thscore 8+9 med = 5.5th score = = = 8.5 2 2 (g) mode = 9 (h) range = highest value - lowest value = 13 - 2 = 11 s 3.43 (i) coefficient of variation (CV) = = = 0.44 x 7.8 4. (a) n = 20 (b) ∑ = 1079 x ∑ x 1079 (c) x = = = 53.95 n 20 (d) s = ∑ 58395− = = x − nx = 2 58212 2 9.63 3.1 n− 1 19 2 (e) s = 9.63 n +1 20 +1 (f) P(med) = th score = th score = 10.5th score 2 2 Copyright Philip Doty, University of Texas at Austin, August 2004 15
  • 16. 53 + 54 median = =53.5 2 (g) mode = 50, 51, 57 (trimodal) (h) range = 59 - 50 = 9 s 3.1 (i) CV = = = 0.06 x 53.95 10. (a) Pos, Pos, Neg (b) n = 28, 221, 217 (c) mode = 26, 5, 12 P(median) = 14.5th, 111th, 109th observations median = 26.5, 4, 10 mode1 < median1 ; mode2 > median2; mode3 > median3 ∑ ; 2058 = (d) mean = x = = x 776 907 ; 27.7,4.1,9.5 n 28 221 217 mean1 > mode1; mode2 > mean2; mode3 > mean3 (e) s 2 ∑ = x − 22218 − nx2 = 2 (28)(27.7) 733.88 = = 2 27.18 1 n− 1 28 − 1 27 2 10887 − (221)(4.1)2 7171.99 s2 = = = 32.60 220 220 2 21948 − (217)(9.5) 2 2363.75 s3 = = = 10.94 216 216 (f) s1,s2 ,s3 = s2 = 27.18, 32.60 , 10.94 = 5.21,5.71,3.31 (g) Range = highest observation - lowest observation = H - L = 53 -25 = 28; 85 - 1 = 84; 14 - 2 = 12 s 5.21 5.71 3.31 (h) CV = , = , = 0.19,1.39,0.35 x 27.7 4.1 9.5 Copyright Philip Doty, University of Texas at Austin, August 2004 16
  • 17. 13. (a) x freq x cum freq x rel freq x cum rel freq x 9 3 25 0.12 1.00 8 9 22 0.36 0.88 6 5 13 0.20 0.52 3 2 8 0.08 0.32 2 6 6 0.24 0.24 25 1.00 (b) N = 25 (c) Range = H - L = 9 - 2 = 7 (d) P(median) = N + 1 th position = 13th observation; median = 6 2 (e) mode = 8 (f) µ = ∑ = x 147 = 5.88 N 25 Nµ ∑ 1041− σ x −= (25)(5.88) 2 2 2 (g) 174.64 = = == 7.07 2.66 N 25 25 (h) P(Q2) = 13th observation; Q2 = 6 ↓ 1+ 2 ) 1+ P(Q 13 P(Q1 ) = = = 7th observation from the beginning; Q1= 2 2 3 ↓ 1+ 2 ) 1+ P(Q 13 P(Q 3 ) = = = 7th observation from the end; Q3= 8 2 2 σ = 2.66 (i) CV == 0.45 µ5.88 (j) IQR = Q3 - Q1 = 8 - 3 = 5 (k) PR = % lower + 1/2 % of that score PR(8) = 0.52 + (0.36) = 0.70 2 Copyright Philip Doty, University of Texas at Austin, August 2004 17
  • 18. PR(2) = 0 + (0.24) = 0.12 2 PR(3) = 0.24 + (0.08) = 0.28 2 Copyright Philip Doty, University of Texas at Austin, August 2004 18
  • 19. x−µ (l) z = (for populations) σ z6 = µ 5.88 = x − 6− = 8− 0.05;z 8 = 5.88 = 2− 0.80;z 2 = 5.88 = −1.46 σ 2.66 2.66 2.66 3− 5.88 9− 5.88 z3 = = − 9= 1.08;z = 1.17 2.66 2.66 16. ND(50, 2.5) x− − µ 45 50 (a) PR(45); z45 = = = ⇒ − − σ 2.5 2.00 0.4772 0.5000 - 0.4772 = 0.0228, 2.28%, 2.28th percentile 52.6 − 50 (b) z52.6 = = 1.04 2.5 58 − 50 (c) PR(58); z58 = =⇒ 3.2 0.4993 2.5 0.5000 + 0.4993 = 0.9993, 99.93%, 99.93rd percentile (d) find the 29.12th percentile 0.5000 - 0.2912 = 0.2088 ⇒ z = - 0.55 x− 50 − = 0.55 2.5 x − =1.375 50 − x= 48.63 (e) find the 89.74th percentile 0.8974 - 0.5000 = 0.3974 ⇒ z = 1.27 x− 50 z=1.27 = 2.5 x= 53.175 45 − 50 (f) z45 = = −2.00 2.5 Copyright Philip Doty, University of Texas at Austin, August 2004 19
  • 20. 49 − − 50 1 (g) PR(49); z49 = = − − =⇒ 0.4 0.1554 2.5 2.5 0.5000 – 0.1554 =0.3446, 34.46%, 34.46th percentile 19. 2, 5, 9, 5, 3, 6, 6, 3, 1, 13 (a) E(x ) = µ= 4.1 σ2.93 0.93 (b) SE µσ == = = x n 10 20. σ=3.9 years, n = 90, x =20.3 years (a) 95% C.I. on µ x− µ µ zSE µ ≤ ≤ zSE x+ SE µ σ == = = σ 3.9 0.41 x n 90 0.95 z95% = 1.96 [Remember that = 0.4750 ⇒ z ] 1.96 = 2 x− µ µ zSE µ ≤ ≤ zSE x+ 20.3 - (1.96)(0.41) ≤ µ ≤ 20.3 + (1.96)(0.41) 19.5 ≤ µ ≤ 21.1 interval width = 1.6 (b) 90% C.I. on µ z90%= 1.65 x− µ µ zSE µ ≤ ≤ zSE x+ 20.3 - (1.65)(0.41) ≤ µ ≤ 20.3 + (1.65)(0.41) 19.62 ≤ µ ≤ 20.98 interval width = 1.36 (c) 99% C.I. on µ z99%= 2.58 x− µ µ zSE µ ≤ ≤ zSE x+ 20.3 - (2.58)(0.41) ≤ µ ≤ 20.3 + (2.58)(0.41) 19.25 ≤ µ ≤ 21.35 interval width = 2.1 Copyright Philip Doty, University of Texas at Austin, August 2004 20
  • 21. Our best estimate of µ is x , 20.3 years Copyright Philip Doty, University of Texas at Austin, August 2004 21
  • 22. 21. N = 145, x = 20.3 years, σ = 3.9 years (a) 95% C.I. on µ 19.67 ≤ µ ≤ 20.93 interval width = 1.26 22. 95% C.I. on µ, ∴α=0.05 df = n - 1 n = 20, 9, ∞ ,1 t = 2.093, 2.306, 1.960, ? 23. 99% C.I. on µ, ∴α=0.01 df = n - 1 t = 2.861, 3.355, 2.576, ? ∑ x 1844 24. (a) x = = = 37.63 n 49 (b) s = ∑= − = x − nx 2 2 70048 69385 3.72 n− 1 48 (c) SE µσ s 3.72 =x = = = 0.537 n −6.93 1 (d) E(x ) = µ n+ 1 (e) P(Q2 ) = th position = 25th observation; Q2 = 36 2 ↓ 1+ 2 ) 1+ P(Q 25 P(Q1 ) = = = 13th observation from the beginning; Q1= 34 2 2 ↓ 1+ 2 ) 1+ P(Q 25 P(Q 3 ) = = = 13th observation from the end; Q3= 42 2 2 (f) mode = 34 Copyright Philip Doty, University of Texas at Austin, August 2004 22
  • 23. s 3.72 (g) CV = = = 0.10 x 37.63 (h) IQR = Q3 – Q1 = 42 - 34 = 8 Copyright Philip Doty, University of Texas at Austin, August 2004 23
  • 24. (i) 95% C.I. on µ, µ, ∴α=0.05 SEµ = 0.537 (see (c) above) df = 48, t = 2.021 x− µ µ zSE µ ≤ ≤ zSE x+ 37.63 - (2.021)(0.537) ≤ µ ≤ 37.63 + (2.021)(0.537) 37.63 - 1.085 ≤ µ ≤ 37.63 + 1.085 36.5 ≤ µ ≤ 38.7 (m) PR42 = % below + 1/2 % of that score = 0.65 + (0.22) = 0.76 2 Copyright Philip Doty, University of Texas at Austin, August 2004 24
  • 25. 27. TIME (MINS) ≤5 >5, ≤10 >10 EXPERTISE Novice 14 (21.12) 20 (19.46) 19 (12.42) 53 Intermediate 15 (15.94) 16 (14.69) 9 (9.38) 40 Expert 22 (13.95) 11 (12.85) 2 (8.20) 35 51 47 30 128 (O − )2 E RxC χ=2 E= E n χ − (15 − 2 (22 − 2 (20 − 2 (16 − 2 2 (14 21.12) 15.94) 13.95) 19.46) 14.69) = 2 + + + + 21.12 15.94 13.95 19.46 14.69 (11 −2 (19 −2 (9 − (2 − 12.85) 12.42) 9.38) 2 8.20)2 + + + + 12.85 12.42 9.38 8.20 χ 2 = 2.40 + 0.06 + 4.65 + 0.01 + 0.12 + 0.27 + 3.49 + 0.02 + 4.69 = 15.71 df = (R - 1)(C - 1) = 2 x 2 = 4 χ 2 (4, 0.10) = 7.78 15.71 > 7.78, reject H0 28. χ 2 (4, 0.05) = 9.49 15.71 > 9.49, reject H0 Copyright Philip Doty, University of Texas at Austin, August 2004 25
  • 26. 27. TIME (MINS) ≤5 >5, ≤10 >10 EXPERTISE Novice 14 (21.12) 20 (19.46) 19 (12.42) 53 Intermediate 15 (15.94) 16 (14.69) 9 (9.38) 40 Expert 22 (13.95) 11 (12.85) 2 (8.20) 35 51 47 30 128 (O − )2 E RxC χ=2 E= E n χ − (15 − 2 (22 − 2 (20 − 2 (16 − 2 2 (14 21.12) 15.94) 13.95) 19.46) 14.69) = 2 + + + + 21.12 15.94 13.95 19.46 14.69 (11 −2 (19 −2 (9 − (2 − 12.85) 12.42) 9.38) 2 8.20)2 + + + + 12.85 12.42 9.38 8.20 χ 2 = 2.40 + 0.06 + 4.65 + 0.01 + 0.12 + 0.27 + 3.49 + 0.02 + 4.69 = 15.71 df = (R - 1)(C - 1) = 2 x 2 = 4 χ 2 (4, 0.10) = 7.78 15.71 > 7.78, reject H0 28. χ 2 (4, 0.05) = 9.49 15.71 > 9.49, reject H0 Copyright Philip Doty, University of Texas at Austin, August 2004 25
  • 27. 27. TIME (MINS) ≤5 >5, ≤10 >10 EXPERTISE Novice 14 (21.12) 20 (19.46) 19 (12.42) 53 Intermediate 15 (15.94) 16 (14.69) 9 (9.38) 40 Expert 22 (13.95) 11 (12.85) 2 (8.20) 35 51 47 30 128 (O − )2 E RxC χ=2 E= E n χ − (15 − 2 (22 − 2 (20 − 2 (16 − 2 2 (14 21.12) 15.94) 13.95) 19.46) 14.69) = 2 + + + + 21.12 15.94 13.95 19.46 14.69 (11 −2 (19 −2 (9 − (2 − 12.85) 12.42) 9.38) 2 8.20)2 + + + + 12.85 12.42 9.38 8.20 χ 2 = 2.40 + 0.06 + 4.65 + 0.01 + 0.12 + 0.27 + 3.49 + 0.02 + 4.69 = 15.71 df = (R - 1)(C - 1) = 2 x 2 = 4 χ 2 (4, 0.10) = 7.78 15.71 > 7.78, reject H0 28. χ 2 (4, 0.05) = 9.49 15.71 > 9.49, reject H0 Copyright Philip Doty, University of Texas at Austin, August 2004 25
  • 28. 27. TIME (MINS) ≤5 >5, ≤10 >10 EXPERTISE Novice 14 (21.12) 20 (19.46) 19 (12.42) 53 Intermediate 15 (15.94) 16 (14.69) 9 (9.38) 40 Expert 22 (13.95) 11 (12.85) 2 (8.20) 35 51 47 30 128 (O − )2 E RxC χ=2 E= E n χ − (15 − 2 (22 − 2 (20 − 2 (16 − 2 2 (14 21.12) 15.94) 13.95) 19.46) 14.69) = 2 + + + + 21.12 15.94 13.95 19.46 14.69 (11 −2 (19 −2 (9 − (2 − 12.85) 12.42) 9.38) 2 8.20)2 + + + + 12.85 12.42 9.38 8.20 χ 2 = 2.40 + 0.06 + 4.65 + 0.01 + 0.12 + 0.27 + 3.49 + 0.02 + 4.69 = 15.71 df = (R - 1)(C - 1) = 2 x 2 = 4 χ 2 (4, 0.10) = 7.78 15.71 > 7.78, reject H0 28. χ 2 (4, 0.05) = 9.49 15.71 > 9.49, reject H0 Copyright Philip Doty, University of Texas at Austin, August 2004 25