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DCE 5950Assignment 2

Univariate Statistics

1. The following table displays test scores for 25 students enrolled in Statistics.Based on this data, solve for the questions below.

Table 1: Test Scores

78 86 50 75 59 60 90

65 65 67 88 56 72 87

61 81 77 65 83 56 7155 87 97 90

a. What is the scale of measurement of the above test scores?

The scale of measurement for the above test scores is Interval for

Mean.

b. Calculate measures of central tendencies mode, median and mean

Mode

x f

50 1

55 1

56 259 1

60 1

61 1

65 3

67 1

71 1

72 1

75 1

77 1

78 1

81 1

83 186 1

87 2

88 1

90 2

97 1

Mode = 65

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Median ML = n + 1

2

= 25 + 1

2

= 26

2

= 13

Median = 72

Mean x = xn

= 50+55+56+56+59+60+61+65+65+65+67+71+72+75+77+78

+81+83+86+87+87+88+90+90+97

25

= 1821

25

= 72.84

c. Calculate measures of dispersion range, variance and standard deviation

Range = 97 - 50

= 47

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Variance

x = 78 86 50 75 59 60 9065 65 67 88 56 72 87

61 81 77 65 83 56 71

55 87 97 90

x2 = 6084 7396 2500 5625 3481 3600 8100

4225 4225 4489 7744 3136 5184 75693721 6561 5929 4225 6889 3136 5041

3025 7569 9409 8100

x = 1821 Formula S2 = x2 - (x) 2x2 = 136963 nn = 25 n - 1

S2 = 136963 - (1821) 2

25

25 - 1

= 136963 - 3316041

25

24

= 136963 132641.64

24

= 4321.36

24

= 180.057

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Standard Deviation

S = x2 - (x) 2

n

n - 1

= S2

= 180.057

= 13.42

d. Does the distribution of the test scores meet the assumption of normality?

Use the appropriate statistic to test on the assumption.

Tests of Normality

Kolmogorov-Smirnova Shapiro-Wilk

Statistic df Sig. Statistic df Sig.

score .120 25 .200* .952 25 .283

a. Lilliefors Significance Correction

*. This is a lower bound of the true significance.

Kolmogorov-Smirnov is a appropriate statistic to test on the assumption of

normality.

Data is considered normal because Skewness is between -1 and +1. Skewness

is 0.042.

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2. Summary of data on frequency of accessing internet per week among the studyrespondents is presented in Table 2. Use the data to respond to the following

questions.

Table 2: Frequency of Accessing Internet

Frequency 5 7 8 10 12 13 15 20

n 7 10 15 33 45 59 18 13

a. What is the sample size for this data set?

Number of cases is the sample size. Number of cases is 200.

b. Calculate mode, median and mean for the frequencies of accessing

internet per week.

Mode

n f

7 5

10 7

15 8

33 10

45 12

59 13

18 1513 20

Mode = 13

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Median ML = n + 1

2

= 200 + 1

2

= 2012

= 100.5

Median = 12+122

= 12

Mean x = xn

= 2392

200

= 11.96

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c. Calculate standard deviation for the frequencies of accessing internet

x = 2392 Formula S2 = x2 - (x) 2

x2 = 91428 nn = 200 n - 1

Standard Deviation

S = x2 - (x) 2

n

n - 1

= 91428 - (2392) 2

200

200 - 1

= 91428 63574199

= 27854

199

= 139.97

= 11.83

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3. In a study, commitment is measured on a scale between 1 to 15. Summary

distribution of respondents by the commitment scores is presented in Table 3.

Table 3: Distribution of Commitment Scores

Commitment scores Freq Percent1 3 8 6.2

4 6 17 13.1

7 9 35 26.910 12 51 39.2

13 15 19 14.6

a. Calculate median and mean of the commitment scores.

Commitment scores f m fm m2 m2f

1 3 8 6.2 49.6 38.44 307.52

4 6 17 13.1 222.7 171.61 2917.377 9 35 26.9 941.5 1225 42875

10 12 51 39.2 1999.2 2601 13265113 15 19 14.6 277.4 361 6859

Total 130 100 3490.4 185609.89

L = 7, i= 3, n = 130, F = 17, fmd = 35

Median ML = n + 1

2

= 130 + 1

2

= 131

2

= 65.5

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The 65.5 value is located in the 7 - 9 class

Md = L + i n/2 Ffmd

= 7 + 3 130/2 1735

= 7 + 3 (1.37)

= 7 + 4.11

= 11.11

Median = 11.11

Mean n = 130 , fm = 3490.4

X = fmn

= 3490.4

130

= 26.85

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c. Calculate variance and standard deviation of the commitment scores

Variance

Commitment scores f m fm m2 m2f

1 3 8 6.2 49.6 38.44 307.524 6 17 13.1 222.7 171.61 2917.37

7 9 35 26.9 941.5 1225 42875

10 12 51 39.2 1999.2 2601 13265113 15 19 14.6 277.4 361 6859

Total 130 100 3490.4 185609.89

S2 = 185609.89 - (3490.4)2

130

130 - 1

= 185609.89 12182892.16130

129

= 185609.89 - 93714.55129

= 91895.34

129

= 712.37

Standard deviation

Varian = 712.37

S = S2

= 712.37

= 26.69

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