A shoe manufacturer is field testing the durability of a leather sole made by a

ID: 2927301 • Letter: A

Question

A shoe manufacturer is field testing the durability of a leather sole made by a new process, and is comparing it with the type of leather sole currently in use. Fifteen pairs of shoes are constructed, with one shoe in each pair having the new type of leather sole. After six months of daily use by 15 randomly selected supermarket employees, the shoes are examined and measured for the percentage of wear remaining in each sole.
a. Compute the difference scores on the wear between the two types of soles, and create a stem-and-leaf plot for the difference scores.
b. Assess the normality of these difference scores.
c. Based upon these results, use either the Wilcoxon Signed Ranks test or the t-test to determine whether there is ant difference in the wear between the two types of soles. Justify your selection of test method.
d. If you decide that the wear of the two types of soles are indeed different, and you have done the parametric test in part (c.), estimate this difference with a 95% confidence interval.

This is a test question can you please show as much work as possible? The data is below:

New Old
73 64
43 41
47 43
53 41
58 47
47 32
52 24
38 43
61 53
56 52
56 57
34 44
55 57
65 40
75 68

Explanation / Answer

a. Differences in mean score of wear between two group is analysed in software STATA , Output is given below

Mean score of sole wear in new sole = 54.2

Mean score of sole wear in old sole = 47.06

. tabstat scoreon_wear, stats (n mean sd) by( group)

Summary for variables: scoreon_wear
by categories of group

group | N mean sd
---------+------------------------------
New | 15 54.2 11.57707
Old | 15 47.06667 1 1.67088
---------+------------------------------
Total | 30 50.63333 11.98414

Stem and leaf plot is given below

. by the group, sort: stem scoreon_wear

------------------------------------------------------------------------------------------------------------------------------------------------
-> group = 1 (New)

Stem-and-leaf plot for scoreon_wear

3* | 48
4* | 377
5* | 235668
6* | 15
7* | 35

------------------------------------------------------------------------------------------------------------------------------------------------
-> group = 2 (Old)

Stem-and-leaf plot for scoreon_wear

2* | 4
3* | 2
4* | 0113347
5* | 2377
6* | 48

b Normality assessment for New shoe sole group data

for New shoe sole group using Shapiro-Wilk P value = 0.96 suggest that data meets assumption of normal distribution. STATA output is given below

sktest scoreon_wear

Skewness/Kurtosis tests for Normality
------- joint ------
Variable | Obs Pr(Skewness) Pr(Kurtosis) adj chi2(2) Prob>chi2
-------------+---------------------------------------------------------------
scoreon_wear | 15 0.7994 0.9750 0.07 0.9677

For Old sole group, test for normally Shapiro-Wilk, P = 0.97 suggest, data is normally distributed (rejected the null hypothesis ( normal distribution)

c. In this condition independent sample t-test is an ideal test because the data is normally distributed in both the groups. Wilcoxon Signed Ranks test is not possible because samples are not paired

STATA output of independent sample t test is given below

Mean and SD in new sole group is 54.2± 11.5

Mean and SD in old sole group is 47.06±11.6

P value of difference was = 0.10, This finding suggests that although high sole score mean in the new sole group as compared with old sole group but statistically non-significant.

. ttest score_wear, by( group)

Two-sample t test with equal variances
------------------------------------------------------------------------------
Group | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval]
---------+--------------------------------------------------------------------
New | 15 54.2 2.989187 11.57707 47.78883 60.61117
Old | 15 47.06667 3.013409 11.67088 40.60355 53.52979
---------+--------------------------------------------------------------------
combined | 30 50.63333 2.187994 11.98414 46.15838 55.10828
---------+--------------------------------------------------------------------
diff | 7.133333 4.244511 -1.561153 15.82782
------------------------------------------------------------------------------
diff = mean(1) - mean(2) t = 1.6806
Ho: diff = 0 degrees of freedom = 28

Ha: diff < 0 Ha: diff != 0 Ha: diff > 0
Pr(T < t) = 0.9480 Pr(|T| > |t|) = 0.1040 Pr(T > t) = 0.0520

d. wear of the two types of soles are not significantly different as observed in the independent sample t-test between the old and new sole

95 % CI of difference = -1.561153 15.82782

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