Size of Sample (N) = 11

Are women’s feet getting bigger? Retailers in the last 20 years have had to increase their stock of larger sizes. Wal-Mart Stores, Inc., and Payless ShoeSource, Inc., have been aggressive in stocking larger sizes, and Nordstrom’s reports that its larger sizes typically sell out first. Assuming equal variances, at a = .025, do these random shoe size samples of 12 randomly chosen women in each age group show that women’s shoe sizes have increased?
ShoeSize1
Born in 1980: 8 7.5 8.5 8.5 8 7.5 9.5 7.5 8 8 8.5 9
Born in 1960: 8.5 7.5 8 8 7.5 7.5 7.5 8 7 8 7 8

Solution:

Size of sample (n) = 11

Sample Mean for Born in 1980(m1): 8.227273

Standard Deviation for Born in 1980(s1): 0.64667

Sample Mean for Born in 1960 (m2): 7.681818

Standard Deviation for Born in 1960(s2): 0.462208

Degree of freedom = 11+11-2 = 22-2 = 20

We can define two-tailed statistics for above observations as follows:

Null Hypothesis: H0: (m1 - m2) =0 vs. Ha: (m1 - m2) > 0

Formula for test statistics,

Using above two-tailed statistics,

Rejection region can be defined as

z < -za/2 or z > za/2

Here significance level is 0.025,

So, z0.025 = 2.4231 (Using Statistical Ratio Calculator

From http://www.graphpad.com/quickcalcs/DistMenu.cfm for calculating z with 0.025 significance with DF = 20)

I am assuming the sample is selected independently and randomly from population. Population size is sufficiently large in both samples.

Calculating test statistics for both samples,

z =

Calculating p-value:

Degree of freedom = DF = 22-2 = 20

P(|z20| < 2.276) = 0.033994

(Using http://www.danielsoper.com/statcalc/calc08.aspx with t=2.276 and DF = 20)

Since calculated test-statistics is equal to 2.276 which is less than 2.4231and calculated probability of 3.4% is greater than significance level of 2.5%. Hence, we fail to reject the null hypothesis as there is not enough evidence at 2.5% level of significance that there is a difference in the mean number of women’s shoe size between 1980 and 1960.