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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(1): 8.227273 Standard Deviation for Born in 1980(s1): 0.64667 Sample Mean for Born in 1960 (2): 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: (1 - 2) =0 vs. Ha: (1 - 2) > 0 Formula for test statistics, Using above two-tailed statistics, Rejection region can be defined as z < -z/2 or z > z/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= 8.227273 7.681818 0.64667 2 0.462208 2 11 11 2.276 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.