Question

Retailers who sell women’s shoes such as Wal-Mart, Nordstrom’s, and Payless Shoes have all reported that, more and more often, their larger shoe sizes are the ones that typically sell out first. These businesses have all therefore been aggressively stocking larger shoe sizes under the assumption that women’s feet have been getting bigger over time. Someone wants to test whether this hypothesis is true and has randomly sampled 12 women who were born in 1960 plus 12 who were born in 1980, asking each woman their shoe size. Those data are below.

Born in 1980:
8	7.5	8.5	8	7.5	9.5
7.5	8	8	8.5	9	8.5
Born in 1960:
8.5	7.5	8	8	7.5	7.5
7.5	8	7	8	7	8

1. What is the average shoe size for women born in 1960?
2. What is the average shoe size for women born in 1980?
3. Using symbols and not words, state the null and alternate hypothesis for the statistical test in this case, that women’s feet have grown on average.
4. Calculate the z-value test statistic. Round it to 2 decimal places.
5. Calculate the p-value for this hypothesis test. Round it to 4 decimal places.
6. At the 90% confidence level, do you Reject H₀ or Fail to Reject H₀?
7. At the 95% confidence level, do you Reject H₀ or Fail to Reject H₀?
8. At the 99% confidence level, do you Reject H₀ or Fail to Reject H₀?
1. If you are satisfied with the 95% confidence level, would you say women’s feet have gotten bigger?

Answer 1

a)

average shoe size for born in 1960 =92.5 / 12 = 7.7083

b)

average shoe size for born in 1980 = 98.5/12 = 8.2083

c)

Ho :   µ1 - µ2 =   0          
Ha :   µ1-µ2 > 0          
                  
Level of Significance ,    α =    0.05          
                  
sample #1   ------->   born in 1980          
mean of sample 1,    x̅1=   8.208333333          
population std dev of sample 1,   σ1 =    0.620056205          
size of sample 1,    n1=   12          
                  
sample #2   --------->   born in 1960          
mean of sample 2,    x̅2=   7.708333333          
population std dev of sample 2,   σ2 =    0.450168319          
size of sample 2,    n2=   12          
                  
difference in sample means = x̅1 - x̅2 =    8.208333333   -   7.708333333   =   0.5
                  
std error , SE =    √(σ1²/n1+σ2²/n2) =    0.2212          
                  

4)

Z-statistic = ((x̅1 - x̅2)-µd)/SE =    0.5   /   0.2212   =   2.26
                  
Z-critical value , Z* =        1.645   [excel function =NORMSINV(α)]      
                  

e)

p-value =        0.0119   [excel function =NORMSDIST(z)]      

f)

Level of Significance ,    α =    0.1          
z-critical value =    Z α/2 =    1.6449   [excel function =normsinv(α/2) ]      
                  
std error , SE =    √(σ1²/n1+σ2²/n2) =    0.2212          
margin of error, E = Z*SE =    1.6449   *   0.221   =   0.3638
                  
difference of means = x̅1 - x̅2 =    8.208333333   -   7.708333333   =   0.500
confidence interval is                   
Interval Lower Limit= (x̅1 - x̅2) - E =    0.500   -   0.364   =   0.1362
Interval Upper Limit= (x̅1 - x̅2) + E =    0.500   +   0.364   =   0.8638

g)

Level of Significance ,    α =    0.05          
z-critical value =    Z α/2 =    1.9600   [excel function =normsinv(α/2) ]      
                  
std error , SE =    √(σ1²/n1+σ2²/n2) =    0.2212          
margin of error, E = Z*SE =    1.9600   *   0.221   =   0.4335
                  
difference of means = x̅1 - x̅2 =    8.208333333   -   7.708333333   =   0.500
confidence interval is                   
Interval Lower Limit= (x̅1 - x̅2) - E =    0.500   -   0.434   =   0.0665
Interval Upper Limit= (x̅1 - x̅2) + E =    0.500   +   0.434   =   0.9335

h)

Level of Significance ,    α =    0.01          
z-critical value =    Z α/2 =    2.5758   [excel function =normsinv(α/2) ]      
                  
std error , SE =    √(σ1²/n1+σ2²/n2) =    0.2212          
margin of error, E = Z*SE =    2.5758   *   0.221   =   0.5698
                  
difference of means = x̅1 - x̅2 =    8.208333333   -   7.708333333   =   0.500
confidence interval is                   
Interval Lower Limit= (x̅1 - x̅2) - E =    0.500   -   0.570   =   -0.0698
Interval Upper Limit= (x̅1 - x̅2) + E =    0.500   +   0.570   =   1.0698

i)

confidence interval does not contain 0 ,so we can say that women’s feet have gotten bigger