Question

1. A survey of 37 men found that an average amount spent on St. Patrick's day of $91 with a standard deviation of $39. A similar survey of 54 women found they spent an average of $46 with a standard deviation of $17. When testing the hypothesis (at the 5% level of significance) that the variance that men spend more than the variance of what women spend on St. Patrick's day, what is the test statistic? (please round your answer to 2 decimal places) ***PLEASE SHOW WORK***

Answer 1

Data:       

n1 = 37      

n2 = 54      

s1^2 = 1521      

s2^2 = 289      

Hypotheses:      

Ho: σ1^2 = σ2^2      

Ha: σ1^2 > σ2^2      

Decision Rule:      

α = 0.05      

Numerator DOF = 37 - 1 = 36    

Denominator DOF = 54 - 1 = 53    

Critical F- score = 1.638742     

Reject Ho if F > 1.638742     

Test Statistic:      

F = s1^2 / s2^2 = 1521/289 = 5.26

p- value = 3.1E-08      

Decision (in terms of the hypotheses):    

Since 5.262976 > 1.6387 we reject Ho and accept Ha

Conclusion (in terms of the problem):    

There is sufficient evidence that the population variance for men is significantly greater than that for women.