The average expenditure on Valentine\'s Day was expected to be $100.89 (USA Toda

Question

The average expenditure on Valentine's Day was expected to be $100.89 (USA Today, February 13, 2006). Do male and female consumers differ in the amounts they spend? The average expenditure in a sample survey of 43 male consumers was $135.67, and the average expenditure in a sample survey of 36 female consumers was $68.64. Based on past surveys, the standard deviation for male consumers is assumed to be $35, and the standard deviation for female consumers is assumed to be $21.

a) What is the point estimate of the difference between the population mean expenditure for males and the population mean expenditure for females (to 2 decimals)?

b) At 99% confidence, what is the margin of error (to 2 decimals)?

c) Develop a 99% confidence interval for the difference between the two population means (to 2 decimals).

Explanation / Answer

Given that X1 = 135.67   X2 = 68.64 n1 = 43 n2 = 36   S1 = 35 S2 = 21

The null hypothesis is given by

H0 : µ1 - µ2 = 100.89 i.e., there is no difference in the amount spend by males and females

Against the alternative hypothesis

Ha : µ1 - µ2   100.89 (two tailed test) i.e., there is difference in the amount spend by males and females

The test statistic is given by

Z =   X1 - X2 / ((S1)2/n1) + ((S2)2/n2) N(0,1)

Z = 135.67 -68.64/ ((35)2/43) + ((21)2/36)

Z =   67.03/ 28.4884 +12.25

Z =   -4 / 40.7384

Z cal     = 10.5018

Z tab at = 0.01 is 2.58,

Conclusion : Zcal > Z­tab at 1% level of significance . hence the null hypothesis is rejected

99% confidence interval for µ1 - µ2 is x1-bar   -   x2-bar ± 2.58 S.E(x1-bar   -   x2-bar)

67.03 ±2.58 (6.3827)

[50.5626 , 83.4974]

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