Question

A consumer advocacy group received a tip that an air conditioning company has been charging female customers more than male customers. The group's statistical expert decides examine this question at the α=0.10α=0.10 level of significance, by looking at the difference in mean charges between a random sample of female customers and a random sample of male customers. Let μFμF represent the average charges for female customers and μMμM represent the average charges for male customers.(Round your results to three decimal places)

Which would be correct hypotheses for this test?

• H0:μF−μM=0    H1:μF−μM≠0
• H0:μF−μM= 0 :H1:μF−μM<0
• H0:μF−μM=0 H1:μF−μM>0
• HO:μF−μM>0 H1:μF−μM<0

If we are going to test this using a confidence interval, which confidence interval should we construct?

• 80%
• 60%
• 90%
• 95%

A random sample of 33 female customers were charged an average of $915, with a standard deviation of $8. A random sample of 52 male customers were charged an average of $903, with a standard deviation of $17. Construct the confidence interval:

_____________ < μF−μ  < ____________________

Which is the correct result:

• 0 is contained in the confidence interval, so we Do not Reject the Null Hypothesis
• 0 is not contained in the confidence interval, so we Reject the Null Hypothesis
• 0 is not contained in the confidence interval, so we Do not Reject the Null Hypothesis
• 0 is contained in the confidence interval, so we Reject the Null Hypothesis

Which would be the appropriate conclusion?

• There is significant evidence to suggest that the company is charging its female customers more than its male customers.
• There is not significant evidence to suggest that the company is charging its female customers more than its male customers.

Answer 1

Correct hypotheses for this test:

H0: μF - μM = 0 vs H1: μF - μM > 0

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Confidence interval to construct: = 1 - 0.10 = 0.90 = 90%

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For Female : x̅1 = 915, s1 = 8, n1 = 33

For Male : x̅2 = 903, s2 = 17, n2 = 52

df = ((s1²/n1 + s2²/n2)²)/[(s1²/n1)²/(n1-1) + (s2²/n2)²/(n2-1) ] = 77.7205 = 78

90% Confidence interval for the difference :

At α = 0.1 and df = 78, two tailed critical value, t_c = T.INV.2T(0.1, 78) = 1.665

Lower Bound = (x̅1 - x̅2) - t_c*√(s1²/n1 +s2²/n2) = (915 - 903) - 1.665*√(8²/33 + 17²/52) = 7.4421

Upper Bound = (x̅1 - x̅2) + t_c*√(s1²/n1 +s2²/n2) = (915 - 903) + 1.665*√(8²/33 + 17²/52) = 16.5579

7.4421 < µF - µM < 16.5579

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Correct result:

0 is not contained in the confidence interval, so we Reject the Null Hypothesis

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Appropriate conclusion:

There is significant evidence to suggest that the company is charging its female customers more than its male customers.