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Testing of hypothesis and 95% Confidence Interval

According to a survey conducted in October, 2001, consumers were trying to reduce their credit card debt (Margaret Price, "Credit Debts Get Cut Down to Size," Newsday, November 25, 2001, F3). Based on a sample of 1,000 consumers in October 2001 and in October 2000, the mean credit card debt was $2,411 in October 2001 as compared to $2,814 in October 2000. Suppose that the standard deviation was $847.43 in October 2001 and $976.93 in October 2000.

a) Assuming that the population variances from both years are equal, is there evidence that the mean credit card debt is lower in October 2001 than in October 2000? (use =0.05 level of significance).
b) Find the p-value in (a) and interpret its meaning.
c) Assuming that the population variances from both years are equal, construct and interpret a 95 % confidence interval estimate of the difference between the population means in October 2001 and October 2000.

Solution Preview

Please see the attached file.

Given that

N Mean SD
October 2001 1000 $2411 $847.43
October 2000 1000 $2814 $976.93

The null hypothesis under consideration is

H0: There is no significant difference in the mean credit card debt of October 2001 & October 2000
H1: The mean credit card debt of October ...

Solution Summary

The solution gives step by step procedure for the calculation of testing of hypothesis and 95% Confidence Interval for population mean of credit card debt. The solution contains null hypothesis, alternative hypothesis, critical value, P value, and decision rule with interpretations.

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