Question

For a random sample of 64 Iowa homes, average weekly food expenditure turns out to be $160, with a standard deviation of $64. Let μ denote the mean weekly food expenditure for Iowa families.

Find a 95% confidence interval for μ, when it is additionally known that the standard deviation of weekly food expenditure for all Iowa families is $64.

Find the p-value of the appropriate test of hypotheses for checking if µ exceeds $145, when it is additionally known that the standard deviation of weekly food expenditure for all Iowa families is $64.

Answer 1

Given:

n=64, = 160, S = 64, = 0.05

Find: 95% Confidence interval

Where,

Critical value:

Z_/2 = Z_0.05/2 = 1.96

95% Confidence interval:

Hypothesis:

Ho: = 145

Ha: > 145 (Mean weekly food expenditure for Iowa families exceeds 145)

Test statistic:

P-value: 0.0304 (From Normal table)

Conclusion:

P-value < , i.e 0.0304 < 0.05, That is Reject Ho at 5% level of significance.

Therefore, Mean weekly food expenditure for Iowa families exceeds 145