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Chapter 9: Q T9.13 (page 616)

A government report says that the average amount of money spent per U.S. household per week on food is about \(158. A random sample of50 households in a small city is selected, and their weekly spending on food is recorded. The sample data have a mean of \)165 and a standard deviation of \(20. Is there convincing evidence that the mean weekly spending on food in this city differs from the national figure of \)158?

Short Answer

Expert verified

There is enough proof to help the claim that the mean weekly spending on food in this city differs from the national figure of $158.

Step by step solution

01

Step 1:Given information

n=50

x炉=165.00

s=20.00

02

Step 2:Explaination

The hypotheses are

H0:渭=$158

H1:渭鈮�$158

The value of the test statistic

t=x炉-渭0s/n

=165.00-15820.00/50

=2.475

The P-value is the probability of getting the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Table B containing the t-value in the row

n-1=50-1=49>40:

0.01=2脳0.005<P<2脳0.01=0.02

If the P-value is lesser than the significance level, then the null hypothesis is rejected.

P<0.05=5%鈬�RejectH0

There is enough proof to help the claim that the mean weekly spending on food in this city differs from the national figure of$158

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