A study is performed in San Antonio to determine whether the average weekly grocery bill per five-person family in the town is significantly different from the national average. A random sample of 50 five-person families in San Antonio showed a mean of $133.474 and a standard deviation of $11.193.
(A) (A) Assume that the national average weekly grocery bill for a five-person family is $131. Does the sample result provide convincing evidence that the mean weekly grocery bill of a five-person family in San Antonio is different from the national average?
?
(B) For which values of the sample mean (i.e., average weekly grocery bill) would you decide to reject the null hypothesis at the 
significance level? For which values of the sample mean would you decide to reject the null hypothesis at the 10% level of significance?
What will be an ideal response?
(A) 
vs. 
?
n= 50, 
=133.474, s =11.193
t-value =
?
?
Using the Excel® function =T DIST(1.563, 49, 2), we get the p-value = 0.1245. Because the p-value is not less than 0.05, we do not have convincing evidence to conclude thatthat the mean weekly grocery bill of a five-person family in San Antonio is different from the national average. The sample mean is not significantly different from 131 at even the 10% level because the p-value is greater than 0.10. We cannot reject the null hypothesis.
?
(B) For either p-value (0.01 or 0.10), we find the t-value that would lead to the rejection of the null hypothesis, and then solve the equation 
on either side of $131. This leads to the following results:
?
?
?
| t-value | Lower limit | Upper limit |
0.01 | 2.680 | 126.758 | 135.242 |
0.10 | 1.677 | 128.346 | 133.654 |
-value?
?
For example, at the 10% level, if 
we would reject 
.
?
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