The Virginia Department of Social Services collected information on 25
Virginia Supplemental Nutrition Assistance Program (SNAP, formerly Food
Stamps) beneficiaries and found that their average monthly food expenditure
was $160. The average monthly food expenditure for SNAP beneficiaries
nationwide is assumed to be $136. Assume that the SNAP beneficiaries’
expenditures are normally distributed and have a standard deviation of
$15.40. If you were to test the claim that Virginia SNAP beneficiaries have a
monthly expenditure different from that of $136, using a 0.05 significance
level, what would be the test statistic?
a) 1.56
b) 7.79
c) 1.96
d) -7.79
e) -1.56
I used a t-test and got b)7.79
To calculate the t-test statistic for testing the claim that Virginia SNAP beneficiaries have a monthly expenditure different from $136, you can follow these steps:
Step 1: Set up the null and alternative hypotheses:
- Null hypothesis (H0): The average monthly food expenditure for Virginia SNAP beneficiaries is equal to $136.
- Alternative hypothesis (Ha): The average monthly food expenditure for Virginia SNAP beneficiaries is different from $136.
Step 2: Determine the significance level, which is given as 0.05 in this case.
Step 3: Calculate the test statistic using the following formula:
t = (x̄ - μ) / (s / √n)
- x̄: Sample mean (average monthly food expenditure for Virginia SNAP beneficiaries) = $160
- μ: Population mean (assumed average monthly food expenditure for SNAP beneficiaries nationwide) = $136
- s: Standard deviation = $15.40
- n: Sample size = 25
Plugging the values into the formula:
t = (160 - 136) / (15.40 / √25)
t = 24 / (15.40 / 5)
t = 24 / 3.08
t ≈ 7.792
Step 4: Compare the calculated t-test statistic with the critical t-value for a two-tailed test at a significance level of 0.05. Since the sample size is small (n < 30), you would use a t-distribution rather than a z-distribution.
The critical t-value for a two-tailed test at a significance level of 0.05 with 24 degrees of freedom is approximately 2.064.
Since the calculated t-test statistic (7.792) is greater than the critical t-value (2.064), we reject the null hypothesis.
Therefore, the correct answer is b) 7.79.