Question

Question 3: 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, which is $131. 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) Formulate the hypotheses. b) What is the value of the test statistic? c) What are the critical values of the test? d) Find the p-value of the test. e) What is your conclusion? Explain it in the context of the problem.

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis: μ = 131
Alternative Hypothesis: μ > 131

Rejection Region
This is right tailed test, for α = 0.05 and df = 49
Critical value of t is 1.677.
Hence reject H0 if t > 1.677

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (133.474 - 131)/(11.193/sqrt(50))
t = 1.563

P-value Approach
P-value = 0.0622
As P-value >= 0.05, fail to reject null hypothesis.

There is not significant evidence to conclude that the mean weekly grocery is not equals to 131.
It is equal to 131