Answer:
[tex]H_o: \mu \geq 13.04\\H_{\alpha}: \mu \geq 13.04[/tex]
b) 0.0751
c) We cannot afirm whether eating out or dinning in is cheaper
d) do not reject neither.
Questions:
(a). Develop appropriate hypotheses for a test to determine whether the sample data support the conclusion that the mean cost of a restaurant meal is less than fixing a comparable meal at home.
(b). Using the sample from the 100 restaurants, what is the p -value?
(c). At α = .05 ,what is your conclusion?
(d). Repeat the preceding hypothesis test using the critical value approach
Explanation:
(b)
[tex]\frac{X-\mu}{\sigma \sqrt{n} } = P_v\\\frac{12.75-13.04}{2 \sqrt{100} } = -1.45[/tex]
We use excel function to get the Pz which is .0751
(c). 0.0751 > .05 , We cannot reject the hypotesis. Thus, we can neither say dinning in is more cheap nor eating out is cheaper.
(d). With = 99 , t .05 = −1.66 Reject H 0 if ? ≤ −1.66 −1.45 > −1.66 , do not reject
Degrees of freedom = n − 1 = 99 and Pz of 0.05 = -1.66
Reject when Pz < of -1.66
-1.45 > 1.66 threfore we cannot reject the hypotesis.
