Question

Which is cheaper: eating out or dining in? The mean cost of a flank steak, broccoli, and rice bought at the grocery store is $13.04. A sample of 100 neighborhood restaurants showed a mean price of $12.60 and a standard deviation of $2 for a comparable restaurant meal. (a) Choose the 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. H0: - Select your answer - Ha: - Select your answer - (b) Using the sample from the 100 restaurants, what is the p value? If required, round your answer to four decimal places. (c) At α = 0.05, what is your conclusion? We - Select your answer - the null hypothesis. We - Select your answer - conclude that the cost of a restaurant meal is significantly cheaper than a comparable meal fixed at home.

Answer 1

Answer:

a) The null and alternative hypothesis are:

$H_0: \mu=13.04\\\\H_a:\mu< 13.04$

b) P-value = 0.0151

c) We reject the null hypothesis.

We conclude that the cost of a restaurant meal is significantly cheaper than a comparable meal fixed at home.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the mean cost of a restaurant meal is significantly less than fixing a comparable meal at home.  

Then, the null and alternative hypothesis are:

$H_0: \mu=13.04\\\\H_a:\mu< 13.04$

The significance level is 0.05.

The sample has a size n=100.

The sample mean is M=12.6.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=2.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{2}{\sqrt{100}}=0.2$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{12.6-13.04}{0.2}=\dfrac{-0.44}{0.2}=-2.2$

The degrees of freedom for this sample size are:

$df=n-1=100-1=99$

This test is a left-tailed test, with 99 degrees of freedom and t=-2.2, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t$

As the P-value (0.0151) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the mean cost of a restaurant meal is significantly less than fixing a comparable meal at home.