Question

A restaurant chain has two locations in a medium-sized town and, believing that it has oversaturated the market for its food, is considering closing one of the restaurants. The manager of the restaurant with a downtown location claims that his restaurant generates more revenue than the sister restaurant by the freeway. The CEO of this company, wishing to test this claim, randomly selects 36 monthly revenue totals for each restaurant. The revenue data from the downtown restaurant have a mean of $360,000 and a standard deviation of $50,000, while the data from the restaurant by the freeway have a mean of $340,000 and a standard deviation of $40,000. Assume there is no reason to believe the population standard deviations are equal, and let ?1 and ?2 denote the mean monthly revenue of the downtown restaurant and the restaurant by the freeway, respectively.
Refer to Exhibit 10.3. At the 5% significance level, does the evidence support the manager's claim?
No, since the test statistic value is less than the critical value.
Yes, since the test statistic value is less than the critical value.
No, since the test statistic value is greater than the critical value.
Yes, since the test statistic value is greater than the critical value.

Answer 1

Answer:

Yes, since the test statistic value is greater than the critical value.

Step-by-step explanation:

Data given and notation

$\bar X_{1}=360000$ represent the mean for the downtown restaurant

$\bar X_{2}=340000$ represent the mean for the freeway restaurant

$s_{1}=50000$ represent the sample standard deviation for downtown

$s_{2}=40000$ represent the sample standard deviation for the freeway

$n_{1}=36$ sample size for the downtown restaurant

$n_{2}=36$ sample size for the freeway restaurant

t would represent the statistic (variable of interest)

$\alpha=0.05$ significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the true mean of revenue for downtown is higher than for freeway restaurant, the system of hypothesis would be:

Null hypothesis:$\mu_{1}\leq \mu_{2}$

Alternative hypothesis:$\mu_{1} > \mu_{2}$

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Determine the critical value.

Based on the significance level$\alpha=0.05$ and $\alpha/2=0.025$ we can find the critical values from the t distribution dith degrees of freedom df=36+36-2=55+88-2=70, we are looking for values that accumulates 0.025 of the area on the right tail on the t distribution.

For this case the value is $t_{1-\alpha/2}=1.66$

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

$t=\frac{360000-340000}{\sqrt{\frac{50000^2}{36}+\frac{40000^2}{36}}}}=1.874$

What is the p-value for this hypothesis test?

Since is a right tailed test the p value would be:

$p_v =P(t_{70}>1.874)=0.033$

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the true mean for the downtown revenue restaurant seems higher than the true mean revenue for the freeway restaurant.

The best option would be:

Yes, since the test statistic value is greater than the critical value.