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.

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:[tex]\mu_{1}\leq \mu_{2}[/tex]

Alternative hypothesis:[tex] \mu_{1} > \mu_{2}[/tex]

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:

[tex]t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}[/tex] (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[tex]\alpha=0.05[/tex] and [tex]\alpha/2=0.025[/tex] 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 [tex]t_{1-\alpha/2}=1.66[/tex]

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:

[tex]t=\frac{360000-340000}{\sqrt{\frac{50000^2}{36}+\frac{40000^2}{36}}}}=1.874[/tex]

What is the p-value for this hypothesis test?

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

[tex]p_v =P(t_{70}>1.874)=0.033[/tex]

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

Comparing the p value with the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] 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.

astha8579 astha8579 · Answer 2 · 2025-03-23T23:00:43-04:00

You can use t test for two means to test the hypotheses.

The answer is

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

How to form the hypotheses?

There are two hypotheses. First one is called null hypothesis and it is chosen such that it predicts nullity or no change in a thing. It is usually the hypothesis against which we do the test. The hypothesis which we put against null hypothesis is alternate hypothesis.

Null hypothesis is the one which researchers try to disprove.

Thus, for given condition, we get the hypotheses as

Null hypothesis [tex]H_0 : \mu_1 \leq \mu_2\\[/tex]

Alternative hypothesis(the hypothesis we want to test): [tex]H_1 : \mu_1 > \mu_2[/tex]
The samples have these statistics:

For sample 1:

[tex]n_1 = 36\\s_1 = 50,000\\\overline{x}_1 = 360,000[/tex]

For sample 2( freeway):

[tex]n_2 = 36\\s_2 = 40,000\\\overline{x}_2 = 340,000[/tex]

Using the t test statistic for two means, we have:

[tex]t = \dfrac{\overline{x}_1 - \overline{x}_2 }{\sqrt{\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}}}[/tex]

Using the values given, we get:

[tex]t = \dfrac{\overline{x}_1 - \overline{x}_2 }{\sqrt{\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}}} = \dfrac{360000 - 340000}{\sqrt{\dfrac{50000^2}{36} + \dfrac{40000^2}{36}}} = \dfrac{120000}{10000\times \sqrt{42}} \approx 1.8519[/tex]

The degree of freedom is [tex]n_1 + n_2 - 2 = 70[/tex]

At 5% level of significance, with 70 degree of freedom, at t = 1.85 and single tailed test, we get the p-value as

The p-value is 0.034267.

The p value obtained is less than the level of significance as 5% 0.05 and p value is 0.0342 approx.

Thus, we cannot accept the null hypothesis and we accept the alternative hypothesis (null hypothesis would've been accepted if p value would not be less than the level of significance).

The manager's claim is supported as the test statistic was obtained higher(which made the p value lower, the more higher the t test statistic for right tailed test go, the less the probability area will be left) than the critical value.

Thus,

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

Learn more about t test here:

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