Question

Extra Credit: Tests of hypothesis for a mean rarm & Day spa A report states that the mean yearly salary offer for studen Submit Answer -15 points POoStats 10.E.050. My Notes Ask Your A magazine collects data each year on the price of a hamburger in a certain fast food restaurant in various countries around the world. The price of this hamburger for a sample of restaurants in Europe in January resulted in the following hamburger prices (after conversion to U.S. dollars). 5.18 4.92 4.09 4.68 5.27 4.69 4.13 4.98 5.12 5.52 5.36 4.60 The mean price of this hamburger in the U.S. in January was $4.61. For purposes of this exercise, assume it is reasonable to regard the sample as representative of these European restaurants. Does the sample provide convincing evidence that the mean January price of this hamburger in Europe is greater than the reported U.S. price? Test the relevant hypotheses using a- 0.05. (Use a statistical computer package to calculate the P-value. Round your test statistic to two decimal places and your P-value to three decimal places.) P-value = State your conclusion. O Reject Ho- We have convincing evidence that the mean price of a hamburger in a certain fast food restaurant in Europe is greater than $4.61. O Do not reject Ho. We do not have convincing evidence that the mean price of a hamburger in a certain fast food restaurant in Europe is greater than $4.61. Reject Ho. We do not have convincing evidence that the mean price of a hamburger in a certain fast food restaurant in Europe is greater than $4.61. Do not reject Ho. We have convincing evidence that the mean price of a hamburger in a certain fast food restaurant in Europe is greater than $4.61. You may need to use the appropriate table in Appendix A to answer this question. Need Help? Talk to a Tutor Read it

Answer 1

We need to calculate the average and standard deviation of the sample first in order to compute t statistic and P-value.

The sample size is n=12. The provided sample data along with the data required to compute the sample mean Xbar and sample variance s² are shown in the table below:

	Price	Price2
	5.18	26.8324
	4.92	24.2064
	4.09	16.7281
	4.68	21.9024
	5.27	27.7729
	4.69	21.9961
	4.13	17.0569
	4.98	24.8004
	5.12	26.2144
	5.52	30.4704
	5.36	28.7296
	4.60	21.16
Sum =	58.54	287.87

The sample mean is computed as follows:

X 58.54 -= 4.878 =- A; = 12

Also, the sample variance s² is

58.542 *(-+ (2x)) = (2737 - SBF) = 0.206 28 7.87 - = 0.208 n - 11 12 - 1 12

Therefore, the sample standard deviation s is

S=V92 = V0.208 = 0.457

Now we will compute t statistic and P-value and critical region on the basis of null and alternative hypothesis.

The provided sample mean is Xbar=4.878 and the sample standard deviation is s=0.457, and the sample size is n=12.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 4.61, The mean price of a hamburger in a certain fast-food restaurant in Europe is $4.61

Ha: μ > 4.61, The mean price of a hamburger in a certain fast-food restaurant in Europe is greater than $4.61

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a right-tailed test is c​=1.796.

The rejection region for this right-tailed test is R = {t: t > 1.796}

Here t value is calculated using t disrtibution table with 5% level of significance with 11 degrees of freedom.

(3) Test Statistics

The t-statistic is computed as follows:

t-X-Mo _ 4.878 – 4.61 = = 2.031 s/n 0.457/V12

(4) A decision about the null hypothesis

Since it is observed that t=2.031>tc​=1.796, it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

We need to calculate p-value using t distribution with 11 degrees of freedom at t = 2.031

The number of degrees of freedom is df=11. We need to find P[T≥2.031] = P-value

The following is obtained graphically:

$normaldistributiongrapher.php?mean=0&sig$

The p-value is p = 0.0336 , and since p=0.0336<0.05, it is concluded that the null hypothesis is rejected.

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is greater than 4.61, at the 0.05 significance level. i.e.

The mean price of a hamburger in a certain fast-food restaurant in Europe is $4.61