Answer to Question #342558 in Statistics and Probability for nana

Question

For this test, do the following.

(a) Identify the claim and state H0 and Ha.

(b) Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed, and whether to use a z-test, a t-test, or a chi-square test. Explain your reasoning.

(c) Choose one of the options. If convenient, use technology.

Option 1: Find the critical value(s), identify the rejection region(s), and find the appropriate standardized test statistic.

Option 2: Find the appropriate standardized test statistic and the P-value.

(d) Decide whether to reject or fail to reject the null hypothesis.

(e) Interpret the decision in the context of the original claim.

3. A travel agent says that the mean hotel room rate for a family of 4 in a certain resort town is at most $170. A random sample of 33 hotel room rates for families of 4 has a mean of $179 and a standard deviation of $19. At a = 0.01, is there enough evidence to reject the agent’s claim?

Answer 1

(a) The following null and alternative hypotheses need to be tested:

$H_0:\mu\le170$

$H_1:\mu>170$

(b) This corresponds to a rightt-tailed one-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

(c) Based on the information provided, the significance level is $\alpha = 0.05,$ $df=n-1=32$ degrees of freedom, and the critical value for a right-tailed test is $t_c = 1.693889.$

The rejection region for this right-tailed test is $R = \{t:t>1.693889\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{179-170}{19/\sqrt{33}}\approx2.7211$$

(d) Since it is observed that $t=2.7211>1.693889=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for right-tailed, $df=32$ degrees of freedom, $t=2.7211,$ is $p=0.005218 ,$ and since $p= 0.005218 <0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

(e) Therefore, there is enough evidence to claim that the population mean $\mu$

is greater than 170, at the $\alpha = 0.05$ significance level.