Question

1. (CO 7) A travel analyst claims that the mean room rates at a three-star hotel...

1.

(CO 7) A travel analyst claims that the mean room rates at a three-star hotel in Chicago is greater than $152. In a random sample of 36 three-star hotel rooms in Chicago, the mean room rate is $163 with a standard deviation of $41. At α=0.10, what type of test is this and can you support the analyst’s claim using the p-value?

Claim is the alternative, fail to reject the null as p-value (0.054) is less than alpha (0.10), and can support the claim
Claim is the null, fail to reject the null as p-value (0.054) is less than alpha (0.10), and cannot support the claim
Claim is the alternative, reject the null as p-value (0.054) is less than alpha (0.10), and can support the claim

Claim is the null, reject the null as p-value (0.054) is less than alpha (0.10), and cannot support the claim

2.

(CO7) A car company claims that the mean gas mileage for its luxury sedan is at least 24 miles per gallon. A random sample of 7 cars has a mean gas mileage of 23 miles per gallon and a standard deviation of 1.2 miles per gallon. At α=0.05, can you support the company’s claim?

Yes, since the test statistic is not in the rejection region defined by the critical value, the null is not rejected. The claim is the null, so is supported
No, since the test statistic is not in the rejection region defined by the critical value, the null is not rejected. The claim is the null, so is supported
Yes, since the test statistic is not in the rejection region defined by the critical value, the null is not rejected. The claim is the null, so is supported

No, since the test statistic is in the rejection region defined by the critical value, the null is rejected. The claim is the null, so is not supported

3.

(CO7) A state Department of Transportation claims that the mean wait time for various services at its different location is more than 6 minutes. A random sample of 16 services at different locations has a mean wait time of 9.5 minutes and a standard deviation of 7.6 minutes. At α=0.05, can the department’s claim be supported?

Yes, since p of 0.043 is greater than 0.05, fail to reject the null. Claim is null, so is supported
No, since p of 0.043 is greater than 0.05, reject the null. Claim is null, so is not supported
No, since p of 0.043 is greater than 0.05, fail to reject the null. Claim is alternative, so is not supported

Yes, since p of 0.043 is less than 0.05, reject the null. Claim is alternative, so is supported

4.

(CO7) A used car dealer says that the mean price of a three-year-old sport utility vehicle in good condition is $18,000. A random sample of 20 such vehicles has a mean price of $18,450 and a standard deviation of $1140. At α=0.08, can the dealer’s claim be supported?

No, since the test statistic of 1.77 is close to the critical value of 1.85, the null is not rejected. The claim is the null, so is supported
Yes, since the test statistic of 1.77 is in the rejection region defined by the critical value of 1.46, the null is rejected. The claim is the null, so is supported
Yes, since the test statistic of 1.77 is not in the rejection region defined by the critical value of 1.85, the null is not rejected. The claim is the null, so is supported
No, since the test statistic of 1.77 is in the rejection region defined by the critical value of 1.85, the null is rejected. The claim is the null, so is not supported

Homework Answers

Answer #1

Solution:-

1) Claim is the alternative, reject the null as p-value (0.054) is less than alpha (0.10), and can support the claim.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u < 152
Alternative hypothesis: u > 152

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 6.8333

z = (x - u) / SE

z = 1.61

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a z statistic test statistic of 1.61.

Thus the P-value in this analysis is 0.054.

Interpret results. Since the P-value (0.054) is less than the significance level (0.10), we have to reject the null hypothesis.

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