Questions: 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 160 with a population standard deviation of 41. At α=0.10, can you support the analyst's claim using the p value? Claim is the alternative hypothesis, fail to reject the null hypothesis as p-value ( 0.121 ) is not less than alpha (0.10), and cannot support the claim

Solution Steps

To determine if the analyst's claim can be supported, we perform a hypothesis test for the mean. The null hypothesis (\(H_0\)) is that the mean room rate is \( \$152 \) or less, and the alternative hypothesis (\(H_a\)) is that the mean room rate is greater than \( \$152 \). We calculate the test statistic using the sample mean, population standard deviation, and sample size. Then, we find the p-value and compare it to the significance level \(\alpha = 0.10\).

We define the null and alternative hypotheses as follows:

• Null Hypothesis: \( H_0: \mu \leq 152 \)
• Alternative Hypothesis: \( H_a: \mu > 152 \)

Using the formula for the z-test statistic: \[ z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} \] where:

• \( \bar{x} = 160 \) (sample mean)
• \( \mu_0 = 152 \) (hypothesized population mean)
• \( \sigma = 41 \) (population standard deviation)
• \( n = 36 \) (sample size)

Substituting the values: \[ z = \frac{160 - 152}{41 / \sqrt{36}} = \frac{8}{6.8333} \approx 1.1707 \]

The p-value is calculated using the z-score: \[ p\text{-value} = 1 - \Phi(z) \approx 1 - \Phi(1.1707) \approx 0.1209 \] where \( \Phi(z) \) is the cumulative distribution function of the standard normal distribution.

We compare the p-value with the significance level:

• \( \alpha = 0.10 \)
• \( p\text{-value} \approx 0.1209 \)

Since \( p\text{-value} > \alpha \), we fail to reject the null hypothesis.

Final Answer