Question

a travel analyst claims that the mean of the room rates for two adults three-star hotels in New Heaven is $134. In a random sample 37 three-star hotels in New Heaven, the mean room rate for two adults is $143. Assume the population standard deviation is $30. At a=0.10, is there enough evidence to reject the analyst's claim

Answer 1

To assess the claim about mean hotel rates, we perform a one-sample z-test with the specified sample data and known population standard deviation. We calculate the test statistic and compare it to the critical z-value for a significance level of 0.10 to determine if we can reject the analyst's claim.

Step-by-step explanation:

The question asks if there is enough evidence to reject a travel analyst's claim about the mean room rates for two adults in three-star hotels in New Haven. To do this, we will conduct a one-sample z-test because the population standard deviation is known. The steps are as follows

1. State the null hypothesis (H0): The population mean (μ) is equal to the claimed mean, which is $134.
2. State the alternative hypothesis (H1): The population mean (μ) is not equal to the claimed mean ($134).
3. Calculate the test statistic using the formula: z = (xbar - μ) / (σ/√n), where xbar is the sample mean, μ is the claimed population mean, σ is the population standard deviation, and n is the sample size.
4. Find the critical z-value for α = 0.10, looking at the z-table for a two-tailed test.
5. Compare the calculated z-value with the critical z-value.
6. If the calculated z-value lies beyond the critical z-value, we reject the null hypothesis. Otherwise, we fail to reject it.

Applying this, we find the z-value for a sample mean of $143, a population mean of $134, a population standard deviation of $30, and a sample size of 37. Comparing against the critical z-value for α = 0.10, we will be able to conclude whether there is enough evidence to reject the claim.