Question

Condé Nast Traveler conducts an annual survey in which readers rate their favorite cruise ship. All ships are rated on a 100-point scale, with higher values indicating better service. A sample of 37 ships that carry fewer than 500 passengers resulted in an average rating of 85.36, and a sample of 44 ships that carry 500 or more passengers provided an average rating of 81.40. Assume that the population standard deviation is 4.55 for ships that carry fewer than 500 passengers and 3.97 for ships that carry 500 or more passengers.

At a α = 0.01, what is your conclusion?

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Answer 1

The provided sample means are shown below:

$\bar X_1 = 85.36$

$\bar X_2 = 81.40$

Also, the provided population standard deviations are:

$\sigma_1 = 4.55$

$\sigma_2 = 3.97$

and the sample sizes are

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

$Ho: \mu_1 = \mu_2$​

$Ha: \mu_1 \ne \mu_2$​

This corresponds to a two-tailed test, for which a z-test for two population means, with known population standard deviations, will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.01, and the critical value for a two-tailed test is zc​=2.58.

The rejection region for this two-tailed test is R={z:∣z∣>2.58}

(3) Test Statistics

The z-statistic is computed as follows:

$z = \frac{\bar X_1 - \bar X_2}{\sqrt{\sigma_1^2/n_1 + \sigma_2^2/n_2}} = \frac{ 85.36 - 81.40}{\sqrt{ 4.55^2/37 + 3.97^2/44}} = 4.134$

(4) Decision about the null hypothesis

Since it is observed that ∣z∣=4.134>zc​=2.58, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0, and since p=0<0.01, it is concluded that the null hypothesis is rejected. There we can conclude at the 1% level of significance that there is significant difference between the ratings of the two ships.

Confidence Interval

The 99% confidence interval for μ1​−μ2​ is 1.492 1.492<μ1​−μ2​<6.428.

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