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Hypothesis testing is a statistical method used to make decisions about population parameters. It involves making an assumption, called the null hypothesis \( H_0 \), and then determining whether the data provides enough evidence to reject this assumption in favor of an alternative hypothesis \( H_a \).

In our exercise, the null hypothesis \( H_0: \mu_1 - \mu_2 = 0 \) suggests that there is no difference in average room rates between the Wyndham and Radisson hotels. If this hypothesis is rejected, it implies that a significant difference does exist, as stipulated by the alternative hypothesis \( H_a: \mu_1 - \mu_2 \eq 0 \).

To test this hypothesis, a two-sample t-test is employed, which compares the sample means from two different populations. Based on the p-value obtained from the t-test and the chosen significance level \( \alpha \), which is often 0.05, we can either reject \( H_0 \) or fail to reject it. Rejecting \( H_0 \) suggests that the observed difference in sample means is not likely due to random chance. However, without sample sizes for our data, it's important to note that the t-test cannot be completed, and thus hypothesis testing cannot be performed.

A confidence interval provides a range of values within which we can be a certain percentage confident that the population parameter, such as the mean difference, falls. In the context of our exercise, a 95% confidence interval gives us a range that, according to the data, contains the true average difference between room rates 95% of the time.

To construct this interval for the difference in average room rates, we would typically use the formula given in the solution, incorporating the sample statistics and t-value associated with \( \alpha/2 \) for a two-tailed test. However, since the sample sizes \( n_1 \) and \( n_2 \) aren't available in the provided data, it's not possible to calculate this interval accurately.

Understanding the 95% Confidence Interval

If we were able to calculate the interval, and it did not contain 0, this would imply that the difference in room rates is statistically significant at the 5% significance level, which would likely lead us to reject the null hypothesis from our hypothesis testing. Therefore, the confidence interval can serve as a visual and numerical way to judge the hypothesis test's result.

Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean \( \mu \), while a high standard deviation indicates that the values are spread out over a wider range.

In hypothesis testing and specifically in a two-sample t-test, the sample standard deviations \( s_1 \) and \( s_2 \) are used to estimate the variability within each group being compared. This information is crucial as it directly influences the test statistic calculation, which in turn affects the p-value and confidence interval construction.

Role of Standard Deviation in the t-Test

The Wyndham and Radisson hotel chains in our exercise have sample standard deviations of 16.5 and 10, respectively. These values express the amount of variability in the room rates around each sample mean. When comparing two means, the larger the standard deviations, the more overlap there might be between the two datasets, which could make it harder to detect a real difference between the populations' means. It's essential to note that we need sample sizes to pair with the standard deviations for a complete analysis – without them, our test outcome remains inconclusive.