Question

Perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A researcher wishes to see if there is a difference between the mean number of hours per week that a family with no children participates in recreational activities and a family with children participates in recreational activities. She selects two random samples and the data are shown. At \(\alpha=0.10\), is there a difference between the means? $$ \begin{array}{lrcl} & \bar{X} & \sigma & n \\ \hline \text { No children } & 8.6 & 2.1 & 36 \\ \text { Children } & 10.6 & 2.7 & 36 \end{array} $$

Short answer

The means are significantly different; reject the null hypothesis.

Step-by-step solution

State the Hypotheses

The first step is to clearly state the null hypothesis, which represents the situation where no change or effect is presumed, and the alternative hypothesis, which challenges this status quo. For this problem, we define the hypotheses regarding the mean number of hours spent on recreational activities: - Null hypothesis (H_0): \(\mu_1 = \mu_2\) (The means are equal)- Alternative hypothesis (H_1): \(\mu_1 eq \mu_2\) (The means are not equal)The claim by the researcher is that there is a difference between the means.

Find the Critical Value(s)

Next, we find the critical value(s) based on the chosen level of significance (\alpha=0.10) for a two-tailed test. Using a standard normal distribution table (Z-table) because both population standard deviations are known:- Critical values are approximately \(\pm 1.645\) for \alpha/2 = 0.05 in each tail.

Compute the Test Value

Now, we calculate the test value using the formula for the test statistic in a two-sample z-test:\[ z = \frac{(\bar{X}_1 - \bar{X}_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \]Substituting the values provided:- \(\bar{X}_1 = 8.6, \sigma_1 = 2.1, n_1 = 36\)- \(\bar{X}_2 = 10.6, \sigma_2 = 2.7, n_2 = 36\)The test statistic calculation is:\[ z = \frac{(8.6 - 10.6)}{\sqrt{\frac{2.1^2}{36} + \frac{2.7^2}{36}}} = \frac{-2}{\sqrt{\frac{4.41}{36} + \frac{7.29}{36}}} = \frac{-2}{\sqrt{\frac{11.7}{36}}} = \frac{-2}{0.570} \approx -3.51 \]

Make the Decision

Compare the calculated test value (\approx -3.51) to the critical value (\pm 1.645). Since -3.51 falls outside the range of \([-1.645, 1.645]\), we reject the null hypothesis \(H_0\).

Summarize the Results

Given that we rejected the null hypothesis, we have sufficient evidence to support the researcher's claim that there is a significant difference in the mean number of hours families with and without children participate in recreational activities per week.

Key concepts

Two-sample Z-test

A two-sample Z-test is a statistical method used to compare the means of two independent groups. This test helps to determine if there is a significant difference between these groups' means. In our case, the researcher wants to know if families with children spend a different amount of time on recreational activities compared to families without children. To perform a two-sample Z-test, we need:

• The means of both groups (\(\bar{X}_1\) and \(\bar{X}_2\))
• The standard deviations of both groups (\(\sigma_1\) and \(\sigma_2\))
• The size of each sample (\(n_1\) and \(n_2\))

The formula for the test statistic is:\[ z = \frac{(\bar{X}_1 - \bar{X}_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \]This equation helps compute the Z-value, which can then be compared with the critical values to make a decision on our hypothesis. This is how we utilize the characteristics of Z-distribution for hypothesis testing.

Critical Value

The critical value is a benchmark in hypothesis testing that indicates the threshold beyond which we reject the null hypothesis. It depends on the significance level \(\alpha\), which represents the probability of making a Type I error (rejecting a true null hypothesis).For a two-tailed test, the significance level gets split between the two tails of the standard normal distribution. In this exercise, the chosen \(\alpha\) is 0.10, meaning we have 0.05 in each tail. Therefore, the critical values correspond to \(\pm 1.645\) as determined by the Z-table. This means that if our calculated test statistic falls beyond \(-1.645\) or \(+1.645\), we reject the null hypothesis. The critical value provides a boundary that helps us understand the extremity of our test statistic result relative to the null hypothesis.

Null Hypothesis

The null hypothesis, often denoted as \(H_0\), is an assumption in hypothesis testing that there is no effect or no difference. It serves as a starting point for testing and claims that any observed effect is due to sampling fluctuations. In the given exercise, the null hypothesis is \(\mu_1 = \mu_2\), which suggests there is no difference between the mean number of hours spent on recreational activities by families with and without children. The null hypothesis is crucial as it provides a baseline risk-free claim. If we find enough evidence against it (using our test statistic and the critical value), we can reject it in favor of the alternative hypothesis. However, unless this evidence is overwhelming, we refrain from rejecting the null hypothesis.

Alternative Hypothesis

Contrary to the null hypothesis, the alternative hypothesis, denoted as \(H_1\), suggests that there is indeed an effect or difference. In this context, it's a challenge to the status quo proposed by the null. For the exercise in question, the alternative hypothesis is \(\mu_1 eq \mu_2\). This means the researcher posits a claim that a significant difference exists in the average recreational hours for families with and without children.The goal of hypothesis testing is often to gather enough evidence from sample data to support the alternative hypothesis. If the test statistic in our Z-test falls outside the critical range, then the evidence is considered substantial enough, and the null hypothesis can be rejected, lending support to the alternative hypothesis. This step is fundamental in hypothesis testing as it directs our conclusions regarding the original research question.