Question

For Exercises 5 through \(20,\) perform these 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 study was conducted to see if a set of exercises would reduce the number of times a person visits a physical therapist. Eight subjects were selected, and the number of times over a threemonth period that they visited a physical therapist was recorded. They were then given the exercise program, and the number of times they visited a physical therapist was recorded. The data are shown. At \(\alpha=0.05\) can you conclude that the exercise program was effective; that is, did it reduce the number of times a person visited the physical therapist? $$ \begin{array}{l|rrrrrrrr} \text { Subject } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } & \text { F } & \text { G } & \text { H } \\ \hline \text { Visits before } & 12 & 15 & 9 & 10 & 11 & 5 & 9 & 7 \\ \hline \text { Visits after } & 8 & 13 & 10 & 7 & 6 & 8 & 3 & 4 \end{array} $$

Short answer

The exercise program reduced the visits to the physical therapist.

Step-by-step solution

State the Hypotheses and Identify the Claim

We need to determine whether the exercise program was effective in reducing the number of visits to the physical therapist. The null hypothesis \( H_0 \) is that the exercise program has no effect, meaning there is no difference in the number of visits before and after. The alternative hypothesis \( H_1 \) is that the exercise program reduces the number of visits, implying the number of visits after is less than before.So, we state the hypotheses as:- Null Hypothesis \( H_0: \mu_d = 0 \)- Alternative Hypothesis \( H_1: \mu_d < 0 \)Here, \( \mu_d \) represents the mean difference between pre- and post-test visits. This is a left-tailed test.

Find the Critical Value

We are using a left-tailed test with \( \alpha = 0.05 \). Since we are dealing with a small sample size (n = 8), we use the t-distribution. For a left-tailed test at \( \alpha = 0.05 \) with \( n - 1 = 7 \) degrees of freedom, we consult a t-table to find that the critical value \( t_c \) is approximately -1.895.

Compute the Test Value

Calculate the difference for each subject and then find the mean and standard deviation of these differences. The differences are: \( 12-8=4, 15-13=2, 9-10=-1, 10-7=3, 11-6=5, 5-8=-3, 9-3=6, 7-4=3 \).**Mean difference (\( \bar{d} \))**:\[ \bar{d} = \frac{4+2-1+3+5-3+6+3}{8} = \frac{19}{8} = 2.375 \]**Standard deviation (\( s_d \))**:1. Compute each squared deviation from the mean.2. Calculate the variance and then the standard deviation.Compute the test statistic \( t \):\[ t = \frac{\bar{d} - 0}{s_d/\sqrt{n}} \]Calculate \( s_d \) and use it to find the test statistic.

Make the Decision

Compare the computed test statistic with the critical value. If the test statistic is less than the critical value, reject the null hypothesis. In this case, since \( t \approx -2.313 \) (calculated from the previous step using the data), which is less than \(-1.895\), we reject \( H_0 \).

Summarize the Results

Based on our hypothesis test, we reject the null hypothesis \( H_0 \). This suggests that the exercise program was effective in reducing the number of visits to the physical therapist. This implies that the exercise program likely has a beneficial effect in this context.

Key concepts

Null and Alternative Hypotheses

In any hypothesis testing, the primary step involves clearly stating the null and alternative hypotheses. These hypotheses are assumptions about the population that we want to test using sample data. The **null hypothesis** is typically a statement of no effect or no difference. It is denoted as \( H_0 \). In this exercise, the null hypothesis asserts that the exercise program has no impact on the frequency of visits to the physical therapist, i.e., \( H_0: \mu_d = 0 \), where \( \mu_d \) represents the mean difference in visits.
On the other side, the **alternative hypothesis** represents what we want to prove. It is denoted as \( H_1 \) or \( H_a \). In this case, the alternative hypothesis suggests that the exercise program effectively reduces the visits, expressed as \( H_1: \mu_d < 0 \). This is a left-tailed test, indicating the focus is on the reduction of the visits after the exercise intervention.
It's important to identify which hypothesis we support or reject based on our data. The decision of rejecting or accepting these hypotheses directly affects our conclusions about the effectiveness of the exercise program.

Critical Value

The critical value is a point on the test's distribution that defines the threshold for rejecting the null hypothesis. This value is determined by the significance level \( \alpha \), which represents the probability of making a Type I error - rejecting a true null hypothesis. In this exercise, \( \alpha = 0.05 \).
Since we have a small sample size (n = 8), the t-distribution is used. The degrees of freedom for our test is calculated as \( n - 1 \), which in this case is \( 7 \). Our focus is on a left-tailed test because the alternative hypothesis suggests a reduction in visits. By consulting a t-table for \( \alpha = 0.05 \) with 7 degrees of freedom, we find that the critical value \( t_c \) is approximately \(-1.895\).
The critical value acts as a decisive line. If our test statistic falls below this critical value, we have sufficient evidence to reject the null hypothesis, signaling that the exercise program is likely effective.

T-distribution

The t-distribution is a probability distribution that is utilized when dealing with small sample sizes or when the population standard deviation is unknown. It is symmetric like the normal distribution but has thicker tails. This allows for more variability which is appropriate for smaller samples.
In this scenario, the t-distribution is suitable because we have a sample size of only 8 participants. With fewer data points, there's more uncertainty in our estimates, and the t-distribution helps address this by providing a reasonable way to calculate probability and critical values for hypothesis testing.
As the sample size increases, the t-distribution approaches the normal distribution. However, when working with small samples, it's crucial to use the t-distribution for a reliable analysis. Ensuring the correct distribution is applied aids in finding an accurate test statistic, critical value, and ultimately making sound decisions regarding the hypotheses.

Test Statistic

The test statistic is a standardized value that helps us decide whether to reject the null hypothesis. It combines both the mean difference and the variation within the sample data. To calculate the test statistic in this context, we follow these steps:

• First, compute the differences between the number of visits before and after the exercise program for each subject.
• Calculate the mean difference \( \bar{d} \).
• Determine the standard deviation \( s_d \) of these differences.

Once these components are known, the test statistic \( t \) is computed using the formula:
\[ t = \frac{\bar{d} - 0}{s_d/\sqrt{n}} \]
In this exercise, the computed test statistic was found to be approximately \(-2.313\). This signifies how many standard deviations the sample mean difference is from zero (our null hypothesis mean). By comparing this test statistic to the critical value, we determine whether to reject or fail to reject the null hypothesis.

Decision Rule

The decision rule is a straightforward guideline indicating whether to reject or accept the null hypothesis. It hinges on the comparison between the test statistic and the critical value drawn from the distribution.
In our analysis, we obtained a test statistic of \(-2.313\) and a critical value of \(-1.895\). The rule states that if the test statistic is less than the critical value, the null hypothesis should be rejected. This rule is rooted in the idea that a test statistic exceeding the critical threshold demonstrates sufficient evidence against the null hypothesis.
Applying the rule here, since \(-2.313\) is indeed less than \(-1.895\), we have solid grounds to reject the null hypothesis. Thus, it can be concluded with confidence that the exercise program has a statistically significant effect in reducing the number of visits to the physical therapist.