Problem 8 - An Introduction to Psychological Statistics

Question

Lots of people claim that having a pet helps lower their stress level. Use the following summary data to test the claim that there is a lower average stress level among pet owners (group 1) than among non-owners (group 2) at the .05 level of significance. $$\overline{X_{1}}=16.25, \overline{X_{2}}=20.95, s_{1}=4.00, s_{2}=5.10, n_{1}=29, n_{2}=25$$

Step-by-Step Solution

Answer

Reject the null hypothesis; pet owners have lower stress levels.

Step 1: Define the Hypotheses

Start by setting up the null and alternative hypotheses. The null hypothesis \( H_0 \) is that there is no difference in stress levels between pet owners and non-owners: \( \mu_1 = \mu_2 \). The alternative hypothesis \( H_a \) is that pet owners have a lower stress level: \( \mu_1 < \mu_2 \).

Step 2: Choose the Significance Level

The significance level \( \alpha \) is given as 0.05, which will be used to determine whether to reject the null hypothesis.

Step 3: Calculate the Test Statistic

Use the formula for the t-test statistic for independent samples:\[ t = \frac{\overline{X_1} - \overline{X_2}}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]Substitute the given values: \( \overline{X_1} = 16.25 \), \( \overline{X_2} = 20.95 \), \( s_1 = 4.00 \), \( s_2 = 5.10 \), \( n_1 = 29 \), \( n_2 = 25 \). Calculate:\[ t = \frac{16.25 - 20.95}{\sqrt{\frac{4.00^2}{29} + \frac{5.10^2}{25}}} = \frac{-4.70}{\sqrt{0.5517 + 1.0416}} = \frac{-4.70}{1.2364} \approx -3.80\]

Step 4: Determine the Critical Value

Since it's a one-tailed test with \( \alpha = 0.05 \), find the critical t-value from the t-distribution table for \( df = min(n_1 - 1, n_2 - 1) = 24 \). The critical value for \( t \) at \( df = 24 \) and \( \alpha = 0.05 \) is approximately \( -1.711 \).

Step 5: Compare Test Statistic to Critical Value

Compare the calculated test statistic from Step 3, \( t = -3.80 \), to the critical value from Step 4, \( -1.711 \). Since \( -3.80 < -1.711 \), we reject the null hypothesis.

Step 6: Conclusion

Given that the test statistic is less than the critical value, we reject the null hypothesis. There is sufficient evidence at the \( 0.05 \) level of significance to support the claim that pet owners have a lower average stress level than non-owners.

Key Concepts

Hypothesis Testing

Hypothesis testing is a fundamental aspect of statistical analysis. It helps us make decisions or inferences about a population based on sample data. In this context, we're examining whether owning a pet reduces stress levels. We do this by creating two hypotheses:

• Null Hypothesis ( \( H_0 \) ): Suggests there's no difference in stress levels. Mathematically, it's expressed as \( \mu_1 = \mu_2 \) , indicating that the average stress level for pet owners and non-pet owners is the same.
• Alternative Hypothesis ( \( H_a \)): Indicates there's a difference, specifically that pet owners have lower stress levels. It is expressed as \( \mu_1 < \mu_2 \).

By testing these hypotheses, we can determine if there's statistical evidence to support the claim that pets reduce stress. The key point is that hypothesis testing relies on data from a sample to infer conclusions about a larger population.

Significance Level

The significance level, denoted as \( \alpha \) , is a critical concept in hypothesis testing. It represents the threshold we set to decide whether to reject the null hypothesis. In essence, it measures the probability of making a Type I error, which occurs if we wrongly reject the null hypothesis when it is actually true. Commonly used significance levels are 0.05, 0.01, or 0.10.

In our exercise, the significance level is set to \(0.05\). This means we are willing to accept a 5% chance of incorrectly rejecting the null hypothesis. Setting \( \alpha \) to 0.05 strikes a balance between being strict enough to avoid false claims but not so strict that it becomes difficult to detect true effects. This value guides our decision on whether the results are statistically significant, which means we have sufficient evidence to support the alternative hypothesis.

Critical Value

In the context of hypothesis testing, finding the critical value is an essential step. The critical value acts as a cutoff point separating the region where we reject the null hypothesis from the region where we do not. It is derived from a statistical distribution, which depends on the test being used.

For a t-test like in our exercise, we determine the critical value from the t-distribution table. Because we have a one-tailed test with \( \alpha = 0.05 \) , and degrees of freedom calculated as \( ext{min}(n_1 - 1, n_2 - 1) = 24 \) , the critical value is found to be approximately \(-1.711.\)

When our calculated t-statistic, which is \(-3.80\), falls in the critical region (i.e., it is less than \(-1.711\)), we reject the null hypothesis. Thus, the critical value is a vital reference point, helping us to make informed decisions based on our statistical analysis.