91Ó°ÊÓ

A Michigan study concerning preference for outdoor activities used a questionnaire with a six-point Likert-type response in which 1 designated "not important" and 6 designated "extremely important." \(\mathrm{A}\) random sample of \(n_{1}=46\) adults were asked about fishing as an outdoor activity. The mean response was \(\bar{x}_{1}=4.9 .\) Another random sample of \(n_{2}=51\) adults were asked about camping as an outdoor activity. For this group, the mean response was \(\bar{x}_{2}=4.3 .\) From previous studies, it is known that \(\sigma_{1}=1.5\) and \(\sigma_{2}=1.2 .\) Does this indicate a difference (either way) regarding preference for camping versus preference for fishing as an outdoor activity? Use a \(5 \%\) level of significance. Note: A Likert scale usually has to do with approval of or agreement with a statement in a questionnaire. For example, respondents are asked to indicate whether they "strongly agree," "agree," "disagree," or "strongly disagree" with the statement.

Step by step solution

01

Understand the Hypotheses

We are tasked with determining whether there is a statistically significant difference in preference for fishing versus camping using a two-sample z-test. Our null hypothesis \(H_0\) is that there is no difference in the mean preferences (\(\mu_1 = \mu_2\)). The alternative hypothesis \(H_1\) states that there is a difference in mean preferences (\(\mu_1 eq \mu_2\)).

02

Set Up the test statistics

We'll use a two-sample z-test to compare the means since the population standard deviations are known. The test statistic \(z\) is calculated as: \[ z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \] Substituting in the given values for our variables.

03

Calculate the Test Statistic

First, calculate the standard error: \(SE = \sqrt{\frac{1.5^2}{46} + \frac{1.2^2}{51}}\). This simplifies to \(SE \approx 0.2976\). Now, calculate the z-value: \[ z = \frac{4.9 - 4.3}{0.2976} \approx \frac{0.6}{0.2976} \approx 2.016 \]

04

Determine the Critical Value and Decision

For a two-tailed test with significance level \(\alpha = 0.05\), the critical values for z are approximately \(\pm 1.96\). Since our calculated z-value 2.016 is greater than 1.96, we reject the null hypothesis.

05

State the Conclusion

Since the calculated z-value lies beyond the critical value in both tails, we reject the null hypothesis and conclude that there is a statistically significant difference in the preference for fishing versus camping.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hypothesis Testing

Hypothesis testing is a fundamental concept in statistics that allows us to make decisions or draw conclusions about populations based on sample data. In this exercise, hypothesis testing helps determine if there is a genuine difference in preferences for fishing and camping among adults.
A hypothesis is an assumption about a parameter or distribution. In our case:

• The **null hypothesis** (\( H_0 \)) posits that there is no difference in the fishing and camping preferences, meaning the mean preference scores are equal, \( \mu_1 = \mu_2 \).
• The **alternative hypothesis** (\( H_1 \)) suggests there is a difference, represented as \( \mu_1 eq \mu_2 \).

The goal of hypothesis testing is to determine whether we have enough statistical evidence to reject \( H_0 \) in favor of \( H_1 \) after analyzing the sample data.

Significance Level

The significance level, denoted by \( \alpha \), is a critical component of hypothesis testing and represents the probability of rejecting the null hypothesis when it is actually true, also known as a Type I error. In our exercise, we use a common significance level of 5%, or \( \alpha = 0.05 \).
Choosing a significance level is crucial because it determines the threshold for our decision-making process. A lower significance level means stricter criteria for rejecting the null hypothesis, minimizing chances of false positives, whereas a higher level makes it easier to detect a true effect but might lead to more false positives.
In our example, a 5% significance level means we are willing to accept a 5% risk of falsely concluding there is a difference in preferences between fishing and camping.

Likert Scale

A Likert scale is a common tool used in surveys to gauge people's attitudes, opinions, or preferences. In this exercise, a six-point Likert-type scale was employed to measure the importance of fishing and camping as outdoor activities.
Ranging from 1 ("not important") to 6 ("extremely important"), participants are asked to express their level of importance for each activity on this scale. This structured format provides quantitative data that can be used for statistical analysis, such as hypothesis testing.
The responses on Likert scales are often treated as interval data, which allows for the use of various statistical tests, including the two-sample z-test employed in our exercise. It's important to ensure that participants understand the scale thoroughly to gather accurate data.

Statistical Significance

Statistical significance is a crucial outcome of hypothesis testing that helps determine whether the results observed in a study are likely attributable to a specific cause rather than chance.
In our example, after conducting the two-sample z-test, we find a calculated z-value that exceeds the critical value set by our chosen significance level (5%).
This finding leads us to reject the null hypothesis, implying a statistically significant difference in preferences for fishing versus camping. Statistical significance means there is sufficient evidence from the sample to conclude a real effect or difference exists in the population. However, it does not mean the findings are practically significant, which involves assessing the real-world relevance of the results.

Test Statistic

A test statistic is a standardized value calculated from sample data during a hypothesis test. It helps determine the distance between your sample observations and the null hypothesis. The two-sample z-test is an ideal choice here because the population standard deviations are known.
In this exercise, the test statistic, denoted by \( z \), is computed as:\[ z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \]By plugging in the values from the problem, we find that \( z \approx 2.016 \).
The test statistic is then compared against critical values determined by the chosen significance level to decide whether to reject or fail to reject the null hypothesis. In our case, since \( z \) exceeds the critical value of approximately \( \pm 1.96 \), we conclude that there is a statistically significant difference in preferences.