91Ó°ÊÓ

For Exercises 2 through \(12,\) perform each of these steps. Assume that all variables are normally or approximately normally distributed. 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. Cholesterol Levels A medical researcher wishes to see if he can lower the cholesterol levels through diet in 6 people by showing a film about the effects of high cholesterol levels. The data are shown. At \(\alpha=0.05,\) did the cholesterol level decrease on average? $$ \begin{array}{lllllll}\hline \text { Patient } & {1} & {2} & {3} & {4} & {5} & {6} \\ \hline \text { Before } & {243} & {216} & {214} & {222} & {206} & {219} \\ \hline \text { After } & {215} & {202} & {198} & {195} & {204} & {213}\end{array} $$

Step by step solution

01

State the Hypotheses

To determine whether the film effectively lowers cholesterol levels, we formulate the hypotheses. The null hypothesis \(H_0\) is that there is no decrease in cholesterol levels on average, expressed as \(\mu_d = 0\). The alternative hypothesis \(H_1\) is that the cholesterol levels do decrease on average, expressed as \(\mu_d > 0\). Here, \(\mu_d\) represents the mean of the differences between the 'Before' and 'After' measurements.

02

Identify the Claim

The researcher's claim is that the cholesterol levels decreased after showing the film, which aligns with the alternative hypothesis \(H_1: \mu_d > 0\).

03

Find the Critical Value

Using a significance level \(\alpha = 0.05\) for a one-tailed test, we find the critical value for \(t\) with \(n-1 = 5\) degrees of freedom. From the t-distribution table, the critical value is approximately 2.015.

04

Compute the Test Value

Calculate the differences between the cholesterol levels before and after: \(d = \text{Before} - \text{After} = [28, 14, 16, 27, 2, 6]\). The mean of the differences \(\bar{d}\) is \(93/6 = 15.5\). The standard deviation of the differences \(s_d\) is calculated as follows: \[s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n - 1}} = \sqrt{\frac{(12.5)^2 + (-1.5)^2 + (0.5)^2 + (11.5)^2 + (-13.5)^2 + (-9.5)^2}{5}} \approx 9.5026\]The test statistic \(t\) is computed as:\[t = \frac{\bar{d} - 0}{s_d / \sqrt{n}} = \frac{15.5 - 0}{9.5026 / \sqrt{6}} \approx 3.868\]

05

Make the Decision

Compare the test statistic with the critical value. Since our calculated \(t\) value of 3.868 is greater than the critical value of 2.015, we reject the null hypothesis \(H_0\).

06

Summarize the Results

There is sufficient evidence at the \(\alpha = 0.05\) significance level to support the claim that showing the film decreases cholesterol levels on average.

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Key Concepts

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

t-test

The t-test is a statistical method used to determine if there is a significant difference between the means of two groups. In our scenario, we used a paired t-test because we are comparing cholesterol levels before and after a specific intervention (showing a film). Here, the scores are paired naturally because they come from the same individuals measured at two different times.

The test helps us to understand if the difference we observe in the sample data is due to chance or if it's statistically significant. To conduct a t-test, we calculate the mean difference, the standard deviation of differences, and use these in a formula to find our test statistic. This calculated test value helps us make a decision about our hypotheses.

significance level

The significance level, denoted by \(\alpha\), is a threshold set by the researcher to decide when to reject the null hypothesis. In this exercise, \(\alpha = 0.05\), meaning we are willing to accept a 5% chance of being wrong when we claim there is an effect when there isn't one.

A smaller \(\alpha\) value means we are more conservative and strict when assessing evidence against the null hypothesis. However, in most standard tests like ours, 0.05 is commonly used as it strikes a balance between overstating or understating statistical results.

critical value

The critical value is a point on the test distribution that is compared with the test statistic to determine the outcome of the test. Depending on the test and significance level (uffff0.05f), we find this value using a t-distribution table based on the degrees of freedom, which is c3"n-1" for a t-test.

In our example, the degrees of freedom is 5 (from 6 samples, or \(6-1\)). With a one-tailed test approach, as we are interested in decreases only, the critical value is 2.015. If our calculated statistic exceeds this critical value, we have enough evidence to reject the null hypothesis.

null hypothesis

In hypothesis testing, the null hypothesis, denoted as \(H_0\), serves as a starting point for our statistical test. It represents the default position that there is no effect or no difference. For the cholesterol level exercise, the null hypothesis states that the mean difference between the 'Before' and 'After' cholesterol measurements is zero, i.e., \(\mu_d = 0\).

Essentially, the null hypothesis suggests that the intervention (watching the film) does not affect cholesterol levels. Our goal is to test this hypothesis and determine whether it can be rejected in favor of the alternative hypothesis.

alternative hypothesis

The alternative hypothesis (\(H_1\)) is contrary to the null hypothesis. It indicates the presence of an effect or a difference. In our scenario, the alternative hypothesis states that the mean difference is greater than zero (i.e., \(\mu_d > 0\)), suggesting that the film does indeed help in lowering cholesterol levels.

This hypothesis is what the researcher aims to prove. If our test statistic surpasses the critical value, we reject the null hypothesis in favor of the alternative hypothesis. Therefore, in this exercise, concluding that \(H_1\) is true means there is significant evidence to suggest that the film showing lowers cholesterol levels on average.