Question

Testing Hypotheses. In Exercises 13–24, assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Garlic for Reducing Cholesterol In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg>dL) have a mean of 0.4 and a standard deviation of 21.0 (based on data from “Effect of Raw Garlic vs Commercial Garlic Supplements on Plasma Lipid Concentrations in Adults with Moderate Hypercholesterolemia,” by Gardner et al., Archives of Internal Medicine, Vol. 167, No. 4). Use a 0.05 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. What do the results suggest about the effectiveness of the garlic treatment?

Short answer

The hypotheses are as follows.

$$\begin{gathered} H_0:\mu =0 \\ {\mathrm{H}}_1:\mu >0 \end{gathered}$$

The test statistic is 0.133.

The critical value is 1.677.

The hypothesis is failed to be rejected.

There is insufficient evidence to support the claim that the population mean change in LDL cholesterol is greater than 0.

The garlic treatment is not effective in reducing the LDL level.

Step-by-step solution

Given information

A sample is taken for the changes in the level of LDL cholesterol from 49 subjects with a mean of 0.4 and a standard deviation of 21.0.

It is required to test the claim that the population mean change in LDL cholesterol is greater than 0 with a 0.05 level of significance.

Check the requirements

The requirements for the test are verified, as shown below.

1. The sample, satisfies the condition of a simple random sampling, as the sample is randomly collected from the change in LDL cholesterol.
2. The sample size of 49 is greater than 30. Thus, there is no need to check for normality of the sample, as the condition for the sample size is satisfied.

As the population standard deviation is unknown, the t-test applies.

State the hypotheses 

The null hypothesis $H_0$represents the population mean change in LDL cholesterol, which is less than or equal to 0. As the original claim is strictly greater than 0, the alternate hypothesis $H_1$represents the population mean change in the LDL cholesterol, which is greater than 0.

Let $\mu$be the population mean change in LDL cholesterol.

State the null and alternate hypotheses.

$$\begin{gathered} H_0:\mu ⩽0 \\ {H}_1:\mu >0 \end{gathered}$$

The test is one-tailed.

State the critical value

The degree of freedom is obtained by using the formula $df=n-1$, where . So,

$$\begin{gathered} df=49-1 \\ =48 \end{gathered}$$

The critical value can be obtained using the t-distribution table with 48 degrees of freedom and a significance level of 0.05.

From the t-table, the value is obtained as 1.677, corresponding to the row with a value of 48 and column 0.05 (one tail).

Thus, the critical value is 1.677 $\left(t_{0.05}\right)$.

Compute the observed test statistic

Apply the t-test to compute the test statistic using the formula $t=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}$.

Substitute the respective value in the above formula and simplify.

$$\begin{gathered} t=\frac{0.4-0}{\frac{21}{\sqrt{49}}} \\ =\frac{0.4}{3} \\ =0.133 \end{gathered}$$

.

State the decision rule 

Reject the null hypothesis when the absolute value of the observed test statistic is greater than the critical value. Otherwise, fail to reject the null hypothesis.

In this case,

$$\begin{gathered} \left|t\right|=\left|0.133\right| \\ =0.133 \\ <1.677\left(t_{0.05}\right) \end{gathered}$$

The absolute value of the observed test statistic is less than the critical value. This implies that the null hypothesis is failed to be rejected.

Conclusion

As the null hypothesis is failed to be rejected, it can be concluded that there is not sufficient evidence to conclude that the population mean change in the LDL cholesterol is greater than 0.

As the change is not greater than 0, the garlic treatment may not be effective in lowering the LDL cholesterol level.