Questions: In a test of the effectiveness of garlic for lowering cholesterol, 81 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 0.6 and a standard deviation of 1.92. Use a 0.10 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? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. What are the null and alternative hypotheses? A. H0: μ>0 mg / dL B. H0: μ=0 mg / dL H1: μ<0 mg / dL H1: μ ≠ 0 mg / dL C. H0: μ=0 mg / dL D. H0: μ=0 mg / dL H1: μ<0 mg / dL H1: μ>0 mg / dL

In a test of the effectiveness of garlic for lowering cholesterol, 81 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 0.6 and a standard deviation of 1.92. Use a 0.10 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? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. What are the null and alternative hypotheses? A. H0: μ>0 mg / dL B. H0: μ=0 mg / dL H1: μ<0 mg / dL H1: μ ≠ 0 mg / dL C. H0: μ=0 mg / dL D. H0: μ=0 mg / dL H1: μ<0 mg / dL H1: μ>0 mg / dL
Transcript text: In a test of the effectiveness of garlic for lowering cholesterol, 81 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 $\mathrm{mg} / \mathrm{dL}$ ) have a mean of 0.6 and a standard deviation of 1.92 . Use a 0.10 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? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P -value, and state the final conclusion that addresses the original claim. What are the null and alternative hypotheses? A. $\mathrm{H}_{0}: \mu>0 \mathrm{mg} / \mathrm{dL}$ B. $\mathrm{H}_{0}: \mu=0 \mathrm{mg} / \mathrm{dL}$ $\mathrm{H}_{1}: \mu<0 \mathrm{mg} / \mathrm{dL}$ $\mathrm{H}_{1}: \mu \neq 0 \mathrm{mg} / \mathrm{dL}$ C. $\mathrm{H}_{0}: \mu=0 \mathrm{mg} / \mathrm{dL}$ D. $\mathrm{H}_{0}: \mu=0 \mathrm{mg} / \mathrm{dL}$ $\mathrm{H}_{1}: \mu<0 \mathrm{mg} / \mathrm{dL}$ $H_{1}: \mu>0 \mathrm{mg} / \mathrm{dL}$
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Solution

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Solution Steps

Step 1: Hypotheses

We set up the null and alternative hypotheses as follows:

  • Null Hypothesis: \( H_0: \mu = 0 \, \text{mg/dL} \)
  • Alternative Hypothesis: \( H_1: \mu > 0 \, \text{mg/dL} \)
Step 2: Calculate Standard Error

The standard error \( SE \) is calculated using the formula: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{1.92}{\sqrt{81}} = 0.2133 \]

Step 3: Calculate Test Statistic

The test statistic \( Z_{\text{test}} \) is calculated as follows: \[ Z_{\text{test}} = \frac{\bar{x} - \mu_0}{SE} = \frac{0.6 - 0}{0.2133} = 2.8125 \]

Step 4: Calculate P-value

For a right-tailed test, the p-value is calculated as: \[ P = 1 - T(z) = 0.0025 \]

Step 5: Conclusion

Given that the p-value \( 0.0025 \) is less than the significance level \( \alpha = 0.10 \), we reject the null hypothesis. This provides sufficient evidence to support the claim that the mean change in LDL cholesterol is greater than \( 0 \, \text{mg/dL} \).

Final Answer

The null and alternative hypotheses are:

  • \( H_0: \mu = 0 \, \text{mg/dL} \)
  • \( H_1: \mu > 0 \, \text{mg/dL} \)

The test statistic is \( Z_{\text{test}} = 2.8125 \).

The p-value is \( 0.0025 \).

Thus, we conclude that there is sufficient evidence to support the claim that the mean change in LDL cholesterol is greater than \( 0 \, \text{mg/dL} \).

\(\boxed{H_0: \mu = 0 \, \text{mg/dL}, \, H_1: \mu > 0 \, \text{mg/dL}, \, Z_{\text{test}} = 2.8125, \, P = 0.0025}\)

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