In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects w

Question

In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes in their levels of LDL cholesterol(in mg/dL) have a mean of 3.2 and a standard deviation of 17.2. Complete parts (a) and (b) Click here to view atdistribution table, Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table, a. What is the best point estimate of the population mean net change in LDL cholesterol after the garlic treatment? The best point estimate is mg/dL. IType an integer or a decimal.) b. Construct a 99% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol? What is the confidence interval estimate of the population mean w? (Round to two decimal places as needed.) What does the confidence interval suggest about the effectiveness of the treatment? O A. The confidence interval limits do not contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels. O B. The confidence interval limits contain 0, suggesting that the garlic treatment did affect the LDL cholesterol levels. O c. The confidence interval limits do not contain 0, suggesting that the garlic treatment did affect the LDL oholesterol levels. O D. The confidence interval limits contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels. Click to select your answer(s)

Explanation / Answer

Solution:-

A point estimate of a population parameter is a single value used to estimate the population parameter. For example, the sample mean x is a point estimate of the population mean . So, it will be same as 3.2 mg/DL.

With 3.2 as mean, 42 sample size and 17.2 as standard deviation, using the formula,

xbar + z*(s/sqrt(n))

where z* is calculated for 99% CI = 2.58

Therefore, 99% C.I. is -3.96898 < x < 10.36898

If the confidence interval (with your chosen level of confidence) includes 0, that implies you think 0 is a reasonable possibility for the true value of the difference. In general, by 'significant' people usually mean that they no longer believe the null hypothesis (0) is a reasonable possibility. Note that if a 99%CI doesn't include 0, the p-value would be <.01, which is the conventional cutoff for 'significance'.

So, we select option (B).

Using the formula,

xbar + z*(s/sqrt(n))

z* value for 80% is 1.28

So, we are 80% confident the mean value is between 98.12 and 98.28. It is safe to conculde that SD is less than 2.40 F, because (B).

For the last question,

We are 90% confident that the mean value is between 60,318 and 67,482.

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