Much discussion has appeared in the medical literature in recent years on the ro

Question

Much discussion has appeared in the medical literature in recent years on the role of diet in the development of heart disease. The serum-cholesterol levels of a group of people who eat a primarily macrobiotic diet are measured. Among 24 of them, aged 20—39, the mean cholesterol level was found to be 175 mg/dL with a standard deviation of 35 mg/dL.

a. If the mean cholesterol level in the general population in this age group is 230 mg/dL, then test the hypothesis that the group of people on a macrobiotic diet have cholesterol levels different from those of the general population. Let = 0.01.

b. Construct a 99% confidence interval estimate of the mean cholesterol level.

c. Suppose the standard deviation of the cholesterol levels in the general population was 29 mg/dL. Test the hypothesis that the group of people on a macrobiotic diet have cholesterol levels that vary more than those of the general population. Let = 0.05.

d. Construct a 95% confidence interval estimate of the standard deviation of cholesterol levels.

Explanation / Answer

Solution:

Part a

We are given

Sample size = n = 24

Sample mean = Xbar = 175

Sample SD = 35

Population mean = µ = 230

Level of significance = = 0.01

df = n - 1 = 24 – 1 = 23

Critical values = -2.8073 and 2.8073

(By using t-table or excel)

Here, we have to use one sample t test for population mean.

Null and alternative hypotheses for this test are given as below:

H0: µ = 230

Ha: µ 230

This is a two tailed test.

Test statistic is given as below:

t = (Xbar - µ) / [SD/sqrt(n)]

t = (175 – 230) / [35/sqrt(24)]

t = -7.6984

P-value = 0.00

P-value < = 0.01

So, we reject the null hypothesis that population mean is 230.

There is sufficient evidence to conclude that the groups of people on a macrobiotic diet have cholesterol levels different from those of the general population.

Part b

Here, we have to find 99% confidence interval for mean cholesterol level.

Confidence interval = Xbar -/+ t*SD/sqrt(n)

We are given

Sample size = n = 24

Sample mean = Xbar = 175

Sample SD = 35

Population mean = µ = 230

df = n - 1 = 24 – 1 = 23

Confidence level = 99%

Critical t value = 2.8073

Confidence interval = 175 -/+ 2.8073*35/sqrt(24)

Confidence interval = 175 -/+ 2.8073*7.144345083

Confidence interval = 175 -/+ 20.0566

Lower limit = 175 - 20.0566 = 154.94

Upper limit = 175 + 20.0566 = 195.06

Confidence interval = (154.94, 195.06)

Part c

Here, we have to use chi square test for population variance or standard deviation.

H0: = 29

Ha: > 29

This is a one tailed test. This is an upper tailed or right tailed test.

We are given

Sample size = n = 24

Sample SD = 35

Population SD = 29

Level of significance = = 0.05

df = n - 1 = 24 – 1 = 23

Test statistic formula is given as below:

Chi square = (n – 1)*S2 / 2 = (24 - 1)*35^2 / 29^2

Chi square = 33.50178

Upper critical value = 35.1725

(By using Chi square table or excel)

P-value = 0.0727

P-value > = 0.05

So, we do not reject the null hypothesis

There is insufficient evidence to conclude that the group of people on a macrobiotic diet has cholesterol levels that vary more than those of the general population.

Part d

Confidence interval for population standard deviation is given as below:

(n – 1)*S^2/Chi square lower < ^2 < (n – 1)*S^2/Chi square upper

We are given

Sample size = n = 24

Sample SD = 35

df = n - 1 = 24 – 1 = 23

Confidence level = 95%

Lower chi square value = 11.6886

Upper chi square value = 38.0756

(By using Chi square table or excel)

By plugging all values in the above formula, we have

(n – 1)*S^2/Chi square upper = (24 - 1)*35^2/11.6886 = 2410.468

(n – 1)*S^2/Chi square lower = (24 - 1)*35^2/38.0756 = 739.9747

So, 739.9747 < 2 < 2410.468

Sqrt(739.9747) < < sqrt(2410.468)

27.2025 < < 49.0966

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