Question

Anderson et al. (1990) studied the effect of diet on the level of low-density lipoprotein (LDL, the “bad” cholesterol) for a group of men with hypercholesterolemia. Half the subjects were given a diet that included corn flakes and the other half were given a diet that included oat bran. LDL was then measured after two weeks. Subjects were then crossed-over to the alternative diet for an additional two weeks. LDL was measured again. The LDL measurements are shown below. Test the null hypothesis that the mean difference in LDL points to a population in which the difference is zero, suggesting that the one diet is no better than the other for controlling LDL. What does the result suggest? (this can be answered in one or two sentences)

LDL (in mmol/L) -------- Corn Flakes 4.61 6.42 5.40 4.54 3.98 3.82 5.01 4.34 3.80 4.56 5.35 3.89

Oat Brab 3.84 5.57 5.85 4.80 3.68 2.96 4.41 3.72 3.49 3.84 5.26 3.73 -----------------------

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁ u ₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 0.3479
DF = 22
t = [ (x₁ - x₂) - d ] / SE

t = 1.095

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 22 degrees of freedom is more extreme than -1.095; that is, less than -1.095 or greater than 1.095.

Thus, the P-value = 0.285.

Interpret results. Since the P-value (0.285) is greater than the significance level (0.10), we have to accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that that the one diet is no better than the other for controlling LDL.