Question

A dietitian wishes to see if a person’s cholesterol level will change if their diet is supplemented by a certain mineral. Six randomly selected subjects had their cholesterol level measured before and then took the mineral supplement for 5 weeks. The same six subjects had their cholesterol level measured again after the 5-week supplement. The results are shown below.  Assume the two population distributions are normal and the subjects were selected independently.

Can it be concluded that the cholesterol level has changed at 5% significance level?    

Subject	1	2	3	4	5	6
Before	210	235	208	190	172	244
After	190	170	210	188	173	228

Question:Which of the following is the correct calculator command to use?

2-SamTTest

ANOVA

2-SampZint

Z-Test

Z-Interval

T-Test

T-Interval

2-SampTInt

2-SampZTest

Answer 1

Solution:- The correct calculator command to use T-Test.

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

Null hypothesis: u_d = 0

Alternative hypothesis: u_d ≠ 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

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

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

s = 25.3903

SE = s / sqrt(n)

S.E = 10.3655

DF = n - 1 = 6 -1

D.F = 5

t = [ (x₁ - x₂) - D ] / SE

t = 1.61

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 5 degrees of freedom is more extreme than 2.667; that is, less than - 1.61 or greater than 1.61

P-value = P(t < - 1.61) + P(t > 1.61)

P-value = 0.084 + 0.084

Thus, the P-value = 0.168

Interpret results. Since the P-value (0.168) is greater than the significance level (0.05), we failed to reject the null hypothesis.

Do not Reject H0. The mean difference appears to be zero.