Question

Eleven employees were put under the care of the company nurse because of high cholesterol readings. The nurse lectured them on the dangers of this condition and put them on a new diet. You have the cholesterol readings of the 11 employees both before the new diet and one month after use of the diet began.

(1). Given alpha = 0.01, the average of differences between before sample and after sample group is 28.09, the average of differences between after sample and before sample group is -28.09, the standard deviation of the differences between these two sample groups is 25.81, manually determine whether there is any significant difference in cholesterol readings between before and after groups of employees taking the diet. Assume differences in cholesterol readings are normally distributed in the population. What is the statistical decision and what is the business decision?

(2). Use R command to construct a 98% confidence interval to estimate the population mean difference of cholesterol readings before and after for people who are involved in this program. Assume differences in cholesterol readings are normally distributed in the population.

(3). Import dataset cholesterolReadings.csv to R script file, and write the correct R commands to solve this problem. What is the p value? What is the statistical decision?

Employee_ID	Before	After
1	255	197
2	230	225
3	290	215
4	242	215
5	300	240
6	250	235
7	215	190
8	230	240
9	225	200
10	219	203
11	236	223

Answer 1

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μD​ = 0

Ha: μD​ ≠ 0

This corresponds to a two-tailed test, for which a t-test for two paired samples be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.01, and the degrees of freedom df=n-1=11-1= 10.

Hence, it is found that the critical value for this two-tailed test is tc​=3.169, for α=0.01 and df=10.

The rejection region for this two-tailed test isR={t:∣t∣>3.169}.

(3) Test Statistics

The t-statistic is computed as shown in the following formula:

(4) Decision about the null hypothesis

Since it is observed that ∣t∣=3.438 >tc​=3.169, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=.006353, and since 0.006353<0.01, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1​ is different than μ2​, at the 0.01 significance level.

NOTE: I HAVE DONE THE FIRST QUESTION STEP BY STEP. PLEASE REPOST 2 AND 3 QUESTIONS ALONG WITH THE DATA . THANK YOU :)