Question

16. Cholesterol levels were measured before and after patients were given a drug designed to lower their cholesterol. Test to see if cholesterol dropped with α = 0.01.

Patient	1	2	3	4	5	6	7	8	9	10	11	12	13
Before	238	240	220	246	202	222	210	233	204	229	244	220	219
After	235	241	219	235	198	208	202	211	188	201	235	211	207

a) Determine the null and alternative hypotheses for this scenario.

H0: H1:

b) Determine the level of significance.
c) Check the assumptions for this problem.

d) Determine the test statistic.
e) Determine the p-value.
f) Interpret the p-value.

g) What decision do we conclude based on this information? Interpret this in the context of the problem.

Answer 1

It is to be determined whether the cholesterol levels were measured before and after patients were given a drug designed to lower their cholesterol or not.

In the given study, paired t test used cholesterol levels were measured before and after patients were given a drug.

The calculation shown in the table:

Patient	Before	After	d= Before - After	d^2
1	238	235	3	9
2	240	241	-1	1
3	220	219	1	1
4	246	235	11	121
5	202	198	4	16
6	222	208	14	196
7	210	202	8	64
8	233	211	22	484
9	204	188	16	256
10	229	201	28	784
11	244	235	9	81
12	220	211	9	81
13	219	207	12	144
Total			136	2238

a) Let be the cholesterol levels were measured before and after patients.

The null and the alternative hypothesis can be stated as:

It is a left tailed test.

b). The level of significance is 0.01

c). Assumption needs to check:

*I: A variable must be measured for two different situation.

*II: A variable must be of continuous nature.

*III: Observations needs to be independent for each participant.

In the study, the same subjects measured for two different occasion such as before and after taking the drugs. So, the assumption 1 satisfied.

The cholesterol level can be any value, so the assumption II satisfied.

Also, the cholesterol levels were measured before and after patients were given a drug. So, the cholesterol level of each patient is different from other subjects. Therefore, the assumption III satisfied.

d). The appropriate test statistic to be used is:

Therefore, the test statistic is 4.5757.

e). The degrees of freedom is n – 1 = 13 – 1 = 12

Using t-table and the degrees of freedom 12, the p-value obtained is,

Therefore, the p-value is 0.00032.

f). The p-value (0.00032) can be interpreted as that if the drug has no effect on the cholesterol levels of patients, then the observed difference between the cholesterol levels of patients in less than or equal to 0.032% of studies arises because of the random sampling error.

g). since, the p-value (0.00032) is less than the significance level (0.01); the decision is to reject the null hypothesis. In conclusion, there is sufficient evidence to support that the cholesterol levels were measured before and after patients were given a drug designed to lower their cholesterol.