A group of eight individuals with high cholesterol levels were given a new drug that was designed to lower cholesterol levels. Cholesterol levels, in mg/dL, were measured before and after treatment for each individual, with the following results: | | | | Subject | 1 | 2 Before 283 299 After 215 206 3 274 187 | 4 284 212 5 248 178 | 6 275 212 7 | 8 293 277 192 196 (a) Is this a matched pairs design/experiment? Explain briefly. (b) Using the appropriate hypothesis/significant test procedure, can you con- clude that the mean cholesterol level after treatment is less than the mean before treatment?

Step-by-step explanation:

The data in the question is not well arranged. The correct arrangement is:

Before After difference

283 215

299 206

274 187

284 212

248 178

275 212

293 192

277 196

Solution:

a) This is a matched pair design/experiment. This is so because the measurement of cholesterol levels were carried out on the same individuals.

b) Corresponding cholesterol levels before and after treatment form matched pairs.

The data for the test are the differences between the cholesterol levels before and after treatment.

μd = cholesterol level before treatment minus cholesterol level after treatment.

Before After difference

283 215 68

299 206 93

274 187 87

284 212 72

248 178 70

275 212 63

293 192 101

277 196 81

Sample mean, xd

= (68 + 93 + 87 + 72 + 70 + 63 + 101 + 81)/8 = 79.4

xd = 79.4

Standard deviation = √(summation(x - mean)²/n

n = 8

Summation(x - mean)² = (68 - 79.4)^2 + (93 - 79.4)^2 + (87 - 79.4)^2 + (72 - 79.4)^2 + (70 - 79.4)^2 + (63 - 79.4)^2 + (101 - 79.4)^2 + (81 - 79.4)^2 = 1253.88

Standard deviation = √(1253.88/8)

sd = 12.52

For the null hypothesis

H0: μd ≥ 0

For the alternative hypothesis

H1: μd < 0

This is a one tailed test(left tailed test)

The distribution is a students t. Therefore, degree of freedom, df = n - 1 = 8 - 1 = 7

The formula for determining the test statistic is

t = (xd - μd)/(sd/√n)

t = (79.4 - 0)/(12.52/√8)

t = 17.94

We would determine the probability value by using the t test calculator.

p < 0.00001

4) Assume alpha = 0.05

Since alpha, 0.05 > than the p value, then we would reject the null hypothesis. Therefore, at 5% level of significance, we cannot conclude that the mean cholesterol level after treatment is less than the mean before treatment.