Question

Ten individuals went on a low-fat diet for 12 weeks to lower their cholesterol. The data are recorded in the table below. Do you think that their cholesterol levels were significantly lowered? Conduct a hypothesis test at the 5% level.

Starting Cholesterol level	ending cholesterol level
150	150
210	240
110	130
240	220
200	190
180	150
190	200
360	300
280	300
260	240

1. In words, state what your random variable X_d  represents.

a. X_d represents the total difference in cholesterol levels before and after the diet.

b. X_d represents the difference in the average cholesterol level before and after the diet.    

c. X_d represents the average difference in the cholesterol level before and after the diet.

d. X_d represents the average cholesterol level of the 10 individuals.

2.State the distribution to use for the test. (Enter your answer in the form z or t_df where df is the degrees of freedom.)

3.What is the test statistic? (If using the z distribution round your answer to two decimal places, and if using the t distribution round your answer to three decimal places.)

4.What is the p-value?

5.Explain how you determined which distribution to use.

a. The standard normal distribution will be used because the samples are independent and the population standard deviation is known.

b. The t-distribution will be used because the samples are dependent.    

c. The t-distribution will be used because the samples are independent and the population standard deviation is not known.

d. The standard normal distribution will be used because the samples involve the difference in proportions.

Answer 1

1)

Xd represents the average difference in the cholesterol level before and after the diet

2)

Ho :   µd=   0
Ha :   µd > 0

test is matched paired t-test because the samples are dependent

sample size ,    n =    10

Degree of freedom, DF=   n - 1 =    9

3)

Sample #1	Sample #2	difference , Di =sample1-sample2
150	150	0
210	240	-30
110	130	-20
240	220	20
200	190	10
180	150	30
190	200	-10
360	300	60
280	300	-20
260	240	20

	sample 1	sample 2	Di
sum =	2180	2120	60
mean=	218	212	6

mean of difference ,    D̅ =   6

std dev of difference , Sd =        27.568

std error , SE =    Sd / √n =    8.7178
      
t-statistic =    (D̅ - µd)/SE = 0.688

4)

p-value =        0.2543 [excel function =t.dist.rt(0.688,9) ]

Conclusion:     p-value>α , Do not reject null hypothesis  

5)

The t-distribution will be used because the samples are dependent