Question

Use Excel to complete the following question. You must submit/attach all work done in Excel

in order to receive full credit.

5. A health advocacy group conducted a study to determine if the nicotine content of a particular

brand of cigarettes was equal to the advertised amount. The cigarette brand advertised that the

average nicotine content per cigarette was 1.4 milligrams. The advocacy group randomly

sampled 24 cigarettes. The nicotine level for each of the sampled cigarettes is given below.

Nicotine mg. Nicotine mg.

1.8          1.9

1.1          1.6

1.2          1.9

1.2          1.9

1.0          2.0

2.0          1.6

1.7          1.1

2.0          1.8

2.3          1.9

1.4          1.4

0.9          1.8

2.4          1.5

a. Use Excel to obtain the sample mean and sample SD. Construct the 95 % confidence

interval. Enter the data into one, single column.

b. Provide an interpretation of the confidence interval from part A.

c. Does the obtained confidence interval contain the average nicotine level per cigarette

suggested by the cigarette maker? Explain what this means.

6. The National Heart, Lung, and Blood Institute completed a large-scale study of cholesterol and

heart disease, and reported that the national average for blood cholesterol level of 50-year old

males was 210 mg/dl. A total of 89 men with cholesterol readings in the average range (200 –

220) volunteered for a low cholesterol diet for 12 weeks. At the end of the dieting period their

average cholesterol reading was 204 mg/dl with a SD of 33 mg/dl.

a. What is the 95% confidence interval for the study described above? (submit all work)

b. Provide an interpretation of your answer to part A.

c. Does the known population mean of 210 mg/dl fall within the interval?

d. Provide an interpretation of your answer to part C.

e. How would (1) decreasing the sample size and (2) decreasing the confidence level

affect the size of the interval calculated in part A? Explain your answer.

Answer 1

5)

Excel formula

Sl.No.	Nicotine mg.
1	1.8
2	1.1
3	1.2
4	1.2
5	1
6	2
7	1.7
8	2
9	2.3
10	1.4
11	0.9
12	2.4
13	1.9
14	1.6
15	1.9
16	1.9
17	2
18	1.6
19	1.1
20	1.8
21	1.9
22	1.4
23	1.8
24	1.5
Total	=SUM(B2:B25)
Mean	=AVERAGE(B2:B25)
Std dev	=STDEV.S(B2:B25)
size	=COUNT(B2:B25)
Confidence	0.95
df	=B30-1
alpha	0.05
critical value	=T.INV.2T(0.05,B32)
Lower confidence limit	=B28-B34*B29/SQRT(B30)
Upper confidence limit	=B28+B34*B29/SQRT(B30)

Mean	1.641667
Std dev	0.407449
size	24
Confidence	95%
df	23
alpha	0.05
critical value	2.068658
Lower 95% confidence limit	1.469616
Upper 95% confidence limit	1.813717

b) We are 95% confident that average nicotine content per cigarette will lie in between (1.5, 1.8) mg.

c) The obtained confidence interval does not contain the average nicotine level per cigarette 1.4 mg suggested by the cigarette maker, which means nicotine content is higher than the advertised content in the cigarettes produced by the company, the advocacy can sue the company.

6) Given population mean = 210 m/dl

sample size n = 89

sample mean = 204 mg/dl

std dev = 33 mg/dl

since the sample size is more than 30 we can use standard normal distribution

a) 95% CI =

alpha = 0.05

Z0.025 = 1.96

95% CI = () = (197.144, 210.856)

b) We are 95% confident that the average cholesterol for 50-year old after a low cholesterol diet will lie in between 197.144 mg/dl and 210.856 mg/dl.

c) Yes,  the known population mean of 210 mg/dl fall within the interval.

d) Which means that the hypothesis for the national average for blood cholesterol level of 50-year old males was 210 mg/dl is fail to be rejected and concludes that the national average for blood cholesterol level of 50-year old males was 210 mg/dl.

e) (1) Decrease in the sample size increases the width of the CI.

(2) Decrease in the confidence level decreases the size or width of the CI.