Question

In: Statistics and Probability

The following data come from a study that examines the efficacy of saliva cotinine as an...

The following data come from a study that examines the efficacy of saliva cotinine as an indicator for exposure to tobacco smoke. In one part of the study, seven subjects – none of whom were heavy smokers and all of whom had abstained from smoking for at least one week prior to the study- were each required to smoke a single cigarette. Samples of saliva were taken from all individuals 2, 12, 24, and 48 hours after smoking the cigarette. The cotinine levels at 12 hours and at 24 hours are shown below:

Subject

Cotinine levels (nmol/l)

After 12 hours

After 24 hours

1

73

24

2

58

27

3

67

49

4

93

59

5

33

0

6

18

11

7

147

43

Let u12  represent the population mean cotinine level 12 hours after the cigarette and u24 represent the mean cotinine level 24 hours after smoking. It is believed that u24 must be lower than u12.

SHOW WORK

What are the null and alternative hypotheses?

What type of test statistic can you calculate to test your hypotheses?

Does the data meet all of the assumptions for that test? (show your work)

Test the null hypothesis at the alpha = 0.05 significance level. What can you conclude?

Solutions

Expert Solution

After 12 hours After 24 hours Difference
73 24 49
58 27 31
67 49 18
93 59 34
33 0 33
18 11 7
147 43 104

Sample mean of the difference using excel function AVERAGE(), x̅d = 39.4286

Sample standard deviation of the difference using excel function STDEV.S(), sd = 31.3946

Sample size, n = 7

Null and Alternative hypothesis: It is believed that u24 must be lower than u12. so u12 - u24 is greater than 0.

Ho : µd = 0 ; H1 : µd > 0

We will conduct a paired t test statistic.

Assumption:

  • The sample is random and independent.
  • The data is approximately normally distributed.

Test statistic:

t = (x̅d)/(sd/√n) = (39.4286)/(31.3946/√7) = 3.3228

df = n-1 = 6

Right tailed p-value = T.DIST.RT(3.3228, 6) = 0.0080

Decision:

p-value < α, Reject the null hypothesis

Conclusion:

There is enough evidence to conclude that the population mean cotinine level 24 hours after is lower than the mean cotinine level 12 hours after smoking.


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