Question

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 [12]. Cotinine Levels (nmol/1) Subject After 12 Hours After 24 Hours 73 24 2 58 27 3 67 49 4 93 59 5 33 0 6 18 11 7 147 43 Let f.llz represent the population mean cotinine level12 hours after smoking the cigarette and f.1z4 the mean cotinine level 24 hours after smoking. It is believed that f.1z4 must be lower than f.llz. (a) Construct a one-sided 95% confidence interval for the true difference in population means f.112 - f.124. (b) Test the null hypothesis that the population means are identical at the a = 0.05 level of significance. What do you conclude?

Please use SPSS (not minitab) or by hand :)

Answer 1

Solution:-

a) 95% confidence interval for the true difference in population means is C.I = (10.393, 68.465).

$C.I=( \bar{x}_{1}- \bar{x}_{2})\pm t_{\alpha /2}\ast \frac{S.D}{\sqrt{n}}$

C.I = (69.85714 - 30.42857) + 2.447*11.866

C.I = 39.4286 + 29.036

C.I = (10.393, 68.465)

b)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: μ_d = 0

Alternative hypothesis: μ_d ≠ 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

s = sqrt [ (Σ(d_i - d)² / (n - 1) ]

s = 31.39457

SE = s / sqrt(n)

SE = 11.866

DF = n - 1 = 7 -1

D.F = 6

t = [ (x₁ - x₂) - D ] / SE

t = 3.32

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 6 degrees of freedom is more extreme than 3.32; that is, less than -3.32 or greater than 3.32.

Thus, the P-value = 0.016

Interpret results. Since the P-value (0.016) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that the population means are identical.