nna cample t-test for the Problem 4 To help assess the health risks of second-ha

Question

nna cample t-test for the Problem 4 To help assess the health risks of second-hand smoke, the levels of cotinine (a metabolite of nicotine) were measured in mmol/I n the urine of seven subjects prior to exposure to second-hand smoke and shortly after a two-hour exposure to secondary cigarette smoke. Assuming the levels of cotinine to be normally distributed, did the exposure significantly increase the cotinine level at a-0.05? us: Use the value Cotinine Levels (mmol/l) Subject Before After d, before after 19 23

Explanation / Answer

Back-up Theory

Let

X = level of cotinine (mmol/l) in the urine prior to exposure to second hand smoking.

Y = level of cotinine (mmol/l) in the urine shortly after a 2-hour exposure to second hand smoking.

Since X and Y correspond to the same subject individual, these are related and hence a paired t-test is appropriate [and not 2-sample t-test]

Let D = Y – X. Then, we assume Y ~ N(µ, 2)

We have a sample of n observations on each of X and Y and hence on D too.

H0: µ = 0   Vs   HA: µ > 0 [> because the aim is to see if there is a significant increase in the cotinine level]

Test Statistic: t = (n) (Dbar)/s

where n = common sample size for X and Y and Dbar and s are sample mean and standard deviation of D respectively.

Under H0, t ~ tn-1.

H0 is rejected/accepted at % level of significance if tcal >< tcrit

where tcal = calculated value of test statistic, t and

tcrit = upper percent point of t-Distribution with n – 1 degrees of freedom.

H0 is rejected/accepted at % level of significance if p-value of tcal </>

Now, to work out solution,

Using Excel function, the following are computed from the given data:

Dbar = 7.714     s = 9.082     tcal = 2.247

Given n = 7 and = 0.05, tcrit = 1.943 [using Excel Function]

H0 is rejected 5% level of significance since tcal| > tcrit.

Conclusion

There is enough evidence to believe that there is a significant increase in the cotinine level due to exposure to second-hand smoking.

[additional input: since test statistic, t ~ t6, p-value = P(t6 > 2.247) = 0.0329 which is less than the level of significance of 0.05. Hence, H0 is rejected 5% level of significance.]

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