Question

2. The extent to which an infant’s health is affected by parent’s smoking is an important public concern. Measurements were taken both from a random sample of infants who had been exposed to household smoke and from a sample of unexposed infants. The following table lists data consists of observations on urinary concentration of cotinine, a major metabolite of nicotine. Unexposed 8 11 12 14 20 43 111 Exposed 35 56 83 92 128 150 176 208 a. Assuming that the population variances are equal, use an appropriate parametric test at ? ? 0.05 , can you conclude that the true average cotinine level is higher for exposed than unexposed infants? b. Construct a 95% confidence interval estimate of the difference between the means of the two populations.

Answer 1

Given:

Un Exposed	Exposed
8	35
11	56
12	83
14	92
20	128
43	150
111	176
	208

a)

Independent two smaples t-test will be used to test the hypothesis

Null hypohtesis:

H0: There is no mean difference of cotinine between exposed and unexposed,

Alternative hypothesis:

H1: The mean cotinine level is higher in the exposed group than the unexposed group,

Level of significance:

Test Statistic:

= Mean of exposed group

= Mean of unexposed group

= Variance of exposed group

= Varaince of unexposed group

= Number of samples in unexposed group = 7

= Number of samples in exposed group = 8

Using excel mean and variance are calculated as below

             

       

          

Critical value:

t-critical value for 5% level of significance with 13 (n_e+n_ue-2 = 8+7-2) degrees of freedom for one-tailed test is given as 1.771

Inference:

t-calculated value of 3.2277 is greater than the t-critical value of 1.771, hence the null hypothesis is rejected.

Conclusion:

Since, the null hypothesis is rejected we conclude that the mena cotinine level is higher in the exposed infants than the unexposed infants

b)

95% CI is given as

From a) we have

is the t-critical value for 5% level of significance with 13 degrees of freedom for two-tailed test which is given as 2.160 (from t-distribution table)

Hence, 95% CI is given as