Question

When people smoke, the nicotine they absorb converts to cotinine, which can be measured. A sample of 40 smokers has a mean cotinine level of 172.5. Assuming that the population deviation of cotinine levels is known to be 119.5, find an estimate of the 90% confidence interval from the mean cotinine level for all smokers.

a) 160.25 < µ < 190.28

b) 141.42 < µ < 198.23

c) 138.24 < µ < 210.25

d) 141.42 < µ < 203.58

Answer 1

Population mean, $\mu$ = 172.5

Standard deviation, $\sigma$ = 119.5

Sample size, n = 40

According to Central Limit Theorem, the distribution of sample mean, is approximately normal.

T

Confidence interval, CI = $\mu$ $\pm$ Z*

0/vn

$\sigma_\bar{x}$ =

0/vn

= $119.5/\sqrt{40}$

= 18.895

For 90% confidence level, Z* = 1.645

CI = 172.5 $\pm$ 1.645 x 18.895

= 172.5 $\pm$ 31.08

= 141.42 < $\mu$ < 203.58