Answer to Question #107531 in Statistics and Probability for jasmine

Question

When people smoke, the nicotine they absorb is converted to cotinine, which can be measured. A sample of 40 smokers has a mean cotinine level of 172.5. Assuming that

Answer 1

When people smoke, the nicotine they absorb is converted to cotinine, which can be measured.

A sample of 40 smokers has a mean cotinine level of 172.5. Assuming that $\sigma$ is known to be 119.5, find a 90% confidence interval estimate of the mean cotinine level of all smokers.

Solution

We need to construct the $90\%$ confidence interval for the population proportion. We have been provided with the following information: $\bar{X}=172.5,\sigma=119.5, n=40, \alpha=0.1.$

The critical value for $\alpha=0.1$  is $z_c=z_{1-\alpha/2}=1.645.$ The corresponding confidence interval is computed as shown below:

$$(CI)=(\bar{X}-z_c\times{\sigma\over \sqrt{n}},\bar{X}+z_c\times{\sigma\over \sqrt{n}})=$$

$$=(172.5-1.645\times{119.5\over \sqrt{40}},172.5+1.645\times{119.5\over \sqrt{40}})=$$

$$=(141.418, 203.582)$$

Therefore, based on the data provided, the $90\%$ confidence interval for the population mean is $(141.418, 203.582),$ which indicates that we are 90% confident that the true population mean $\mu$

is contained by the interval $(141.418, 203.582).$