Question

A study of 420 comma 100 cell phone users found that 133 of them developed cancer of the brain or nervous system. Prior to this study of cell phone​ use, the rate of such cancer was found to be 0.0449​% for those not using cell phones. Complete parts​ (a) and​ (b). a. Use the sample data to construct a 95​% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system.

Answer 1

Answer:

(0.0263%, 0.0370%)

Step-by-step explanation:

Sample size = n = 420,100

Number of users who developed cancer = x = 133

Proportion of users who developed cancer = p = $\frac{133}{420100}$

Proportion of users who didnot develop cancer = q = 1 - p = $1-\frac{133}{420100}=\frac{419967}{420100}$

Confidence Level = 95%

Z value associated with this confidence level = z = 1.96

The formula to calculate the confidence interval is:

$\text{Lower Bound} = p-z\sqrt{\frac{pq}{n}}\\\\ \text{Upper Bound} = p+z\sqrt{\frac{pq}{n}}$

Using the values in above expressions, we get:

$\text{Lower Bound}=\frac{133}{420100}-1.96\sqrt{\frac{\frac{133}{420100}\times\frac{419667}{420100}}{420100}}\\\\\text{Lower Bound}=0.000263$

and

$\text{Upper Bound}=\frac{133}{420100}+1.96\sqrt{\frac{\frac{133}{420100}\times\frac{419667}{420100}}{420100}} \\\\ \text{Upper Bound}=0.000370$

Thus, the bounds of the confidence interval are:

(0.000263, 0.000370)

This can be expressed in percentages as:

(0.0263%, 0.0370%)

Therefore, a 95​% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system is (0.0263%, 0.0370%)