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erastovalidia [21]
3 years ago
7

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.
Mathematics
1 answer:
postnew [5]3 years ago
3 0

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%)

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