Question

A study of 420 comma 034 cell phone users found that 138 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 0.0352​% for those not using cell phones. Complete parts​ (a) and​ (b).

1. Use the sample data to construct a 90% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system.
2. Do cell phone users appear to have a rate of cancer of the brain or nervous system that is different from the rate of such cancer among those not using cell​ phones? Why or why​ not?
A. No, because 0.04120% is not included in the confidence interval.
B. Yes, because 0.04120% is not included in the confidence interval.
C. Yes, because 0.04120% is included in the confidence interval.
D. No, because 0.04120% is included in the confidence interval.

Answer 1

1) The 90% confidence interval would be given by (0.000283;0.000374)

2) On this case 0.0352% is equal to 0.000352. And our interval contian this value. So ont his case the best option is:

D. No, because 0.0352% is included in the confidence interval.

Explanation:

Notation and definitions

`X=138` number of people who developed cancer of the brain or nervous system

`n=420034` random sample taken

`\hat p=(138)/(420034)=0.000329` estimated proportion of people who developed cancer of the brain or nervous system

`p` true population proportion of people who developed cancer of the brain or nervous system

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

`p \sim N(p,\sqrt{(p(1-p))/(n)})`

Part 1

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by
`\alpha=1-0.90=0.1` and
`\alpha/2 =0.05`. And the critical value would be given by:

`z_(\alpha/2)=-1.64, z_(1-\alpha/2)=1.64`

The confidence interval for the mean is given by the following formula:

`\hat p \pm z_(\alpha/2)\sqrt{(\hat p (1-\hat p))/(n)}`

If we replace the values obtained we got:

`0.000329 - 1.64\sqrt{(0.000329(1-0.000329))/(420034)}=0.000283`

`0.000329 + 1.64\sqrt{(0.000329(1-0.000329))/(420034)}=0.000374`

The 90% confidence interval would be given by (0.000283;0.000374)

Part 2

On this case 0.0352% is equal to 0.000352. And our interval contian this value. So ont his case the best option is:

D. No, because 0.0352% is included in the confidence interval.