Question

A study of 420 comma 090 cell phone users found that 130 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 0.0212​% for those not using cell phones. Complete parts​ (a) and​ (b). a. 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. nothing​%less than less than nothing​% ​(Round to three decimal places as​ needed.) b. 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

Answer 1

a)
`0.0003095 - 1.64\sqrt{(0.0003095(1-0.0003095))/(420090)}=0.000265`

`0.0003095 - 1.64\sqrt{(0.0003095(1-0.0003095))/(420090)}=0.000354`

The 90% confidence interval would be given by (0.000265;0.000354) and in % would be (0.0265, 0.0354)

b) For this case the rate given is 0.0212% and since our interval not contains this value we can conclude at 10% of significance that the real parameter is different from the value given.

Explanation:

Previous concepts

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".

Solution to the problem

Part a

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)}`

the estimated proportion for this case is:

`\hat p = (130)/(420090)=0.0003095`

If we replace the values obtained we got:

`0.0003095 - 1.64\sqrt{(0.0003095(1-0.0003095))/(420090)}=0.000265`

`0.0003095 - 1.64\sqrt{(0.0003095(1-0.0003095))/(420090)}=0.000354`

The 90% confidence interval would be given by (0.000265;0.000354) and in % would be (0.0265, 0.0354)

Part b

For this case the rate given is 0.0212% and since our interval not contains this value we can conclude at 10% of significance that the real parameter is different from the value given.