Mathematics, 07.04.2020 19:26 gransazer863

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

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Answer from: samarkanduzbek

a) $0.0003095 - 1.64\sqrt{\frac{0.0003095(1-0.0003095)}{420090}}=0.000265$

$0.0003095 - 1.64\sqrt{\frac{0.0003095(1-0.0003095)}{420090}}=0.000354$

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.

Step-by-step 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:

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

the estimated proportion for this case is:

$\hat p = \frac{130}{420090}=0.0003095$

If we replace the values obtained we got:

$0.0003095 - 1.64\sqrt{\frac{0.0003095(1-0.0003095)}{420090}}=0.000265$

$0.0003095 - 1.64\sqrt{\frac{0.0003095(1-0.0003095)}{420090}}=0.000354$

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.

Answer from: Quest

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Which statement about the frequency table is true?

Answer from: Quest

it is contrapositive

i could not see the question too well so check yourself again too . this is because i want you to get it correct.

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