Question

In a study of 479,908 cell phone​ users, it was found that 213 developed cancer of the brain or nervous system. Assuming that cell phones have no​ effect, there is a 0.000462 probability of a person developing cancer of the brain or nervous system. We therefore expect about 222 cases of such cancer in a group of 479,908 people. Estimate the probability of 213 or fewer cases of such cancer in a group of 479,908 people. What do these results suggest about media reports that cell phones cause cancer of the brain or nervous​ system?

Answer 1

The probability of 213 or fewer cases of such cancer in a group of 479,908 people is 0.2776.

Explanation:

We are given that assuming that cell phones have no​ effect, there is a 0.000462 probability of a person developing cancer of the brain or nervous system. We, therefore, expect about 222 cases of such cancer in a group of 479,908 people.

Let
`\hat p` = sample probability of a person developing cancer of the brain or nervous system.

The z-score probability distribution for the sample proportion is given by;

Z =
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` ~ N(0,1)

where, p = population probability of a person developing cancer of the brain or nervous system = 0.000462

n = sample of cell phone​ users = 479,908

Now, the probability of 213 or fewer cases of such cancer in a group of 479,908 people is given by = P(
`\hat p`
`\leq`
`(213)/(479,908)` )

P(
`\hat p`
`\leq` 0.000444) = P(
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }`
`\leq`
`\frac{0.000444-0.000462}{\sqrt{(0.000444(1-0.000444))/(479,908) } }` ) = P(Z
`\leq` -0.59)

= 1 - P(Z < 0.59) = 1 - 0.7224 = 0.2776

The above probability is calculated by looking at the value of x = 0.59 in the z table which has an area of 0.7224.

These results suggest about media reports that cell phones cause cancer of the brain or nervous​ system that there is around 28% chance that 213 or fewer cases of such cancer in a group of 479,908 people can take place.