Question

A simple random sample of 140 residents in a certain town was polled the week after the Super Bowl. 75 of them reported that they watched the game on TV. a) Construct the 95% CI for the proportion of people that watched Super Bowl. b) The prognosis stated that the proportion of people watching the game is 48.1%. Does this prognosis contradict your findings

Answer 1

a) The 95% CI for the proportion of people that watched Super Bowl is (0.4531, 0.6183).

b) 48.1% = 0.481 is part of the confidence interval, which means that this prognosis does not contradict the findings.

Explanation:

a) Construct the 95% CI for the proportion of people that watched Super Bowl.

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

75 out of 140 watched the Super Bowl:

This means that
`n = 140, \pi = (75)/(140) = 0.5357`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.5357 - 1.96\sqrt{(0.5357*0.4643)/(140)} = 0.4531`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.5357 + 1.96\sqrt{(0.5357*0.4643)/(140)} = 0.6183`

The 95% CI for the proportion of people that watched Super Bowl is (0.4531, 0.6183).

b) The prognosis stated that the proportion of people watching the game is 48.1%. Does this prognosis contradict your findings

48.1% = 0.481 is part of the confidence interval, which means that this prognosis does not contradict the findings.