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

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.

Step-by-step 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 , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

75 out of 140 watched the Super Bowl:

This means that

95% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

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.

answer: the first one

step-by-step explanation: use a calculator and divide the minutes by the dollars.

395.84

step-by-step explanation:

2πrh+2πr²

2 (pie)(radius)(height)+2(pie)(radius)^2