Question

A polling organization contacts 1600 people and of these 14% say that they jog regularly. Construct and interpret a 96% confidence interval for the proportion of people who say they jog regularly. Explain what changes if you instead use a 99% confidence level. What is the smallest number of people we could contact to still have a margin of error of 3% at a 96% confidence level.

Answer 1

The 96% confidence interval for the proportion of people who say they jog regularly is between 0.1222 and 0.1578. This means that we are 96% sure that the true population proportion of people who jog is between 0.1222 and 0.1578.

With a 99% confidence level, the interval would be wider.

The smallest number of people we could contact to still have a margin of error of 3% at a 96% confidence level is 566.

Explanation:

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)`.

The margin of error is:

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

A polling organization contacts 1600 people and of these 14% say that they jog regularly.

This means that
`n = 1600, \pi = 0.14`

Construct and interpret a 96% confidence interval for the proportion of people who say they jog regularly.

96% confidence level

So
`\alpha = 0.04`, z is the value of Z that has a pvalue of
`1 - (0.04)/(2) = 0.98`, so
`Z = 2.056`.

Margin of error:

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

`M = 2.056\sqrt{(0.14*0.86)/(1600)}`

`M = 0.0178`

Lower bound:

Sample proportion subtracted by the margin of error. So

`\pi - M = 0.14 - 0.0178 = 0.1222`

Upper bound:

Margin of error added to the sample proportion. So

`\pi + M = 0.14 + 0.0178 = 0.1578`

The 96% confidence interval for the proportion of people who say they jog regularly is between 0.1222 and 0.1578. This means that we are 96% sure that the true population proportion of people who jog is between 0.1222 and 0.1578.

Explain what changes if you instead use a 99% confidence level.

With a 99% confidence level, the value of Z would be higher(2.575), which means that the interval would be wider.

What is the smallest number of people we could contact to still have a margin of error of 3% at a 96% confidence level.

This is n for which
`M = 0.03`

So

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

`0.03 = 2.056\sqrt{(0.14*0.86)/(n)}`

`0.03√(n) = 2.056√(0.14*0.86)`

`√(n) = (2.056√(0.14*0.86))/(0.03)`

`(√(n))^2 = ((2.056√(0.14*0.86))/(0.03))^2`

`n = 565.5`

Rounding up:

The smallest number of people we could contact to still have a margin of error of 3% at a 96% confidence level is 566.