Mathematics, 26.03.2021 05:20 k3thagoat

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.

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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.

Step-by-step 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.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

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 $1 - \frac{0.04}{2} = 0.98$ , so Z = 2.056 .

Margin of error:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$M = 2.056\sqrt{\frac{0.14*0.86}{1600}}$

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

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

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

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 2.056\sqrt{\frac{0.14*0.86}{n}}$

$0.03\sqrt{n} = 2.056\sqrt{0.14*0.86}$

$\sqrt{n} = \frac{2.056\sqrt{0.14*0.86}}{0.03}$

$(\sqrt{n})^2 = (\frac{2.056\sqrt{0.14*0.86}}{0.03})^2$

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.

Answer from: Quest

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Answer from: Quest

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