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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26-03-2021
Mathematics

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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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joaobezerra joaobezerra · Answer 1 · 2021-03-26T20:10:21+01:00

Answer:

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 [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

The margin of error is:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

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

This means that [tex]n = 1600, \pi = 0.14[/tex]

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

96% confidence level

So [tex]\alpha = 0.04[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.04}{2} = 0.98[/tex], so [tex]Z = 2.056[/tex].

Margin of error:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]M = 2.056\sqrt{\frac{0.14*0.86}{1600}}[/tex]

[tex]M = 0.0178[/tex]

Lower bound:

Sample proportion subtracted by the margin of error. So

[tex]\pi - M = 0.14 - 0.0178 = 0.1222[/tex]

Upper bound:

Margin of error added to the sample proportion. So

[tex]\pi + M = 0.14 + 0.0178 = 0.1578[/tex]

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 [tex]M = 0.03[/tex]

So

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]0.03 = 2.056\sqrt{\frac{0.14*0.86}{n}}[/tex]

[tex]0.03\sqrt{n} = 2.056\sqrt{0.14*0.86}[/tex]

[tex]\sqrt{n} = \frac{2.056\sqrt{0.14*0.86}}{0.03}[/tex]

[tex](\sqrt{n})^2 = (\frac{2.056\sqrt{0.14*0.86}}{0.03})^2[/tex]

[tex]n = 565.5[/tex]

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.