Question

A survey was run by a high school student in order to determine what proportion of mortgage-holders in his town expect to own their house within 10 years. He surveyed 35 mortgage holders and found that the proportion of these that did expect to own their house within 10 years is 0.51. The student decides to construct a 95% confidence interval for the population proportion. a)Calculate the margin of error that the high school student will have. Give your answer as a decimal to 2 decimal places. Margin of error

Answer 1

Answer:

The margin of error is of 0.17.

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:

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

He surveyed 35 mortgage holders and found that the proportion of these that did expect to own their house within 10 years is 0.51.

This means that $n = 35, \pi = 0.51$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

a)Calculate the margin of error that the high school student will have.

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

$M = 1.96\sqrt{\frac{0.51*0.49}{35}}$

$M = 0.1656$

Rounding to 2 decimal places, the margin of error is of 0.17.