Question

A student was asked to find a 90% confidence interval for the proportion of students who take notes using data from a random sample of size n = 79. Which of the following is a correct interpretation of the interval 0.13 < p < 0.34?
Check all that are correct.

a. There is a 90% chance that the proportion of the population is between 0.13 and 0.34.
b. The proportion of all students who take notes is between 0.13 and 0.34, 90% of the time.
c. With 90% confidence, a randomly selected student takes notes in a proportion of their classes that is between 0.13 and 0.34.
d. There is a 90% chance that the proportion of notetakers in a sample of 79 students will be between 0.13 and 0.34.
e. With 90% confidence, the proportion of all students who take notes is between 0.13 and 0.34.

Answer 1

Answer:

After use the formula we got the following result for the 90% confidence interval (0.13 <p<0.34)

And the conclusion for this case would be:

e. With 90% confidence, the proportion of all students who take notes is between 0.13 and 0.34.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Description in words of the parameter p

$p$ represent the real population proportion of students who take notes

$\hat p$ represent the estimated proportion of students who take notes

n is the sample size required  

$z_{\alpha/2}$ represent the critical value for the margin of error  

The population proportion have the following distribution  

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Numerical estimate for p

In order to estimate a proportion we use this formula:

$\hat p =\frac{X}{n}$ where X represent the number of people with a characteristic and n the total sample size selected.

Confidence interval

The confidence interval for a proportion is given by this formula  

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$  

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.64$  

After use the formula we got the following result for the 90% confidence interval (0.13 <p<0.34)

And the conclusion for this case would be:

e. With 90% confidence, the proportion of all students who take notes is between 0.13 and 0.34.