Question

An online site presented this​ question, "Would the recent norovirus outbreak deter you from taking a​ cruise?" Among the 34 comma 630 people who​ responded, 67​% answered​ "yes." Use the sample data to construct a 90​% confidence interval estimate for the proportion of the population of all people who would respond​ "yes" to that question. Does the confidence interval provide a good estimate of the population​ proportion?

Answer 1

Answer:

$0.67 - 1.64\sqrt{\frac{0.67(1-0.67)}{34630}}=0.666$

$0.67 + 1.64\sqrt{\frac{0.67(1-0.67)}{34630}}=0.674$

The 90% confidence interval would be given by (0.666;0.674)

For this case we are confident (90%) that the true proportion of interest is between 0.66 and 0.674 and we can conclide that the confidence interval is a good estimation fo the population proportion at the confidence level used (90%)

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".  

The population proportion have the following distribution

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

Solution to the problem

The estimated proportion for this case is $\hat p = 0.67$ and the sample size is 34630

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$. And the critical value would be given by:

$t_{\alpha/2}=-1.64, t_{1-\alpha/2}=1.64$

The confidence interval for the mean is given by the following formula:  

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

If we replace the values obtained we got:

$0.67 - 1.64\sqrt{\frac{0.67(1-0.67)}{34630}}=0.666$

$0.67 + 1.64\sqrt{\frac{0.67(1-0.67)}{34630}}=0.674$

The 90% confidence interval would be given by (0.666;0.674)

For this case we are confident (90%) that the true proportion of interest is between 0.66 and 0.674 and we can conclide that the confidence interval is a good estimation fo the population proportion at the confidence level used (90%)