Question

A poll was taken of 1000 residents in county. The residents sampled were asked whether they think their local government did a good job overall. 750 responded "yes". Let p denote the proportion of all residents in that county who think their local government did a good job. Construct a 95% confidence interval for p. Round off to two decimal places. a) (0.72, 0.78) b)(0.70, 0.86) c (0.68, 0.92) d) (0.10,1.56) e)(0.79, 0.87)

Answer 1

Answer:

a) (0.72, 0.78)

Step-by-step explanation:

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 all residents in that county who think their local government did a good job

$\hat p$ represent the estimated proportion of all residents in that county who think their local government did a good job

n=1000 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.

$\hat p=\frac{750}{1000}=0.75$ represent the estimated proportion of all residents in that county who think their local government did a good job

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 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.96$  

And replacing into the confidence interval formula we got:  

$0.750 - 1.96 \sqrt{\frac{0.75(1-0.75)}{1000}}=0.72$  

$0.750 + 1.96 \sqrt{\frac{0.75(1-0.75)}{1000}}=0.78$  

And the 95% confidence interval would be given (0.72;0.78).  

We are confident at 95% that the true proportion of people who think their local government did a good job is between (0.72;0.78).

a) (0.72, 0.78)