Respuesta :
Answer:
[tex]0.61 - 1.28\sqrt{\frac{0.61(1-0.61)}{700}}=0.586[/tex]
[tex]0.61 + 1.28\sqrt{\frac{0.61(1-0.61)}{700}}=0.634[/tex]
The 80% confidence interval would be given by (0.586;0.634)
Step-by-step explanation:
A recent survey showed that 61% of U.S. employers were likely to require higher employee contributions for health care coverage. Suppose the survey was based on a sample of 700 companies. Find the 80% confidence interval for the proportion of all companies likely to require higher employee contributions for health care coverage.
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". Â
The margin of error is the range of values below and above the sample statistic in a confidence interval. Â
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". Â
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
Solution to the problem
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 80% of confidence, our significance level would be given by [tex]\alpha=1-0.80=0.2[/tex] and [tex]\alpha/2 =0.1[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.28, z_{1-\alpha/2}=1.28[/tex]
The confidence interval for the mean is given by the following formula: Â
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
If we replace the values obtained we got:
[tex]0.61 - 1.28\sqrt{\frac{0.61(1-0.61)}{700}}=0.586[/tex]
[tex]0.61 + 1.28\sqrt{\frac{0.61(1-0.61)}{700}}=0.634[/tex]
The 80% confidence interval would be given by (0.586;0.634)
