Answer :
Answer:
[tex]0.226 - 1.96 \sqrt{\frac{0.226(1-0.226)}{13000}}=0.219[/tex]
[tex]0.226 + 1.96 \sqrt{\frac{0.226(1-0.226)}{13000}}=0.233[/tex]
And the 95% confidence interval would be given (0.219;0.233).
So then we can conclude a 99% of confidence that the true proportion of people were definitely not willing to pay is between 0.219 and 0.233
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
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
Solution to the problem
The estimated proportion for this case is :
[tex] \hat p = \frac{2938}{13000} = 0.226[/tex]
The confidence interval would be given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex]0.226 - 1.96 \sqrt{\frac{0.226(1-0.226)}{13000}}=0.219[/tex]
[tex]0.226 + 1.96 \sqrt{\frac{0.226(1-0.226)}{13000}}=0.233[/tex]
And the 95% confidence interval would be given (0.219;0.233).
So then we can conclude a 99% of confidence that the true proportion of people were definitely not willing to pay is between 0.219 and 0.233