Respuesta :
Answer:
[tex]0.8 - 2.58\sqrt{\frac{0.8(1-0.8)}{100}}=0.697[/tex]
[tex]0.8 + 2.58\sqrt{\frac{0.8(1-0.8)}{100}}=0.903[/tex]
The 99% confidence interval would be given by (0.697;0.903)
And the best option is : 0.697 to 0.903
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".
Solution to the problem
For this case the conditions we have the conditions to assume a normal distribution for the population proportion since:
np=100*0.8=80 >10 , n(1-p) = 10*(1-0.8) =20 >10
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 99% of confidence, our significance level would be given by [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2 =0.005[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
The estimated proportion is given by [tex] \hat p =\frac{80}{100} =0.8[/tex]. If we replace the values obtained we got:
[tex]0.8 - 2.58\sqrt{\frac{0.8(1-0.8)}{100}}=0.697[/tex]
[tex]0.8 + 2.58\sqrt{\frac{0.8(1-0.8)}{100}}=0.903[/tex]
The 99% confidence interval would be given by (0.697;0.903)
And the best option is : 0.697 to 0.903
