Respuesta :
Answer:
[tex]\hat p =0.11[/tex] represent the proportion estimated of subjects reported with dizziness
n = 268 represent the random sample selected
[tex] \hat q = 1-\hat p = 1-0.11= 0.89[/tex] represent the proportion of subjects No reported with dizziness
E= 0.04 = 4% represent the margin of error for the confidence interval
[tex] \alpha= 1-0.90 =0.1[/tex] and this value represent the significance level of the test or the probability of error type I
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". 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
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]
The margin of error is given by:
[tex] ME= z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
[tex]Lower= 0.11-0.04 = 0.07[/tex]
[tex]Upper= 0.11+0.04 = 0.15[/tex]
And for this case we have the following info:
[tex]\hat p =0.11[/tex] represent the proportion estimated of subjects reported with dizziness
n = 268 represent the random sample selected
[tex] \hat q = 1-\hat p = 1-0.11= 0.89[/tex] represent the proportion of subjects No reported with dizziness
E= 0.04 = 4% represent the margin of error for the confidence interval
[tex] \alpha= 1-0.90 =0.1[/tex] and this value represent the significance level of the test or the probability of error type I
