Respuesta :
Answer:
Assuming a confidence level of 95%.
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]
The standard error for this case is given by: [tex]SE= 0.154[/tex]
[tex](0.667-0.25) - 1.96*0.154=0.115[/tex]
[tex](0.667-0.25) + 1.96*0.154=0.719[/tex]
And the 95% confidence interval for the difference of the two proportions would be given (0.115;0.719).
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".
[tex]p_A[/tex] represent the real population proportion for brand A
[tex]\hat p_A =\frac{12}{18}=0.667[/tex] represent the estimated proportion for male
[tex]n_A=18[/tex] is the sample size required for male
[tex]p_B[/tex] represent the real population proportion for female
[tex]\hat p_B =\frac{6}{24}=0.25[/tex] represent the estimated proportion for female
[tex]n_B=24[/tex] is the sample size required for female
[tex]z[/tex] represent the critical value for the margin of error
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 difference of two proportions would be given by this formula
[tex](\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}[/tex]
Assuming a confidence level of 95%.
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]
The standard error for this case is given by: [tex]SE= 0.154[/tex]
And replacing into the confidence interval formula we got:
[tex](0.667-0.25) - 1.96*0.154=0.115[/tex]
[tex](0.667-0.25) + 1.96*0.154=0.719[/tex]
And the 95% confidence interval for the difference of the two proportions would be given (0.115;0.719).
