Respuesta :
Answer:
[tex](0.467-0.441) - 2.33 \sqrt{\frac{0.467(1-0.467)}{1800} +\frac{0.441(1-0.441)}{1800}}=-0.0127[/tex]
[tex](0.467-0.441) + 2.33 \sqrt{\frac{0.467(1-0.467)}{1800} +\frac{0.441(1-0.441)}{1800}}=0.0647[/tex]
And the 98% confidence interval would be given (-0.0127;0.0647).
We are confident at 98% that the difference between the two proportions is between [tex]-0.0127 \leq p_A -p_B \leq 0.0647[/tex]
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
[tex]p_A[/tex] represent the real population proportion for women
[tex]\hat p_A =\frac{841}{1800}=0.467[/tex] represent the estimated proportion for women
[tex]n_A=1800[/tex] is the sample size required for women
[tex]p_B[/tex] represent the real population proportion for men
[tex]\hat p_B =\frac{793}{1800}=0.441[/tex] represent the estimated proportion for men
[tex]n_B=1800[/tex] is the sample size required for Brand B
[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]
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]
For the 98% confidence interval the value of [tex]\alpha=1-0.98=0.02[/tex] and [tex]\alpha/2=0.01[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=2.33[/tex]
And replacing into the confidence interval formula we got:
[tex](0.467-0.441) - 2.33 \sqrt{\frac{0.467(1-0.467)}{1800} +\frac{0.441(1-0.441)}{1800}}=-0.0127[/tex]
[tex](0.467-0.441) + 2.33 \sqrt{\frac{0.467(1-0.467)}{1800} +\frac{0.441(1-0.441)}{1800}}=0.0647[/tex]
And the 98% confidence interval would be given (-0.0127;0.0647).
We are confident at 98% that the difference between the two proportions is between [tex]-0.0127 \leq p_A -p_B \leq 0.0647[/tex]
