Respuesta :
Answer:
[tex](0.13-0.09) - 1.96 \sqrt{\frac{0.13(1-0.13)}{500} +\frac{0.09(1-0.09)}{500}}=0.00129[/tex]
[tex](0.13-0.09) + 1.96 \sqrt{\frac{0.13(1-0.13)}{500} +\frac{0.09(1-0.09)}{500}}=0.0787[/tex]
And the 95% confidence interval would be given (0.00129;0.0787).
We are confident at 95% that the difference between the two proportions is between [tex]0.00129 \leq p_B -p_A \leq 0.0787[/tex]
And as we can see the confidence interval for the difference on this case not contains the 0.
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 the gift certificate
[tex]\hat p_A =\frac{65}{500}=0.13[/tex] represent the estimated proportion for the gift certificate
[tex]n_A=500[/tex] is the sample size required for the gift certificate
[tex]p_B[/tex] represent the real population proportion for without the gift certificate
[tex]\hat p_B =\frac{45}{500}=0.09[/tex] represent the estimated proportion for without the gift certificate
[tex]n_B=500[/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 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](0.13-0.09) - 1.96 \sqrt{\frac{0.13(1-0.13)}{500} +\frac{0.09(1-0.09)}{500}}=0.00129[/tex]
[tex](0.13-0.09) + 1.96 \sqrt{\frac{0.13(1-0.13)}{500} +\frac{0.09(1-0.09)}{500}}=0.0787[/tex]
And the 95% confidence interval would be given (0.00129;0.0787).
We are confident at 95% that the difference between the two proportions is between [tex]0.00129 \leq p_B -p_A \leq 0.0787[/tex]
And as we can see the confidence interval for the difference on this case not contains the 0.
