Answer :
Answer:
[tex] ME= 2.58 \sqrt{\frac{0.8(1-0.8)}{125} +\frac{0.625 (1-0.625)}{120}}= 0.1465[/tex]
The best option would be:
d. 0.1465
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 18-25 users
[tex]\hat p_A =\frac{100}{125}=0.8[/tex] represent the estimated proportion for 18-25 users
[tex]n_A=125[/tex] is the sample size
[tex]p_B[/tex] represent the real population proportion for 26-35 users
[tex]\hat p_B =\frac{75}{120}=0.625[/tex] represent the estimated proportion for 26-35 users
[tex]n_B=120[/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]
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]
For the 99% confidence interval the value of [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=2.58[/tex]
And the margin of error is given by:
[tex] ME= 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]
And if we replace we got:
[tex] ME= 2.58 \sqrt{\frac{0.8(1-0.8)}{125} +\frac{0.625 (1-0.625)}{120}}= 0.1465[/tex]
The best option would be:
d. 0.1465