Respuesta :
Answer:
a) [tex]\hat p_1 -\hat p_2= 0.2-0.35= -0.15[/tex]
b) [tex] ME= 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.169[/tex]
c) [tex](0.2-0.35) - 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =-0.319[/tex]
[tex](0.2-0.35) + 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.0185[/tex]
And the 99% confidence interval would be given (-0.319;0.0185).
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_1[/tex] represent the real population proportion 1
[tex]\hat p_1=0.2[/tex] represent the estimated proportion 1
[tex]n_1=60[/tex] is the sample size required 1
[tex]p_2[/tex] represent the real population proportion for 2
[tex]\hat p_2 =0.35[/tex] represent the estimated proportion 2
[tex]n_2=100[/tex] is the sample size required for Brand B
[tex]z[/tex] represent the critical value for the margin of error
Solution to the problem
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_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_2)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}[/tex]
Part a
The best estimate is given by:
[tex]\hat p_1 -\hat p_2= 0.2-0.35= -0.15[/tex]
Part b
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]
The margin of error is given by:
[tex] ME= 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.169[/tex]
Part c
And replacing into the confidence interval formula we got:
[tex](0.2-0.35) - 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =-0.319[/tex]
[tex](0.2-0.35) + 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.0185[/tex]
And the 99% confidence interval would be given (-0.319;0.0185).
