Respuesta :
Answer: 0.17 ± 0.096
Step-by-step explanation:
Confidence interval for difference between the population proportion :
[tex]p_1-p_2\pm z^*\sqrt{\dfrac{p_1(1-p_1)}{n_1}+\dfrac{p_2(1-p_2)}{n_2}}[/tex]
, where [tex]p_1[/tex] = sample proportion for population 1.
 [tex]p_2[/tex] = sample proportion for population 2.
[tex]n_1[/tex] = Sample size from population 1.
[tex]n_2[/tex] = Sample size from population 1.
As per given , we have
Population
[tex]p_1=0.65\ \ \& \ \ p_2=0.48[/tex]
[tex]n_1=300\ \ \ \&\ \ n_2=400[/tex]
Critical z-value for 99% confidence level is z*=2.576 Â [By z-table]
Now , the 99% confidence interval estimate for the difference between the percentages of men and women who prefer Coca Cola over Pepsi :
[tex](0.65-0.48)\pm (2.576)\sqrt{\dfrac{(0.65)(1-0.65)}{300}+\dfrac{(0.48)(1-0.48)}{400}}\\\\= 0.17\pm(2.576)\sqrt{0.0007583+0.000624}\\\\=0.17\pm(2.576)\sqrt{0.0013823}\\\\=0.17\pm(2.576)(0.03718)\\\\=0.17\pm0.09577568\\\\\approx0.17\pm0.096[/tex]\
Hence, the correct answer is 0.17 ± 0.096.
Answer:
0.17 [tex]\pm[/tex] 0.096
Step-by-step explanation:
We are given that Researchers are interested in determining whether more men than women prefer Coca Cola to Pepsi.
For this, In a random sample of 300 men, 65% prefer Coca Cola, whereas in a random sample of 400 women, 48% prefer Coca Cola.
The pivotal quantity for confidence interval is given by;
       P.Q. = [tex]\frac{(\hat p_1 - \hat p_2)-(p_1 - p_2)}{\sqrt{\frac{\hat p_1(1- \hat p_1)}{n_1} + \frac{\hat p_2(1- \hat p_2)}{n_2} } }[/tex] ~ N(0,1)
where, Â [tex]\hat p_1[/tex] = 0.65 Â Â Â Â [tex]\hat p_2[/tex] = 0.48
       [tex]n_1[/tex] = 300     [tex]n_2[/tex] = 400
So, 99% confidence interval for the difference between the percentages of men and women who prefer Coca Cola over Pepsi is given by;
 P(-2.5758 < N(0,1) < 2.5758) = 0.99 {At 1% significance level, the z table
                               gives value of 2.5758}
P(-2.5758 < [tex]\frac{(\hat p_1 - \hat p_2)-(p_1 - p_2)}{\sqrt{\frac{\hat p_1(1- \hat p_1)}{n_1} + \frac{\hat p_2(1- \hat p_2)}{n_2} } }[/tex] < 2.5758) = 0.99
P(-2.5758 * [tex]\sqrt{\frac{\hat p_1(1- \hat p_1)}{n_1} + \frac{\hat p_2(1- \hat p_2)}{n_2} }[/tex]< [tex](\hat p_1 - \hat p_2)-(p_1 - p_2)[/tex] < 2.5758 * [tex]\sqrt{\frac{\hat p_1(1- \hat p_1)}{n_1} + \frac{\hat p_2(1- \hat p_2)}{n_2} }[/tex] ) = 0.99
P([tex](\hat p_1 - \hat p_2)[/tex] - 2.5758 * [tex]\sqrt{\frac{\hat p_1(1- \hat p_1)}{n_1} + \frac{\hat p_2(1- \hat p_2)}{n_2} }[/tex] < [tex](p_1 - p_2)[/tex] < [tex](\hat p_1 - \hat p_2)[/tex] + 2.5758 * [tex]\sqrt{\frac{\hat p_1(1- \hat p_1)}{n_1} + \frac{\hat p_2(1- \hat p_2)}{n_2} }[/tex] ) = 0.99
So, 99% confidence interval for [tex](p_1 - p_2)[/tex] =
[ [tex](\hat p_1 - \hat p_2)[/tex] - 2.5758 * [tex]\sqrt{\frac{\hat p_1(1- \hat p_1)}{n_1} + \frac{\hat p_2(1- \hat p_2)}{n_2} }[/tex] , [tex](\hat p_1 - \hat p_2)[/tex] + 2.5758 * [tex]\sqrt{\frac{\hat p_1(1- \hat p_1)}{n_1} + \frac{\hat p_2(1- \hat p_2)}{n_2} }[/tex] ]
= [ (0.65 - 0.48) - 2.5758 * [tex]\sqrt{\frac{0.65(1- 0.65)}{300} + \frac{0.48(1- 0.48)}{400} }[/tex] , Â (0.65 - 0.48) + 2.5758 * [tex]\sqrt{\frac{0.65(1- 0.65)}{300} + \frac{0.48(1- 0.48)}{400} }[/tex] ]
= [ 0.17 - 2.5758 * [tex]\sqrt{\frac{0.65(1- 0.65)}{300} + \frac{0.48(1- 0.48)}{400} }[/tex] , 0.17 + 2.5758 * [tex]\sqrt{\frac{0.65(1- 0.65)}{300} + \frac{0.48(1- 0.48)}{400} }[/tex] ]
= [ 0.17 - 0.096 , 0.17 + 0.096 ] = [0.17 [tex]\pm[/tex] 0.096]
Therefore, 99% confidence interval for the difference between the percentages of men and women who prefer Coca Cola over Pepsi is 0.17 ± 0.096 .
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