Question

Researchers are interested in determining whether more men than women prefer Coca Cola to Pepsi. In a random sample of 300 men, 65% prefer Coca Cola, whereas in a random sample of 400 women, 48% prefer Coca Cola. What is the 99% confidence interval estimate for the difference between the percentages of men and women who prefer Coca Cola over Pepsi? 0.17 ± 0.036 0.17 ± 0.096 0.17 ± 0.067 0.565 ± 0.067 0.565 ± 0.096

Answer 1

0.17
`\pm` 0.096

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. =
`\frac{(\hat p_1 - \hat p_2)-(p_1 - p_2)}{\sqrt{(\hat p_1(1- \hat p_1))/(n_1) + (\hat p_2(1- \hat p_2))/(n_2) } }` ~ N(0,1)

where,
`\hat p_1` = 0.65
`\hat p_2` = 0.48

`n_1` = 300
`n_2` = 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 <
`\frac{(\hat p_1 - \hat p_2)-(p_1 - p_2)}{\sqrt{(\hat p_1(1- \hat p_1))/(n_1) + (\hat p_2(1- \hat p_2))/(n_2) } }` < 2.5758) = 0.99

P(-2.5758 *
`\sqrt{(\hat p_1(1- \hat p_1))/(n_1) + (\hat p_2(1- \hat p_2))/(n_2) }`<
`(\hat p_1 - \hat p_2)-(p_1 - p_2)` < 2.5758 *
`\sqrt{(\hat p_1(1- \hat p_1))/(n_1) + (\hat p_2(1- \hat p_2))/(n_2) }` ) = 0.99

P(
`(\hat p_1 - \hat p_2)` - 2.5758 *
`\sqrt{(\hat p_1(1- \hat p_1))/(n_1) + (\hat p_2(1- \hat p_2))/(n_2) }` <
`(p_1 - p_2)` <
`(\hat p_1 - \hat p_2)` + 2.5758 *
`\sqrt{(\hat p_1(1- \hat p_1))/(n_1) + (\hat p_2(1- \hat p_2))/(n_2) }` ) = 0.99

So, 99% confidence interval for
`(p_1 - p_2)` =

[
`(\hat p_1 - \hat p_2)` - 2.5758 *
`\sqrt{(\hat p_1(1- \hat p_1))/(n_1) + (\hat p_2(1- \hat p_2))/(n_2) }` ,
`(\hat p_1 - \hat p_2)` + 2.5758 *
`\sqrt{(\hat p_1(1- \hat p_1))/(n_1) + (\hat p_2(1- \hat p_2))/(n_2) }` ]

= [ (0.65 - 0.48) - 2.5758 *
`\sqrt{(0.65(1- 0.65))/(300) + (0.48(1- 0.48))/(400) }` , (0.65 - 0.48) + 2.5758 *
`\sqrt{(0.65(1- 0.65))/(300) + (0.48(1- 0.48))/(400) }` ]

= [ 0.17 - 2.5758 *
`\sqrt{(0.65(1- 0.65))/(300) + (0.48(1- 0.48))/(400) }` , 0.17 + 2.5758 *
`\sqrt{(0.65(1- 0.65))/(300) + (0.48(1- 0.48))/(400) }` ]

= [ 0.17 - 0.096 , 0.17 + 0.096 ] = [0.17
`\pm` 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 .

Answer 2

Answer: 0.17 ± 0.096

Explanation:

Confidence interval for difference between the population proportion :

`p_1-p_2\pm z^*\sqrt{(p_1(1-p_1))/(n_1)+(p_2(1-p_2))/(n_2)}`

, where
`p_1` = sample proportion for population 1.

`p_2` = sample proportion for population 2.

`n_1` = Sample size from population 1.

`n_2` = Sample size from population 1.

As per given , we have

Population

`p_1=0.65\ \ \& \ \ p_2=0.48`

`n_1=300\ \ \ \&\ \ n_2=400`

Critical z-value for 99% confidence level is z*=2.576 [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 :

`(0.65-0.48)\pm (2.576)\sqrt{((0.65)(1-0.65))/(300)+((0.48)(1-0.48))/(400)}\\\\= 0.17\pm(2.576)√(0.0007583+0.000624)\\\\=0.17\pm(2.576)√(0.0013823)\\\\=0.17\pm(2.576)(0.03718)\\\\=0.17\pm0.09577568\\\\\approx0.17\pm0.096`\

Hence, the correct answer is 0.17 ± 0.096.