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

Answer: 0.17 ± 0.096

Step-by-step explanation:

Confidence interval for difference between the population proportion :

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

, 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 :

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

Hence, the correct answer is 0.17 ± 0.096.

Answer 2

Answer:

0.17 $\pm$ 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. = $\frac{(\hat p_1 - \hat p_2)-(p_1 - p_2)}{\sqrt{\frac{\hat p_1(1- \hat p_1)}{n_1} + \frac{\hat p_2(1- \hat p_2)}{n_2} } }$ ~ 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 < $\frac{(\hat p_1 - \hat p_2)-(p_1 - p_2)}{\sqrt{\frac{\hat p_1(1- \hat p_1)}{n_1} + \frac{\hat p_2(1- \hat p_2)}{n_2} } }$ < 2.5758) = 0.99

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

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

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

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

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

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

= [ 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 .