$8$ . A random sample of $200$ female high school students in india showed that $110$ are on Facebook A separate random sample of $150$ male high school students in india showed that $69$ are on Facebook. Let $p 1$ and $p 2$ be the proportion of all female and male high school students in India, respectively, who are on Facebook. Which of the following represents a $95 \%$ confidence interval for $p 1$ - p $22$ $\newline$ $\cos =2 \sqrt{\frac{\sin 6}{3}+\frac{\cos x}{\cos }}$ $\newline$ $0.05=1.65 \sqrt{\frac{\operatorname{lan}}{2}}$

Full solution

8

. A random sample of

200

female high school students in india showed that

110

are on Facebook A separate random sample of

150

male high school students in india showed that

69

are on Facebook. Let

p 1

and

p 2

be the proportion of all female and male high school students in India, respectively, who are on Facebook. Which of the following represents a

95 \%

confidence interval for

p 1

- p

22

\newline

\cos =2 \sqrt{\frac{\sin 6}{3}+\frac{\cos x}{\cos }}

\newline

0.05=1.65 \sqrt{\frac{\operatorname{lan}}{2}}

Calculate Sample Proportions: Calculate the sample proportions for females and males. $\newline$ For females: $p_1 = \frac{110}{200} = 0.55$ $\newline$ For males: $p_2 = \frac{69}{150} = 0.46$
Calculate Standard Error: Calculate the standard error (SE) of the difference in proportions. $\newline$ $SE = \sqrt{(p1 \cdot (1 - p1) / 200) + (p2 \cdot (1 - p2) / 150)}$ $\newline$ $= \sqrt{(0.55 \cdot 0.45 / 200) + (0.46 \cdot 0.54 / 150)}$ $\newline$ $= \sqrt{(0.2475 / 200) + (0.2484 / 150)}$ $\newline$ $= \sqrt{0.0012375 + 0.001656}$ $\newline$ $= \sqrt{0.0028935}$ $\newline$ $= 0.0538$
Calculate Confidence Interval: Calculate the $95$ % confidence interval using the Z-value for $95$ % confidence $1.96$ .(\newline\)Lower limit = $p_1 - p_2$ - $1.96 \times SE$ = $0.55 - 0.46$ - $1.96 \times 0.0538$ = $0.09 - 0.1054$ = \$$-0$.$0154$(\newline\)Upper limit = $p_1 - p_2$ + $1.96 \times SE$ = $0.55 - 0.46$ + $1.96 \times 0.0538$ = $p_1 - p_2$$0$ = $p_1 - p_2$$1$