F(pi/3)= 4, f'(pi/3)= -2, g(x)= f(x)sinx, h(x)= cosx/f(x) Find g'(pi/3) Find h'(pi/3)

A survey asked whether respondents favored or opposed the death penalty for people convicted of murder. Software shows the resul

pav-90 [236]

Answer:

95% confidence interval for the proportion of the adults who were opposed to the death penalty is (0.668, 0.704).

Step-by-step explanation:

We are given that a survey asked whether respondents favored or opposed the death penalty for people convicted of murder. Software shows the results below, where X refers to the number of the respondents who were in favor.

X = 1,790

N = 2,610

$\hat p$ = Sample proportion = X/N = 0.6858

Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion = 0.6858

n = sample of respondents = 2,610

p = population proportion

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at

2.5% level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ]