Question

When 1146 subjects were​ asked, "Do you believe in​ heaven?" the proportion who answered yes was 0.84. The standard deviation of this point estimate is 0.01. a. Find and interpret the margin of error for a​ 95% confidence interval for the population proportion of people who believe in heaven. b. Construct the​ 95% confidence interval. Interpret it in context.

Answer 1

Answer:

(a) Margin of error = 0.0212.

(b) 95% confidence interval for the population proportion of people who believe in heaven is [0.82 , 0.86].

Step-by-step explanation:

We are given that when 1146 subjects were​ asked, "Do you believe in​ heaven?" the proportion who answered yes was 0.84. The standard deviation of this point estimate is 0.01.

(a) Margin of error tells us that how many percentage points our results will differ from the true population proportion value.

Margin of error is given by =  $Z_\frac{\alpha}{2} \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

where, $\hat p$ = point estimate or sample proportion = 0.84

n = sample of subjects = 1146

$\alpha$ = significance level = 1 - 0.95 = 0.05

So, value of  $Z_\frac{0.05}{2}$ in z table is given as 1.96

Hence, Margin of error  =  $1.96 \times \sqrt{\frac{0.84(1-0.84)}{1146} }$ = 0.0212

(b) 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 of people who answered yes = 0.84

           n = sample of subjects = 1146

           p = population proportion of people

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

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

 = [ $0.84-1.96 \times {\sqrt{\frac{0.84(1-0.84)}{1146} }}$ , $0.84+1.96 \times {\sqrt{\frac{0.84(1-0.84)}{1146} }}$ ]

 = [0.82 , 0.0.86]

Therefore, 95% confidence interval for the population proportion of people who believe in heaven is [0.82 , 0.86].

Interpretation of above confidence interval is that we are 95% confident that the population proportion of people who believe in heaven will lie between 0.82 and 0.86.