Mathematics, 05.05.2020 21:01 yarbor800592

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.

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Answer from: AM28

(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.

Answer from: Quest

The answers 5 so there’s the you need trust me it’s right

Answer from: Quest

70.30

step-by-step explanation:

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When 1146 subjects were asked, "Do you believe in heaven?" the proportion who answered yes was 0.84....