Question

Since 1936, the Gallup Organization has been asking Americans: "Are you in favor of the death penalty for a person convicted of murder?" The percentage has fluctuated significantly over the years, ranging from a low of 42% in 1966 to a high of 80% in 1994. Here are the results of the most recent survey; in a sample of 3100 females, 62% said that they were in favor of the death penalty for convicted murders. Construct a 98% confidence interval for the proportion of all American females who support the death penalty for convicted murders.

Answer 1

Answer:

98% confidence interval for the proportion of all American females who support the death penalty for convicted murders is [0.59 , 0.64].

Step-by-step explanation:

We are given that the results of the most recent survey; in a sample of 3100 females, 62% said that they were in favor of the death penalty for convicted murders.

Firstly, the pivotal quantity for 98% 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 females who were in favor of the death penalty for convicted murders = 62%

n = sample of females = 3100

p = population proportion of all American females who support the death penalty

Here for constructing 98% confidence interval we have used One-sample z proportion statistics.

So, 98% confidence interval for the population proportion, p is ;

P(-2.33 < N(0,1) < 2.33) = 0.98  {As the critical value of z at 1% level

                                                   of significance are -2.33 & 2.33}  

P(-2.33 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.33) = 0.98

P( $-2.33 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.33 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.98

P( $\hat p-2.33 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.33 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.98

98% confidence interval for p = [ $\hat p-2.33 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.33 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.62-2.33 \times {\sqrt{\frac{0.62(1-0.62)}{3100} } }$ , $0.62+2.33 \times {\sqrt{\frac{0.62(1-0.62)}{3100} } }$ ]

 = [0.59 , 0.64]

Therefore, 98% confidence interval for the proportion of all American females who support the death penalty for convicted murders is [0.59 , 0.64].