Question

Question 2.02 A random sample of 200 voters in a town is selected, and 114 are found to support an annexation suit. Find the 96% confidence interval for the fraction of the voting population favoring the suit.

Answer 1

Answer:

96% confidence interval for the fraction of the voting population favoring the suit is [0.498 , 0.642].

Step-by-step explanation:

We are given that a random sample of 200 voters in a town is selected, and 114 are found to support an annexation suit.

Firstly, the pivotal quantity for 96% confidence interval for the fraction of the voting population favoring the suit is given by;

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

where, $\hat p$ = proportion of voters found to support an annexation suit in a sample of 200 voters = $\frac{114}{200}$ = 0.57

           n = sample of voters = 200

           p = population proportion

Here for constructing 96% confidence interval we have used One-sample z proportion statistics.

So, 96% confidence interval for the population proportion, p is ;

P(-2.0537 < N(0,1) < 2.0537) = 0.96  {As the critical value of z at 2%

                                                  significance level are -2.0537 & 2.0537}

P(-2.0537 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.0537) = 0.96

P( $-2.0537 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.0537 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.96

P( $\hat p -2.0537 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p +2.0537 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.96

96% confidence interval for p =[ $\hat p -2.0537 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p +2.0537 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

 = [ $0.57 -2.0537 \times {\sqrt{\frac{0.57(1-0.57)}{200} } }$ , $0.57 +2.0537 \times {\sqrt{\frac{0.57(1-0.57)}{200} } }$ ]

 = [0.498 , 0.642]

Therefore, 96% confidence interval for the fraction of the voting population favoring the suit is [0.498 , 0.642].