Question

We wish to estimate what percent of adult residents in a certain county are parents. Out of 300 adult residents sampled, 144 had kids. Based on this, construct a 90% confidence interval for the proportion p of adult residents who are parents in this county.

Answer 1

Answer:

90% confidence interval for the proportion p of adult residents who are parents in this county is [0.433 , 0.527].

Step-by-step explanation:

We are given that we wish to estimate what percent of adult residents in a certain county are parents. Out of 300 adult residents sampled, 144 had kids.

Firstly, the pivotal quantity for 90% confidence interval for the proportion p of adult residents who are parents in this county 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 adults residents who are parents in this county in a sample of 300 adults = $\frac{144}{300}$ = 48%

          n = sample of adults residents = 300

          p = population proportion of adults

Here for constructing 90% confidence interval we have used One-sample z proportion statistics.

So, 90% confidence interval for the population proportion, p is ;

P(-1.6449 < N(0,1) < 1.6449) = 0.90  {As the critical value of z at 5%

                                                   significance level are -1.6449 & 1.6449}

P(-1.6449 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.6449) = 0.90

P( $-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

90% confidence interval for p = [ $\hat p-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$, $\hat p+1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

     = [ $0.48-1.6449 \times {\sqrt{\frac{0.48(1-0.48)}{300} } }$ , $0.48+1.6449 \times {\sqrt{\frac{0.48(1-0.48)}{300} } }$ ]

     = [0.433 , 0.527]

Therefore, 90% confidence interval for the proportion p of adult residents who are parents in this county is [0.433 , 0.527].