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.

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. = ~ N(0,1)

where, = proportion of adults residents who are parents in this county in a sample of 300 adults = = 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 < < 1.6449) = 0.90

P( < < ) = 0.90

P( < p < ) = 0.90

90% confidence interval for p = [ , ]

     = [ , ]

     = [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].

i think its a and d

step-by-step explanation:

the answer is 13

step-by-step explanation:

5*5=25

12*12=144

25+144=169

169^2=13