A study was conducted to determine the percent of children that want to grow up work in the same career as a parent. In a sample of 200 children, it was calculated that 43% wanted to eventually work in the same career as a parent. Construct the 95% confidence interval for the population proportion. Solution: phat = 0.43/200 = 0.00215 phat – qnorm(1.95/2)*sqrt(phat*(1-phat)/20 0) = -0.00426926 phat + qnorm(1.95/2)*sqrt(phat*(1-phat)/20 0) = 0.00856926 [-0.0043, 0.0086] What is wrong with this solution?

From the information given, the 95% confidence interval for the population proportion is: (0.3614, 0.4986).

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the z-score that has a p-value of .

In this problem:

95% confidence level

So , z is the value of Z that has a p-value of , so .  

The lower limit of this interval is:

The upper limit of this interval is:

The 95% confidence interval for the population proportion is: (0.3614, 0.4986).

A similar problem is given at link

Step-by-step explanation:

From the information given,

Number of sample, n = 200

Probability of success, p = 43/100 = 0.43

q = 1 - p = 1 - 0.43

q = 0.57

For a confidence level of 95%, the corresponding z value is 1.96.

The formula for determining the error bound for the proportion is

z × √pq/n

= 1.96 ×√(0.43 × 0.57)/200

= 1.96 × 0.035 = 0.0686

The upper boundary of the population proportion is

0.43 + 0.0686 = 0.5

The lower boundary of the population proportion is

0.43 - 0.0686 = 0.4

The error in the solution is

phat = 0.43/200 = 0.00215

Also,

[-0.0043, 0.0086] is wrong

1. 3. can also be expressed as (22/7)

2. -4.5

3. 3.6

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