Question

What do you want to be when you grow up? According to a nationwide survey of a random sample of 1000 kids under the age of 12 some kids want to be a ninja, a dragon keeper, a dancing unicom, and even an octopus. Fifty-five of the 1000 kids want to be a doctor. We would like to use this study to find a 98% confidence interval for the true proportion of kids who want to be a doctor.

a. Identify the parameter of interest.

b. Check if the conditions for constructing a confidence interval for p are met.

C. Find the critical value for a 98% confidence interval. Then calculate the interval

d. Interpret the interval in context

Answer 1

i think it is b i’m so sorry if i’m wrong

Answer 2

Using the z-distribution, we have that:

• a) The parameter of interest is the proportion of kids who want to be a doctor.

• b) Since there are more than 10 successes and more than 10 failures in the sample, the conditions are met.

• c) The critical value is $z = 2.327$. The interval is (0.0339, 0.066).

• d) We are 98% sure that the true proportion of kids who want to be a doctor is between 0.0339 and 0.066.

Confidence interval of a proportion

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

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

• In which z is the z-score that has a p-value of $\frac{1+\alpha}{2}$.

Item a:

We are interested in the confidence interval for the true proportion of kids who want to be a doctor, hence the parameter of interest is the proportion of kids who want to be a doctor.

Item b:

Fifty-five of the 1000 kids want to be a doctor, hence, since there are more than 10 successes and more than 10 failures in the sample, the conditions are met.

Item c:

• Fifty-five of the 1000 kids want to be a doctor, hence $n = 1000, \pi = \frac{55}{1000} = 0.055$.

• 98% confidence level, hence$\alpha = 0.98$, z is the value of Z that has a p-value of $\frac{1+0.98}{2} = 0.99$, so the critical value is $z = 2.327$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.05 - 2.327\sqrt{\frac{0.05(0.95)}{1000}} = 0.0339$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.05 + 2.327\sqrt{\frac{0.05(0.95)}{1000}} = 0.066$

The interval is (0.0339, 0.066).

Item d:

We are 98% sure that the true proportion of kids who want to be a doctor is between 0.0339 and 0.066.

You can learn more about the use of the z-distribution to build a confidence interval at https://brainly.com/question/25649070