Question

An article in Engineering Horizons (Spring 1990, p. 26) reported that 117 of 484 new engineering graduates were planning to continue studying for an advanced degree . Consider this as a random sample of the 1990 graduating class.
(A) Find a 90% confidence interval on the proportion of such graduates planning to continue their education.
(B) Find a 95% confidence interval on the proportion of such graduates planning to continue their education.
(C) Compare your answers to parts (a) and (b) and explain why they are the same or different.
(D) Could you use either of these confidence intervals to determine whether the proportion is actually 0.25? Explain your answer. Hint: Use the normal approximation to the binomial.

Answer 1

Answer:

a) The 90% confidence interval on the proportion of such graduates planning to continue their education is (0.2097, 0.2737).

b) The 95% confidence interval on the proportion of such graduates planning to continue their education is (0.2036, 0.2798).

c) Due to higher value of z, a 95% confidence interval, as in part b, is wider than a 90% confidence interval, found in part a, and is the difference between these two intervals.

d) Yes, because 0.25 is inside both intervals.

Step-by-step explanation:

Question a:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

117 of 484 new engineering graduates were planning to continue studying for an advanced degree.

This means that $n = 484, \pi = \frac{117}{484} = 0.2417$

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2417 - 1.645\sqrt{\frac{0.2417*0.7583}{484}} = 0.2097$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2417 + 1.645\sqrt{\frac{0.2417*0.7583}{484}} = 0.2737$

The 90% confidence interval on the proportion of such graduates planning to continue their education is (0.2097, 0.2737).

Question b:

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2417 - 1.96\sqrt{\frac{0.2417*0.7583}{484}} = 0.2036$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2417 + 1.96\sqrt{\frac{0.2417*0.7583}{484}} = 0.2798$

The 95% confidence interval on the proportion of such graduates planning to continue their education is (0.2036, 0.2798).

(C) Compare your answers to parts (a) and (b) and explain why they are the same or different.

The margin of error of a confidence interval of proportions is:

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

This means that, due to higher value of z, a 95% confidence interval, as in part b, is wider than a 90% confidence interval, found in part a, and is the difference between these two intervals.

(D) Could you use either of these confidence intervals to determine whether the proportion is actually 0.25? Explain your answer. Hint: Use the normal approximation to the binomial.

Yes, because 0.25 is inside both intervals.