Question

An accounting professor is notorious for being stingy in giving out good letter grades. In a large section of 140 students in the fall semester, she gave out only 5% As, 23% Bs, 42% Cs, ad 30% Ds and Fs. Assuming that this was a representative class, compute a 95% confidence interval for the proportion of students that obtain a letter grade of B or better from this professor. Interpret your confidence interval. (16 points)

Answer 1

Answer:

The 95% confidence interval for the proportion of students that obtain a letter grade of B or better from this professor is (0.2056, 0.3544). The interpretation is that we are 95% sure that the true proportion of students who obtain a letter grade of B or better from this professor is between 0.2056 and 0.3544.

Step-by-step explanation:

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 $1 - \frac{\alpha}{2}$.

For this problem, we have that:

140 students, so $n = 140$

B or better are grades of A or B.

5% earn As, 23% earn Bs, so $p = 0.05 + 0.23 = 0.28$

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.28 - 1.96\sqrt{\frac{0.28*0.72}{140}} = 0.2056$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.28 + 1.96\sqrt{\frac{0.28*0.72}{140}} = 0.3544$

The 95% confidence interval for the proportion of students that obtain a letter grade of B or better from this professor is (0.2056, 0.3544). The interpretation is that we are 95% sure that the true proportion of students who obtain a letter grade of B or better from this professor is between 0.2056 and 0.3544.