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)

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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 [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

140 students, so [tex]n = 140[/tex]

B or better are grades of A or B.

5% earn As, 23% earn Bs, so [tex]p = 0.05 + 0.23 = 0.28[/tex]

95% confidence level

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

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.28 - 1.96\sqrt{\frac{0.28*0.72}{140}} = 0.2056[/tex]

The upper limit of this interval is:

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

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.