Question

Suppose that a standardized biology exam has a mean score of 80% correct, with a standard deviation of 3. The school administration believes that the new class of freshmen have a different mean exam score. A sample of 65 students from the freshmen class is used and a mean score of 76% correct is obtained.
Calculate a 99% confidence interval. List the lower bound (rounded to 2 decimal places) and the upper bound (rounded to 2 decimal places).

Answer 1

Answer:

The 99% confidence interval is between 62.36%(lower bound) and 89.64%(upper bound).

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

A sample of 65 students from the freshmen class is used and a mean score of 76% correct is obtained.

This means that $n = 65, \pi = 0.76$

99% confidence level

So $\alpha = 0.005$, z is the value of Z that has a pvalue of $1 - \frac{0.005}{2} = 0.995$, so $Z = 2.575$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.76 - 2.575\sqrt{\frac{0.76*0.24}{65}} = 0.6236$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.76 + 2.575\sqrt{\frac{0.76*0.24}{65}} = 0.8964$

0.6236*100 = 62.36%

0.8964*100 = 89.64%

The 99% confidence interval is between 62.36%(lower bound) and 89.64%(upper bound).