Question

About 217,000 high school students took the AP Statistics exam in 2017. The free-response section of the exam consisted of five open-ended problems and an investigative task. Each free-response question is scored on a 0 to 4 scale (with 4 being the best). For one of the problems, a random sample of 30 student papers yielded a mean score of x =1.267 and a standard deviation of 1.230. a. Find and interpret the standard error of the mean. b. Construct and interpret a 99% confidence interval to estimate the true mean score on this question.

Answer 1

Using the t-distribution, we have that:

a. The standard error is of 0.2246, which means that the mean scores for samples of 30 vary around 0.2246 from the mean.

b. The 99% confidence interval to estimate the true mean score on this question is (0.6479, 1.8861). It means that we are 99% that the true mean score of all students in this question is between 0.6479 and 1.8861.

We have the standard deviation for the sample, hence the t-distribution is used to solve this question.

What is a t-distribution confidence interval?

The confidence interval is:

$\overline{x} \pm t\frac{s}{\sqrt{n}}$

In which:

• $\overline{x}$ is the sample mean.

• t is the critical value.

• n is the sample size.

• s is the standard deviation for the sample.

In this problem, the parameters are:

$n = 30, \overline{x} = 1.267, s = 1.23$

Item a:

The standard error is:

$S_e = \frac{s}{\sqrt{n}} = \frac{1.23}{\sqrt{30}} = 0.2246$

The standard error is of 0.2246, which means that the mean scores for samples of 30 vary around 0.2246 from the mean.

Item b:

Using a t-distribution calculator, considering a confidence level of 0.99 with 30 - 1 = 29 df, the critical value is t = 2.7564.

Hence:

$\overline{x} \pm tS_e$

$\overline{x} - tS_e = 1.267 - 2.7564(0.2246) = 0.6479$

$\overline{x} + tS_e = 1.267 + 2.7564(0.2246) = 1.8861$

The 99% confidence interval to estimate the true mean score on this question is (0.6479, 1.8861). It means that we are 99% that the true mean score of all students in this question is between 0.6479 and 1.8861.

To learn more about the t-distribution, you can take a look at https://brainly.com/question/16313918