In Exercises 13–16, (a) find the margin of error for the value...

Question

In Exercises 13–16, (a) find the margin of error for the values of c, s, and n, and (b) construct the confidence interval for using the t-distribution. Assume the population is normally distributed.

c = 0.98, s = 0.9, n = 12, xbar = 6.8

Accepted Answer

Hello, everyone, let's take a look at this question together. A sample of 20 students' test scores as a mean of 81.5 and a sample standard deviation of 5.3. Construct the 90% confidence interval or the population mean using the T distribution. Assume the population of test scores is normally distributed. Is it answer choice A, mu is between 79.1 and 83.9? Answer choice B, mu is between 79.5 and 83.5. Answer choice C, mu is between 79.8 and 83.2, or answer choice D, mu is between 80.3 and 82.7. So in order to solve this question, we have to recall how to construct a confidence interval so that we can construct the 90% confidence interval for the population mean using the T distribution, where we know that we have a sample of test scores from 20 students that has a mean of 81.5 and a sample standard deviation of 5.3. So the first step in solving this problem is to identify our sample statistics, which includes our sample size N of 20, our sample mean of 81.5, and our sample standard deviation of 5.3. And then we can identify the degrees of freedom, our level of confidence C, and the critical value of TC, starting with our degrees of freedom, which is equal to N minus 1. So our degrees of freedom is 20 minus 1, which is 19, and then our level of confidence C is 90% or 0.90. And then we use the table for T distribution to get our critical value of Tub C, and from the table, we can identify our critical value is equal to 1.729. And then we can find the margin of error E. Using the formula, which is E is equal to our critical value multiplied by our sample standard deviation, divided by the square root of our sample size. And substituting the values into the formula, we have 1.729 multiplied by in parentheses, 5.3 divided by the square root of 20. Which is equal to 2.049065613, or approximately 2.0. And now we can find the left and right endpoints and construct our confidence interval. Starting with our left endpoint, which is calculated as the sample mean subtracted by the margin of error. Which we know is 81.5 minus 2.049065613, which is approximately 79.5 as the value for the left endpoint. And then the right end point is the sample mean, plus our margin of error E. Which is equal to 81.5 plus the value for our margin of error, so the value for our right endpoint is approximately 83.5, and then we can construct our confidence interval, which is the left endpoint is less than mu, which is less than our right endpoint, so substituting the values into the confidence interval. Our confidence interval is m is between 79.5 and 83.5, and we also know that there are other ways to write this confidence interval, including interval notation, which is in parentheses 79.5 and 83.5, or point estimate plus minus margin of error, so 81.5 plus minus 2.0. And based on our calculations, we know that we are 90% confident that the true mean of the population, which is mu, lies between 79.5% and 83.5. Therefore, looking at our answer choices, we can identify that answer choice B is the correct answer, and all other answer choices are incorrect. I hope you found this video to be helpful. Thank you and goodbye.

In Exercises 13–16, (a) find the margin of error for the values of c, s, and n, and (b) construct the confidence interval for using the t-distribution. Assume the population is normally distributed.
c = 0.98, s = 0.9, n = 12, xbar = 6.8

Key Concepts

Margin of Error

Confidence Interval

t-Distribution

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