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 = 0.90, s = 25.6, n = 16, xbar = 72.1

Accepted Answer

Hello everyone. Let's take a look at this question together. A researcher collects a sample of size and equals 25 from a normally distributed population. The sample standard deviation is 18.3, and the sample mean is 105.7. Construct a 95% confidence interval for the population meanmu using the T distribution. Is it answer choice Amu is between 98.1 and 113.3. Answer choice B, mu is between 99.4 and 112. Answer choice C, mu is between 96.6 and 114.8, or answer choice D, mu is between 95.5 and 115.9. So in order to solve this question, we have to recall how to construct a confidence interval so that we can construct a 95% confidence. In for the population mean view, where we have a sample size of N equals 25 from a normally distributed population and a sample standard deviation of 18.3, and a sample mean of 105.7. And so the first step to solve this problem is to identify the sample statistics. Where we know our sample size N is equal to 25, our sample mean is 105.7, and our sample standard deviation is 18.3. And then the second step is to identify the degrees of freedom, the level of confidence C, and the critical value of T sub C. Starting with our degrees of freedom, which is equal to and minus 1, so our degrees of freedom is 25 minus 1, which equals 24. Our level of confidence is that 95% confidence level, so it is equal to 0.95%. And then to get our critical value of TC, we use the table for the T distribution, and we can see that from the table, our critical value of Tub C is equal to 2.064. And now our 3rd step is to find the margin of error. E. Which the margin of error E is calculated as our critical value of TC multiplied by our sample standard deviation, divided by the square root of our sample size. So substituting the values into the formula, we have 2.064, multiplied by 18.3 divided by the square root of 25, which is equal to 7.55424, so approximately 7.6. And now we can find our left and right endpoints and construct the confidence interval. Starting with our left endpoint, which is calculated as our sample mean, subtracted by the margin of error E. Which substituting the values, we have 105.7, subtracted by 7.55424, so our left endpoint is approximately 98.1. And then our right end point is the sample mean, plus our margin of error E. So substituting the values in, we get 105.7 plus 7.55424. So our right endpoint is approximately 113.3. Therefore, we can now construct our confidence. Interval, which is our sample mean subtracted by E or the left endpoint, is less than our population mean, which is less than our sample mean, plus our margin of error E or our right endpoint. So then we can substitute the values for the left and right endpoints. Giving us our confidence interval, which is the population mean mu, is between 98.1 and 113.3, and we should also note that there are other ways to write our confidence interval, such as interval notation, where we have in parentheses 98.1 and 113.3, as well as our point estimate plus minus the margin of error. So we have 105.7 plus minus 7.6. And so based on our calculations, we know that we are 95% confident that the true mean of the population lies between 98.1 and 113.3. And looking at our answer choices, we can identify that answer choice A contains the correct confidence interval for our population mean mute. 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 = 0.90, s = 25.6, n = 16, xbar = 72.1

Key Concepts

Margin of Error

Confidence Interval

t-Distribution

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