Constructing Confidence Intervals In Exercises 13–24, assume t...

Question

Constructing Confidence Intervals In Exercises 13–24, assume the sample is from a normally distributed population and construct the indicated confidence intervals for (a) the population variance σ^2. Interpret the results.

Drive-Thru Times The times (in seconds) spent by a random sample of 28 customers in the drive-thru of a fast-food restaurant have a sample standard deviation of 56.1. Use a 98% level of confidence.

Accepted Answer

Hello everyone. Let's take a look at this question together. The lengths in centimeters of 20 randomly selected pencils at a sample standard deviation of 1.2 centimeters. Assume the data comes from a normally distributed population. Using a 98% confidence level, construct a confidence interval for the population variance. Also interpret the results. So in order to solve this question, we have to recall how to construct a confidence interval for the population variance, so that we can construct one using a 98% confidence level for the population of the lengths in centimeters of 20 randomly selected pencils, which had a sample standard deviation of 1.2 centimeters. And we must also assume that the data comes from a normally distributed population. And once we construct our confidence interval, we must also interpret the results. And so the first step in constructing our confidence interval for the population variance is to identify our given values, which we know that from the information provided, the values that we are given includes a sample size of 20. A sample standard deviation of 1.2 centimeters. We can calculate our degrees of freedom, which is equal to and minus 1, so our degrees of freedom is 19. We know the confidence level is 98%, and so our alpha equals 0.02, therefore, an alpha divided by 2 is equal to 0.01. And then we can use the formula for the confidence interval for the population variance, which is given as the following formula, where we must first compute the square of our. sample standard deviation, so 1.2 squared is 1.44, and then N minus 1 multiplied by Squared is equal to 19 multiplied by 1.44, which is 27.36. And then we must also get our chi square values from the chi square distribution table for degrees of freedom equaling 9. where our right tail critical value is 36.191 and our left tail critical value is equal to 7.633. So then plugging these values into our formula, starting with the lower limit, we have 27.36 divided by 36.191, which is approximately 0.8. And then for the upper limit, we have 27.36 divided by 7.633, which is approximately 3.6. So our lower limit value is 0.8, and our upper limit value is 3.6, which tells us that the confidence interval for the population variance is 0.8 to 3.6. And interpreting these results, we know that we are 98% confident that the true population variance in pencil lengths is between 0.8 and 3.6 centimeters squared, and based on our constructed confidence interval and the interpretation of that interval, we can identify that answer choice C is the correct answer, and all other answer choices are incorrect. I hope you found this video to be helpful. Thank you and goodbye.

Key Concepts

Confidence Interval

Population Variance

Sample Standard Deviation

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