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.

[APPLET] Earnings The annual earnings (in thousands of dollars) of 21 randomly selected level 1 computer hardware engineers are listed. Use a 99% level of confidence. (Adapted from Salary.com)

Table displaying annual earnings (in thousands) of 21 computer hardware engineers, with values ranging from 52.4 to 95.5.

Accepted Answer

Hello, everyone. Let's take a look at this question together. The weights in kilograms of 8 randomly selected sealed sugar packs were recorded as follows, where we have the values for the weights in kilograms of the 8 randomly selected sugar packs. Assuming the weights follow a normal distribution, construct a 99% 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, so that we can construct a 99% confidence interval for the population variance of the weights in kilograms of 8 randomly selected sealed sugar packs, and we are assuming that the weights follow. Normal distribution, then we need to interpret our results. And the first step in solving this problem is organizing our data and computing, which we will first compute the sample mean and the sample variants using the following formula. More specifically, we will use the formula for the Sample mean and the sample variants, which starting with the sample mean, we know is calculated as the sum of our Xi values divided by our sample size N, which we know that the sum of our X values is equal to 8.4 and our sample size N is equal to 8. So 8.4 divided by 8 is 1.05, so our sample mean is equal to 1.05. And then using the value of our sample mean, which is 1.05, we can calculate our sample variance, which is calculated by the square of the X sub I subtracted by the sample mean values, and then the sum of all of those values, which the sum of those values is equal to 0.18, and then 0.18 is divided by our sample size and subtracted by 1, so 0.18. divided by 8 minus 1 is equal to 0.0257, so our sample variance is 0.0257. And now we can find the chi square critical values, which we know we have a degrees of freedom equaling N minus 1. So our degrees of freedom is equal to 7. And then from the chi square table, we can determine our critical values, which are the right chi square critical value is Equal to 20.278, and the left chi square critical value is equal to 0.989. And then using these values, we can use the confidence interval formula for population variance, which is given as the following formula, and then we can substitute the values into the formula, which gives us 7 multiplied by 0.0257 divided by 20.278 and 7. Multiplied by 0.0257 divided by 0.989, which, when simplified, gives us the confidence interval or population variance of approximately 0.01 to 0.18. And then interpreting the results, we know that we are 99% confident that the true variance in sealed sugar pack weights is between 0.01 and 0.18 kg squared. And looking at our answer choices, we can identify that answer choice D 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

Normal Distribution

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