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 (b) the population standard deviation σ. 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)

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, and here we have the values for the 8 randomly selected sugar packs. Assuming the weights follow a normal distribution, construct a 99% confidence interval for the population standard deviation. 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 standard deviation of the weights of 8 randomly selected sealed sugar packs that were recorded in kilograms, assuming that they follow a normal distribution. And then we need to interpret our results. And so the first step in solving this problem is organizing our data and computing. More specifically, we will first compute the sample mean and the sample variants using the following formula, which are the formulas for the sample mean and the sample variants. So starting with the sample mean, which is Calculated as the sum of our X values, divided by the sample size N, we know that the sum of our X values is equal to 8.4, and our sample size N is equal to 8, so our sample mean is calculated as 8.4 divided by 8, which is equal to 1.05. And then using the sample mean value. Of 1.05, we can calculate our sample variants as it is calculated by using the sum of in parentheses, our X values subtracted by the sample mean, which is then squared, which we know that the sum of those values is 0.18, and then it is divided by our sample size and subtracted by 1. So the sample variance is equal. To 0.18 divided by 8 minus 1, which when simplified is 0.0257. Therefore, our sample mean is 1.05, and our sample variance is 0.0257. And then we use the formula for the confidence interval of population variance, which is given by the following formula. And we know that to find the right and left tail areas. We use our degrees of freedom, which we know is calculated as N minus 1, so our degrees of freedom is equal to 7, and then the area to the right of the right tail critical value is calculated as 1 minus C divided by 2, so the area to the right of the right tail critical value is 1 minus 0.99 divided by 2, which is equal to 0.005. And then the area to the right of the left tail critical value is 1 minus 1 minus C divided by 2, which is equal to 0.995. Therefore, we know our right tail critical value is given as our chi square sub 0.005, with a degree of freedom equaling 7, and then the left tail critical value is the chi square value sub 0.995. A degree of freedom equaling 7. And using this information we can find the chi square critical values and from the chi square table we know the right tail critical value is 20.278 and the left tail critical value is 0.989. And then we can plug in the values and calculate the confidence interval for population variance, which plugging the values into the formula, we have 7 multiplied by 0.0257 divided by 20.278, and 7 multiplied by 0.0257 divided by 0.989, which, when simplified, our confidence interval for population variance is approximately 0.01 to 0.18. But since we are looking for the confidence interval of the standard deviation, we have to Take the square roots of our confidence interval or population variance to get the confidence interval of the standard deviation. So we have the square root of 0.00887 and the square root of 0.18190, which is approximately 0.09 and 0.43. So our confidence interval of the standard deviation is 0. 09 to 0.43. And then interpreting these results, we know we are 99% confident that the true standard deviation in sealed sugar pack weights is between 0.09 and 0.43 kg. And looking at our answer choices, we can identify that answer choice A 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 Standard Deviation (σ)

Normal Distribution

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