Large Data Sets from Appendix B. In Exercises 21 and 22, use t...

Question

Large Data Sets from Appendix B. In Exercises 21 and 22, use the data set in Appendix B. Assume that each sample is a simple random sample obtained from a population with a normal distribution.

Birth Weights Refer to Data Set 6 “Births” in Appendix B.

a. Use the 205 birth weights of girls to construct a 95% confidence interval estimate of the standard deviation of the population from which the sample was obtained.

Accepted Answer

A random sample of 10 adult males had their total cholesterol levels in milligrams per DL measured. We have a data set here with 10 values. Use the samples to construct a 95% confidence interval estimate of the population standard deviation of cholesterol levels. We were also given 4 possible answers which are all confidence intervals. Now, to solve this, we'll make a chi squared confidence interval, but we'll first need the mean. So, we'll first add all of our values together. 180. 190 195 2 And so on Divided by our total amount of observations, being 10. Now we can put this in the calculator. To get 1970 divided by 10 or 197.0. Now, we'll find the standard deviation. So we know the standard deviation. We can use the variance formula to find. Which is S2 equals the sum. Of each observation minus the mean squared. Divided by And -1 So in our case, We will have 180. 197 squared. Plus 190 minus 197 squared. And so on Until we get to our final. 190 minus 197 squared. This is divided by 10 minus 1. Now we can simplify this. We'll end up getting 1710 divided by 10 minus 1. Which gives us 710 divided by 9, or approximately 78.89. Now, this is our variance, or the standard deviation, we will take the square root of this value. 78.89. Which gives us a standard deviation of about 8.88. Now we have our mean and serin deviation. Next, let's find our chi square critical value. So, Our squid. Critical value we'll use our degrees of freedom. Our degrees of freedom Is our sample size minus 1, which will give us 9. No. We will make Our critical values. Now a good value here. would be to use an alpha value of 0.05. This means archi squared. We use two values. Being 0.025 with the degrees of freedom of nine. And the one kind of squared of 0.975 with degrees of freedom of 9. You can find these with a table. But if you look at the table, you'll get 19. 023 for the first one. And 2.7 for the second one. Finally, we can use the formula to find the confidence interval for Achi square. This will be The square root Of N minus 1 Multiplied by variants. Divided by each chi square critical value. So we will do Our first chi square critical value as our lower bounds, and our second chi square value as our upper bounds. So then, if we simplify this, We will have the square root. Of 9 multiplied by the variance, 78.89. All divided By 19.023. And we'll do the same on the upper bounds. We have 9 multiplied by 78.89 divided by. 2.7 We can now simplify both of these values. For our lower bounds We get 6.11. And for upper bounds, we will get 16.22. And this is in milligrams per DL. So now, if we look at our possible answers, lower bound is 6.11. Upper bound is 16.22. This will give us answer B. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

Key Concepts

Confidence Interval

Standard Deviation

Simple Random Sample

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