Ages of Prisoners The accompanying frequency distribution summ...

Question

Ages of Prisoners The accompanying frequency distribution summarizes sample data consisting of ages of randomly selected inmates in federal prisons (based on data from the Federal Bureau of Prisons). Use the data to construct a 95% confidence interval estimate of the mean age of all inmates in federal prisons.

Accepted Answer

A researcher collected data on the ages of patients admitted to a city hospital. The frequency distribution below summarizes the ages. We're given a table of values. Using this data, construct a 95% confidence interval estimate for the mean age of all patients admitted to the hospital. Now, we're looking for a confidence interval, which we will use the formula. Mean or X bar plus or minus margin of error. In this case, the marginal error for this type of experiment. Will be critical Multiplied By standard deviation Divided by the square root of sample size N. Let's find all of these values. So, we need to identify some parameters from our values. We need the summer frequencies, the mean of each class. Or the midpoint in our cases. The overall meaning of that is it. And our deviations. So let's take our first class, for instance, 18 to 27. Our class midpoint. will be the sum of our smallest value and our largest value of the group, divided by 2. 18 + 27 divided by 2, which gives us 22.5. We'll then have our frequency multiplied by our midpoint. Which is 10 for the first group, multiplied by 22.5 or 225. We now have our mean. Our expar will then be. The sum Our frequencies multiplied. By their midpoints divided by the sum of frequencies. Now we'll find this in a moment when we make our table. We also have our deviations. Which is our observation minus the mean. And then we'll use that to find our standard deviation. This table down here summarizes all of our data. We have the measurement of frequency for each row. We have each class' midpoint. We have the midpoint multiplied by the frequency, and then we have our deviations. Now if you look there, Our mean X bar is equals to 43.5. This is the sum of our frequency multiplied by our class midpoint, which is 8700. Divided by the sum of frequencies 200. We also have our standard deviation. standard deviation is given by. The square root Of the sum of our frequencies. Multiplied by our square deviations. All divided By the sum of frequencies -1. In this case, we have 25,800. Divided by 200 minus 1 under the square root. This gives us 11.39 for our sarin deviation. Now, finally, we're just missing our T critical value for our confidence interval. We have a 95% interval, which means we have alpha divided by 2 degrees of freedom. Our T critical will be at 0.05 divided by 2, with degrees of freedom of 200 minus 1. Which gives us T critical of 0.025 with 199 degrees of freedom. We can find this from a table, and the table will tell us. That we have 1.972 as our value approximately. We now have everything for our confidence interval. We have 43.5. Plus or minus RT critical, 1.972. Multiplied By serend deviation 11.39 Divided by the square root of sample size 200. This will give us 43.5. Plus or minus our value here. We can simplify that value. And we end up getting an interval. Of 41.9 as our lower bounds with the minus. To 45.14 upper bounds. This is in years, and this will be our confidence interval. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

Key Concepts

Confidence Interval

Mean

Frequency Distribution

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