Age of Death-Row InmatesIn 2002, the mean age of an inmate on ...

Question

Age of Death-Row InmatesIn 2002, the mean age of an inmate on death row was 40.7 years, according to data from the U.S. Department of Justice. A sociologist wondered whether the mean age of a death-row inmate has changed since then. She randomly selects 32 death-row inmates and finds that their mean age is 38.9, with a standard deviation of 9.6. Construct a 95% confidence interval about the mean age. What does the interval imply?

Accepted Answer

Below there. Today we want to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A researcher collects a random sample of 40 city buses and finds that the mean life span is 12.8 years with a standard deviation of 3.5 years. Construct a 90% confidence interval for the true mean lifespanmu of city buses. What does this interval mean? Awesome. So it appears for this particular problem we're asked to solve for two separate answers. Our first answer is we're asked to construct or to create a 90% confidence interval for the true mean life span denoted as mu for these city buses. And then our second answer we're asked to solve for is we're asked to determine what does this interval mean. So, now that we know that we're now that we know what we're ultimately trying to solve for, First off, we need to note that since N is equal to 40 because our sample size is 40, and this value is greater than 30, and that the population standard deviation value sigma. It is unknown, we can therefore recall and use the T distribution. So once again, as we should recall, we can use the T distribution because the sample size in this case N is equal to 40, which is greater than 30, and we also do not know what the value for the population standard deviation. We do not know what the value of sigma is. So with that in mind, Our first step that we need to take to solve this problem is we need to note that N is equal to 40, because once again, that's our sample size. X bar in this case is going to be equal to 12.8, and this X bar represents the mean lifespan, which in this case is 12.8 years. And S in this case, because we're told that the standard deviation of 3.5 years. So once again, we know the standard deviation for the lifespan, but however, we do not know the standard deviation of the population. That is very clear. We have to make that very clear that there's a distinction between the two and they're not the same. So S is equal to 3.5, but we do not know what the value for sigma is because sigma once again is our population standard deviation is unknown to us. Moving right along here, our second step now is we need to identify the degrees of freedom. The level of confidence which will denote SC and the critical value T subscript C which I'll call TC. So let us recall that the degrees of freedom, which we'll call DF, is equal to N minus 1, which in this case, it will be equal to 40, minus 1 when we plug in our known variable, which will give us an answer of 39. And our value for C is equal to 90% in this case, which is equal to 0.90 when we convert our percentage to a decimal. So that's when you take 90 divided by 100. So moving right along here. Our next step now is we need to use our table for the T distribution, which you should have a table for the T distribution in your textbook. I've gone ahead and. Pulled out a, once again, I've already took from a textbook a table for the tea distribution and I've copied and pasted it here. And if we follow our column for DF, which is 39 for this particular case, and then we follow the row over to our next column for our value for the confidence level, or the level of confidence, which is 0.90, and we follow that column down to where they intersect. We will find that our value for TC will be equal to 1.685, which I've circled this value in red in our table. And once again, our value for TC, which is our critical value, is equal to 1.685. Awesome. So moving right along here. Our third step now as we're solving this problem is we need to find the margin of error E. So E once again denotes the margin of error, and as we should recall, this equation states that the margin of error is equal to TC multiply. By S divided by the square root of N, which is equal to when we plug in our known variables, 1.685 multiplied by parentheses, 3.5 divided by the square root of 40. So when we plug it into our calculator, we will find that it's equal to 0.93. 2476625 And that will mean when we round to one decimal place, we will find that E, once again, our margin of error E, is approximately equal to 0.9. Fantastic. So now our 4th and final step is we need to find the left and right endpoints and construct the confidence interval. So the left endpoint, which I'll call LE for short, so once again, our left endpoint. As we should recall, the equation to solve for the left endpoint is X minus E, which is equal to when we plug in our known variables, 12.8 minus 0.932476625. So when we plug that into our calculator, we will find that it's approximately equal to when we round to one decimal place, 11.9. And similarly, for our right endpoint, which I'll call RE, so for our right endpoint, as we should recall, the equation to solve for the right endpoint is X plus E. So instead of rewriting out our equation again, I'm just going to give you the answer for the right endpoint. And that will mean when we plug in our known variables, our right end point will be approximately equal to 13.7 when we round to one decimal place. So that will mean that our confidence interval, which I'll denote as CI, will be, as we should recall, X minus E is less than mu, and mu is less than X plus E. Which will mean when we plug in our known variables, our confidence interval will be 11.9 is less than mu, and mu is less than 13.7. Fantastic. So moving right along here, we need to note that there are other ways to write the confidence interval and the two different there's two different ways we can write our confidence interval. We have interval notation, which we'll call IN for short, and in interval notation we could write our confidence interval as parentheses 11.9.13.7. And our second way we could write our confidence interval is as a point estimate, which I'll call PE for short. And as we should recall, a point estimate for whenever we write it in the point estimate style, we can write it as plus or minus the margin of error, which I'll call MOE for short. So, once again, the point estimate. States plus or minus the margin of error, which in this case, we could write our confidence interval using this style of notation as 12.8 plus or minus 0.9. Fantastic. So now to solve for the second answer we're asked to solve for, we need to note that the confidence interval of 11.9 years is less than mu and mu is less than 1. As it is less than 13.7 years means that we are 90% confident that the true mean life span of all Valleys of new city buses lies between 11.9 years and 13.7 years. If we were to take many random samples and build a confidence interval from each sample, about 90% of those intervals would contain the true mean lifespan. So that will mean going back to look at our problem itself to make a quick recap of everything that we've learned, that would mean that our final two answers, without a doubt, will have to be the following. So once again, our confidence interval will be 11.9 years is less than mu, and mu is less than 13.7 years. And our second answer will have to be this means that we are 90% confident that the true mean lifespan of all city buses lies between 11.9 years and 13.7 years. And that's it. We've officially solved for this problem. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Key Concepts

Confidence Interval

Sample Mean and Standard Deviation

t-Distribution and Degrees of Freedom

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