In Exercises 13–16, (a) find the margin of error for the value...

Question

In Exercises 13–16, (a) find the margin of error for the values of c, s, and n, and (b) construct the confidence interval for using the t-distribution. Assume the population is normally distributed.

c = 0.99, s = 16.5, n = 20, xbar = 25.2

Accepted Answer

Below there. Today we're going 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 scientist measures the concentration of a chemical in 25 water samples and finds a sample mean of 13.6 mg per liter with a sample standard deviation of 3.2 mg per liter. Construct the 98% confidence interval for the population mean using the T distribution. Assume the population of concentrations is normally distributed. Awesome. So it appears for this particular problem, we're ultimately trying to create a 98% confidence interval for the population mean using the T distribution. And we're also asked to assume that the population of concentrations is normally distributed. So now that we know that we're trying to construct a 98% confidence interval for the population mean using a T distribution. Our first step that we need to take in order to solve this problem, as we should recall, is we need to identify the sample statistics. So our sample size N is equal to 25, means we're sampling 25 water samples. XA, our sample mean, as we're told, once again, our sample mean is equal to 13.6 mg per liter. And finally, our standard deviation of our sample, which will denote as S is equal to 3.2 mg per liter. Fantastic. 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 which will denote ST subscript C. So our degrees of freedom, which we'll call DF or short, our degrees of freedom is equal to N minus 1, which is equal to. 25 minus 1, which is equal to 24. And our confidence, our level of confidence is equal to 98%, which is equal to, because we need to convert our percentage into a decimal, so it'll be equal to 0.98 because we divide 98 by 100 and get 0.98. So now, we need to use our table for the T distribution. And you should have a table for the tea distribution in your textbook. And it should look something along the lines of this. And as you can see, when we use our table for the T distribution, when we go down our column to 24 and then go across our row to meet our second column which is 0.98 because we have a 98% confidence interval and we go down to see where they intersect. We have a value of which I circled and read of 2.492. Awesome. So once again, using our table, our value for TC are critical value, so TC is equal to 2.492. Fantastic. So moving right along here. Our third step now is we need to find the margin of error E. So as we should recall, the margin of error margin of error equation denoted as E is equal to TC multiplied by S divided by the square root of N. So when we plug in our known variables, we will find that this is equal to 2.492 multiplied by parentheses 3.2 divided by the square root of 25. And when we plug this into our calculator and we round to one decimal point, you will find that this is approximately equal to 1.6. Fantastic. So now our fourth step now is we need to find the left and right endpoints and construct our confidence interval. So the left endpoint, which I'll call LE, so the left endpoint, the equation for this, as we should recall, is X bar minus E, which is equal to 13.6 minus 1. 59488, which if you're wondering where this value came from, this comes from our margin of error E which I rounded to one decimal place, but if you don't round, you will get 1.59488. Fantastic. So with that in mind, when we perform this calculation, when we subtract 13.6 from 1.5948, so if we subtract these two values, we will find when we round to one decial place that it's approximately equal to 12.0. Fantastic. So now we need to focus our attention on the right endpoint, which I'll call RE. So for our right endpoint, we have X bar plus E. So to save some time, I'm not going to write the equation out again. So setting instead of writing 13.6 minus 1.5948, you'll be adding. 13.6 plus 1.5948, and when we do that, and we round to one decimal place, we will find that it is approximately equal to 15.2. So now to solve for our confidence interval, which I'll call CI for short, so our confidence interval is, as we should recall, X minus E is less than mu, where mu is our mean, which is less than X bar plus E. So that will mean that when we plug in our numbers, we will get 12.0. is less than mu and mu is less than 15.2. Fantastic. So now at this point we need to recall a note that the other way to write the confidence interval is an interval notation such as 12.0. 15.2 in parentheses, that is our interval notation, and our point estimate, which is our plus or minus marginal error or margin of error, so I'll call it MFO, so margin or MO. E, so margin of error, so plus or minus our margin of error, which we could write this as 13.6 plus or minus 1.6. Fantastic. And once again, this is interval notation, which I'll call IN for short. So we have interval notation is a one way to write the confidence interval, and then we have the point estimate, which is plus or minus the margin of error, which is written as 13.6 plus or minus 1.6. So, therefore, without a doubt, we are 98% confident that the true meaning of the population lies between. 12.0 mg per liter and 15.2 mg per liter. So that will mean that we could write our final answer as follows 12.0 mg per liter is less than mu and mu is less than 15.2 mg per liter. And that's it. That is our final answer. Hooray, we did it. So going back to look at the prom itself to recap what we've learned. We have determined Without a doubt that our final answer for the 98% confidence interval for the population mean using our T distribution will have to be once again. 12.0 mg per liter is less than m and mu is less than 15.2 mg per liter. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

In Exercises 13–16, (a) find the margin of error for the values of c, s, and n, and (b) construct the confidence interval for using the t-distribution. Assume the population is normally distributed.
c = 0.99, s = 16.5, n = 20, xbar = 25.2

Key Concepts

Margin of Error

t-Distribution

Confidence Interval

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