In Exercises 35–40, use the standard normal distribution or th...

Question

In Exercises 35–40, use the standard normal distribution or the t-distribution to construct a 95% confidence interval for the population mean. Justify your decision. If neither distribution can be used, explain why. Interpret the results.

In a random sample of 18 months from January 2011 through December 2020, the mean interest rate for 30-year fixed rate home mortgages was 3.95% and the standard deviation was 0.49%. Assume the interest rates are normally distributed. (Source: Freddie Mac)

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says a researcher collected a random sample of 22 weeks of gasoline prices from January 2018 to December 2022. The sample mean price was $2.87 per gallon, with a standard deviation of 38 cents per gallon. Assume the prices are normally distributed. Construct a 95% confidence interval for the population mean price per gallon, and we're given 4 possible choices as our answers. For choice A, we have 2.70 is less than you, which is less than 3.04. For choice B, we have 2.75 is less than mu, which is less than 2.99. For choice C, we have 2.66 is less than mu, which is less than 3.08. And for choice D, we have 2.82 is less than mu, which is less than 2.92. So we're asked to calculate a 95% confidence interval for the population mean price per gallon. And the first thing we want to do is identify the information that we're given in the problem. So the first quantity that we're given is our sample size, which is going to be 22, since we're looking at 22 weeks of gasoline prices. That means N is going to be equal to 22. We're also given our sample mean, which is $2.87. So that means that X bar is going to be equal to. 2.87. And we're also given our sample standard deviation of 38 cents. That means our sample standard deviation S is going to be equal to 0.38. Now, in order to find our confidence interval, we first need to calculate our margin of error. E, and we call that E is going to be equal to. TC multiplied by S divided by the square root of N. Now we have our values for S and N. But we do not have a value for TC, which is our critical value, or our 95% confidence interval. Now, in order to figure out this value, we first need to determine the degrees of freedom. And we call that our degrees of freedom, which we'll label as DF, is going to be equal to N minus 1, which in this case is going to be equal to 22 minus 1, which is equal to 21. So we're going to have 21 degrees of freedom, and we're also looking for a confidence interval for 95%. So that means our confidence level. C is going to be equal to 0.95. So when we convert 95% into a decimal by divided by 100, we get 0.95. And so when we look up our critical value and our table with C equal to 0.95 and degrees of freedom equal to 21, we see that our critical value, E C, is going to be equal to. 2.080. So now we're substitute in all of our values into our equation for the margin of error. That means that E is going to be equal to. 2.080. Multiplied by 0.38, divided by the square root of 22. And when we evaluate this expression, this is going to be approximately equal to. 0.17. So now we use the margin of varior to calculate the left and right endpoints for our confidence interval. So for our left endpoint, We're going to have X minus E. Which is going to be equal to. 2.87. Minus 0.17. Which is going to be equal to 2.70. And for our right end point, We're gonna have XR plus E. Which is going to be equal to 2.87 plus 0.17. Which is going to be equal to 3.04. So, our 95% confidence interval is going to be 2.70 is less than you. Which is less than 3.04. But this here is the answer that we're looking for for this problem. And if we compare that to our answer choices, we see that the correct answer choice corresponds to answer choice A. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Confidence Interval

Normal Distribution

t-Distribution

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