Choosing a Distribution In Exercises 35–40, use the standard n...

Question

Choosing a Distribution 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.
Body Mass Index In a random sample of 50 people, the mean body mass index (BMI) was 27.7 and the standard deviation was 6.12.

Body Mass Index In a random sample of 50 people, the mean body mass index (BMI) was 27.7 and the standard deviation was 6.12.

Accepted Answer

All right, hello, everyone. So this question says, a researcher measures the resting heart rates of a random sample of 36 athletes and finds a sample mean of 62.5 beats per minute, with a sample standard deviation of 7.9 beats per minute. Construct a 95% confidence interval for the population mean. Resting heart rate. And here we've got 4 different entry choices labeled A through D. So for this question, to construct the 95% confidence interval for the population mean, we're going to use the T distribution. So let's talk about the reasons for this. The reason why we're using a T distribution in this case is because the population standard deviation is unknown. And the sample size is small to moderate, because here, N, that's our sample size, is equal to 36. Now, I just want to point out here that N equals 36 is greater than our normal cutoff of 30 for the T distribution. But in this context, it's still appropriate to use the T distribution because sigma is unknown. That is, the population standard deviation is unknown, which makes the T distribution appropriate regardless of N. Now, in addition, because N is greater than 30, the central limit theorem suggests that the sampling distribution of the mean will be approximately normal. So, let's gather the information we have. As mentioned before, N is equal to 36, but we also know the sample mean or X bar. That's equal to 62.5 beats per minute. Now, the sample standard deviation S is equal to 7.9 beats per minute. We know the confidence level is set at 95%. And we can also find the degrees of freedom, which if you recall, is equal to and subtracted by 1. So 36 subtracted by 1 gives you 35 degrees of freedom. So, degrees of freedom are necessary for the T critical value. So we have to find the T score that corresponds to a 95% confidence level with 35 degrees of freedom. Using either the T distribution table or a calculator. T or T critical. Is approximately 2.030. So now we have to calculate T for our data set and compare the critical value to the calculated value. Wait, sorry, I was thinking about a different type of problem. What we have to do now is compute the standard error because we're finding a confidence interval. So let's do that. Right, the standard error or SE is equal to S divided by the square root of N. So plugging in what we know, SE is equal to 7.9, divided by the square root of 36. And this equals approximately 1.3167. So from here you compute the margin of error. And the margin of error is that T critical value. Multiplied by the standard error. So 2.030. Multiplied by 1.3167. Equals approximately 2.6729. So now We can construct the confidence interval. And so the confidence interval. Is equal to X bar. Added to and subtracted by the margin of error. And that's how you get two values, right? One from adding the margin of error, that's the upper bound of the confidence interval, and the other, the lower bound from subtracting the margin of error. So CI is equal to 62.5, added to and subtracted by 2.67 approximately. And so The confidence interval, ultimately. Is 59.8 as the lower bound. And 65.2 as the upper bound after rounding to one decimal place. And this corresponds to. Option C in the multiple choice, making that your final answer. And there you have it. So with that being said, thank you so very much for watching, and I hope you found this helpful.

Key Concepts

Normal Distribution

t-Distribution

Confidence Interval

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