High School

Suppose a random sample of 40 American men is drawn, and the weights of the men are measured. If the sample mean is 182 lbs and the sample standard deviation is 18.4 lbs, construct a 90% confidence interval for the mean weight of the population of American men. Round each of the endpoints of your interval to one decimal place.

Answer :

Final answer:

To create a 90% confidence interval for the mean population weight, we consider the central 90% probability of the normal distribution. The confidence interval, given a sample mean of 182 lbs., a standard deviation of 18.4 lbs., and a z-score for a 90% confidence interval of approximately 1.65, is (177.2 lbs, 186.8 lbs). Therefore, we can assert with 90% confidence that the mean weight of all American men fits within this range.

Explanation:

To find the 90% confidence interval for the mean population weight, we need to include the central 90% probability of the normal distribution. This leaves out 5% in each tail of the distribution. Here our sample mean is 182 lbs., and our sample standard deviation is 18.4 lbs. We usually use a z-score for standard normal distribution. For a 90% confidence interval, the z-score is approximately 1.645.

First, let's calculate the standard error by dividing the standard deviation by the square root of sample size i.e., 18.4/sqrt(40) which gives us approximately 2.91.

Next, we multiply this standard error by the z-score i.e., 1.645*2.91 = approx 4.8

Then, we add and subtract this value to/from the sample mean to get the confidence interval. So, the confidence interval is (182 - 4.8, 182 + 4.8) = (177.2, 186.8).

So, based on this sample, we can say with 90% confidence that the mean weight of all American men is between 177.2 lbs and 186.8 lbs.

Learn more about Confidence Intervals here:

https://brainly.com/question/34700241

#SPJ11