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:
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