Consider a 95% confidence interval for a population mean constructed from a random sample of 250 observations taken from a Normal population. Sigma is known to be 14. If we were to construct another 95% confidence interval, this time using 1,000 observations, how would the length of this interval compare to the length of the original interval?

the sample of selected men have the average weight of 180lb with the standard deviation of 30lb. for 0.95 find the confidence interval for the mean of the whole population

To compare the length of the confidence intervals, we need to consider the sample size and the standard deviation (sigma) in both cases.

For the original confidence interval:
Sample size (n1) = 250
Standard deviation (sigma) = 14

For the second confidence interval:
Sample size (n2) = 1000

Now, the formula to calculate the length of a confidence interval is:

Length = 2 * (Z * (sigma / sqrt(n)))

Where:
Z is the z-score corresponding to the desired confidence level. For a 95% confidence level, Z is approximately 1.96.
sigma is the population standard deviation.
sqrt(n) is the square root of the sample size.

Let's calculate the length of the original confidence interval:

Length1 = 2 * (1.96 * (14 / sqrt(250)))

Now, let's calculate the length of the new confidence interval with 1,000 observations:

Length2 = 2 * (1.96 * (14 / sqrt(1000)))

Comparing the two lengths, we can see that the length of the confidence interval for the 1,000 observations will likely be shorter than the length of the original interval. This is because as the sample size increases, the standard error decreases, leading to a narrower confidence interval. In this case, the larger sample size of 1,000 observations would result in a smaller margin of error and a shorter interval.