Internet connections are often slowed by delays at nodes. Let us determine if the delay time increases during heavy-volume times. Four hundred packets are sent through the same network between 5 pm and 6 pm (sample X ), and four hundred packets are sent between 10 pm and 11 pm (sample Y ). The early sample has a mean delay time of 0.7 sec with a standard deviation of 0.11 sec whereas the second sample has a mean delay time of 0.6 sec with a standard deviation of 0.09 sec.Construct a 99.5% confidence interval for the difference between the mean delay times.

To construct a 99.5% confidence interval for the difference between the mean delay times, we first need to calculate the standard error of the difference.

The formula for the standard error of the difference between two means is:

SE_diff = sqrt[(s1^2/n1) + (s2^2/n2)]

where s1 and s2 are the standard deviations of the two samples, and n1 and n2 are the sizes of the two samples.

Substituting the given values:

SE_diff = sqrt[(0.11^2/400) + (0.09^2/400)] SE_diff = sqrt[(0.0121/400) + (0.0081/400)] SE_diff = sqrt[0.00003025 + 0.00002025] SE_diff = sqrt[0.0000505] SE_diff = 0.0071

Next, we need to find the z-score that corresponds to a 99.5% confidence level. The z-score for a 99.5% confidence level is 2.807 (you can find this value in a standard z-table or using a calculator).

The confidence interval is then calculated as:

CI = mean_diff ± (z-score * SE_diff)

The difference between the means is 0.7 - 0.6 = 0.1.

So, the confidence interval is:

CI = 0.1 ± (2.807 * 0.0071) CI = 0.1 ± 0.0199

Therefore, the 99.5% confidence interval for the difference between the mean delay times is (0.0801, 0.1199) seconds.