Question

Samples of computers are taken from two county library locations and the number of internet tracking spyware programs is counted. The first location hosted 21 computers with a mean of 4.1 tracking programs and a standard deviation of 0.8. The second location hosted 19 computers with a mean of 6.2 tracking programs and a standard deviation of 1.2.

a) Please calculate the appropriate standard error statistic for the scenario provided.

b)Please calculate a confidence interval around your point estimates as appropriate for the scenario. Use a confidence limit of 99% (.01)

c) Please calculate a confidence interval around your point estimates as appropriate for the scenario. Use a confidence limit of 95%.

d) How do your confidence intervals compare? Is this what you expected to see? Why?

Answer 1

(a)

The appropriate standard error statistic for the scenario provided = 0.3195

(b)

= 0.01

ndf = n1 + n2 - 2 = 21 + 19 - 2 = 38

From Table, critical values of t = 2.7116

99% Confidence Interval:

(4.1 - 6.2) (2.7116 X 0.3195)

= - 2.1 0.8665

= (- 2.9665 ,- 1.2335)

99% Confidence Interval:

- 2.9665 < < - 1.2335

(c)

= 0.05

ndf = n1 + n2 - 2 = 21 + 19 - 2 = 38

From Table, critical values of t = 2.0244

99% Confidence Interval:

(4.1 - 6.2) (2.0244 X 0.3195)

= - 2.1 0.6468

= (- 2.7468 ,- 1.4532)

95% Confidence Interval:

- 2.7468 < < - 1.4532

(d)

99% Confidence Interval (- 2.9665 ,- 1.2335) is narrower than 95% confidence Interval (- 2.7468 ,- 1.4532). This is what you expected to see because the greater the confidence level, the wider the confidence interval. As we increase the confidence level, the t - multiplier increases and hence the width of the interval increases.