Question

In: Math

Samples of computers are taken from two county library locations and the number of internet tracking...

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?

Solutions

Expert Solution

(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.


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