Question

A random sample of 100 workers in one large plant took an average of 12 minutes to complete a task, with a standard deviation of 2 minutes. A random sample of 50 workers in a second large plant took an average of 11 minutes to complete the task, with a standard deviation of 3 minutes. Construct a 95% confidence interval for the difference between the two population mean completion times.

Answer 1

The 95% Confidence Interval for the difference between the two population mean completion times =

(0.081, 1.919)

Explanation:

Confidence Interval for difference between two means =

μ1 -μ2 ± z × √ σ²1/n1 + σ²2/n2

Where

μ1 = mean 1 = 12 mins

σ1 = Standard deviation 1 = 2 mins

n1 = 100

μ2= mean 2 = 11 mins

σ2 = Standard deviation 2 = 3 mins

n1 = 50

z score for 95% confidence interval = 1.96

μ1 -μ2 ± z × √ σ²1/n1 + σ²2/n2

= 12 - 11 ± 1.96 × √2²/100 + 3²/50

= 1 ± 1.96 × √4/100 + 9/50

= 1 ± 1.96 × √0.04 + 0.18

= 1 ± 1.96 × √0.22

= 1 ± 1.96 × 0.469041576

= 1 ± 0.9193214889

Confidence Interval

= 1 - 0.9193214889

= 0.0806785111

≈ 0.081

1 + 0.9193214889

= 1.9193214889

≈ 1.919

Therefore, the 95% Confidence Interval for the difference between the two population mean completion times =

(0.081, 1.919)