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In: Statistics and Probability

29. You conducted an independent samples t-test to determine whether there is a difference in productivity...

29. You conducted an independent samples t-test to determine whether there is a difference in productivity between employees who work 5-days per week versus 4-days per week. You find the average units produced for the 5-day work week is 26.6 and the average units produced for the 4-day work week is 24.4. Your standard error (of the differences) is 1.48. Using degrees of freedom of 18 and an alpha of .05, construct a 95% confidence interval. (1 point)

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Expert Solution

Given that, the average units produced for the 5-day work week is 26.6 and the average units produced for the 4-day work week is 24.4. That means, x1-bar = 26.6 and x2-bar = 24.4

The standard error (of the differences) = SE = 1.48

Degrees of freedom = 18 and significance level = 0.05

Using t-table we get, t-critical value at significance level of 0.05 with 18 degrees of freedom is,

The 95% confidence interval for (μ1 - μ2) is,

Therefore, required confidence interval is, (-0.9095, 5.3095).

orchestra answered 2 years ago
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