Question

An ergonomics consultant is engaged by a consumer products company to see what they can do to increase productivity. The consultant recommends an “employee athlete” program, encouraging every employee to devote five minutes an hour to physical activity. The company decides to try this program. To measure the difference in productivity, they measure the average number of keystrokes per hour of 23 employees before and after the five-minutes-per-hour program is instituted. The data follow: Before After Difference (After – Before) Mean 1497.3 1544.8 47.5 SD 155.4 136.7 122.8 n 23 23 23 Table: Keystrokes per Hour Is there evidence to suggest the program increases productivity? Use a 5% level of significance to test it.

Answer 1

Solution:

Here, we have to use paired t test for difference between two population means.

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H₀: The average number of keystrokes before and after the employee athlete program is same.

Alternative hypothesis: H_a: The average number of keystrokes increases after the employee athlete program.

Let µ_d = Population mean difference of number of keystrokes after minus before athlete program.

H₀: µ_d = 0 versus H_a: µ_d > 0

This is an upper/right tailed (One tailed) test.

The test statistic formula is given as below:

t = (Dbar - µ_d)/[S_d/sqrt(n)]

From given data, we have

Dbar = 47.5

S_d = 122.8

n = 23

df = n – 1 = 22

α = 0.05

Upper critical value = 1.7171

(by using t-table)

t = (Dbar - µ_d)/[S_d/sqrt(n)]

t = (47.5 - 0)/[122.8/sqrt(23)]

t = 1.8551

P-value = 0.0385

(by using t-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that the average number of keystrokes increases after the employee athlete program.