A new training program is designed to increase the productivity of workers at Am

Question

A new training program is designed to increase the productivity of workers at American Construction, Inc. To test the effectiveness of the training program, the manager took a random sample of hourly outputs of 49 workers. The sample results showed an average of 75 units per hour with a standard deviation of 5 units. The hourly output of workers without the new training was 73 units. What would be your conclusion at the 1% level of significance

GIVEN:   Find:    

Solution

step 1 Ho  step 2 Ha step 3 CV

step 4 draw the curve with acceptance and rejection regions properly annotated

step 5 compute the statistical test properly annotate on the curve

step 6 decision

step 7 conclusion

actions to be taken

step 8 compute the p value and check

p-value : 0.5000-

Explanation / Answer

The test hypothesis is
Ho:=73
Ha:>73

The test statistic is

Z=(xbar-)/(s/vn)

=(75-73)/(5/7)

= 2.8

Given a=0.01, the critical value is |Z(0.01)|=2.33 (check standard normal table)

Since 2.8 is larger than 2.33, we reject Ho.

So we can conclude that A new training program is designed to increase the productivity of workers at American Construction, Inc.

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