A new operator was recently assigned to a crew of workers who perform a certain

Question

A new operator was recently assigned to a crew of workers who perform a certain job. From the records of the number of units of work completed by each worker each day last month, a sample of size five was randomly selected for each of the two experienced workers and the new worker. At the = .05 level of significance, does the evidence provide sufficient reason to reject the claim that there is no difference in the amount of work done by the three workers?

(a) Find the test statistic. (Give your answer correct to two decimal places.)


(ii) Find the p-value. (Give your answer bounds exactly.)
_______< p < _______

Workers New A B Units of work (replicates) 10 10 9 9 13 13 10 10 11 8 11 9 8 12 11

Explanation / Answer

Here there are three groups.

For more than two groups we use ANOVA.

We have to test the hypothesis that,

H0 : mu1 = mu2 = mu3 Vs H1 : Atleast one of the mean is differ than 0.

where mu1, mu2 and mu3 are three population means.

Assume alpha = 5% = 0.05

ANOVA we can done using EXCEL.

ENTER the data in EXCEL sheet --> Data --> Data Analysis --> Anova : Single factor --> ok --> Input Range : select all the data --> Grouped by : columns --> Alpha : 0.05 --> Output range : select one empty cell --> ok

The test statistic F = 3.53

P-value = 0.06

P-value > alpha

Accept H0 at 5% level of significance.

Conclusion : There is sufficient evidence to say that there is no difference in the amount of work done by the three workers.

Anova: Single Factor SUMMARY Groups Count Sum Average Variance Column 1 5 45 9 1 Column 2 5 56 11.2 1.7 Column 3 5 53 10.6 2.8 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 12.93333 2 6.466667 3.527273 0.062387 3.885294 Within Groups 22 12 1.833333 Total 34.93333 14

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.