The results shown below are mean productivity measurements (average number of assemblies completed per hour) for a random sample of workers at each of three plants.Research question:Are the mean hourly productivity levels the same for workers in these three plants?

Hourly Productivity of Assemblers in PlantsPlantFinished Units Produced per HourA(9 workers)3.65.12.84.64.74.13.42.94.5B(6 workers)2.73.15.01.92.23.2C(10 workers)6.82.55.46.74.63.95.44.97.18.4

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Fill in the missing data.(Round yourp-value to 4 decimal places, mean values to 2 decimal places, and other answers to 3 decimal places.)

GroupMeannStd. DevVariancePlant APlant BPlant CTotal

One-Factor ANOVASourceSSdfMSFp-valueTreatmentErrorTotal

Based on the hypotheses given below, choose the correct option.

H₀:₁=₂=₃

H₁: Not all the means are equal

= .05

multiple choice 1

• Reject the null hypothesis ifp-value > 0.05
• Reject the null hypothesis ifp-value < 0.05

We reject the null hypotheses.

Yes

On the basis of the above findings, we conclude that

all the means are above the overall mean.

Calculate the mean for each group and the Tukey test statisticT_calcfor each pair. Provide the critical values for both= .05 and= .01.(Input the mean values within the input boxes of the first row and input boxes of the first column. InputT_calcin the appropriate boxes in the table. Round all answers to two decimal places.)

Tukey simultaneous comparisont-values (d.f.= 22)Plant BPlant APlant CPlant BPlant APlant CCritical values for experimentwise error rate:0.050.01

Based on the hypothesis given below, choose the correct option.

H₀:²₁=²₂=²₃

H₁: Not all the variances are equal

= .05

multiple choice 4

• Reject the null hypothesis ifH< 6.94
• Reject the null hypothesis ifH> 6.94

Calculate the value ofH_calc.(Round your answer to 2 decimal places.)

H_calc

On the basis of the above-determinedH_calc, we reject the null hypothesis.

False