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 Plants Plant Finished Units Produced per Hour A (9 workers) 3.6 5.1 2.8 4.6 4.7 4.1 3.4 2.9 4.5 B (6 workers) 2.7 3.1 5.0 1.9 2.2 3.2 C (10 workers) 6.8 2.5 5.4 6.7 4.6 3.9 5.4 4.9 7.1 8.4 Click here for the Excel Data File Fill in the missing data. (Round your p-value to 4 decimal places, mean values to 2 decimal places, and other answers to 3 decimal places.) Group Mean n Std. Dev Variance Plant A Plant B Plant C Total One-Factor ANOVA Source SS df MS F p-value Treatment Error Total Based on the hypotheses given below, choose the correct option. H0: μ1 = μ2 = μ3 H1: Not all the means are equal α = .05 Reject the null hypothesis if p-value < 0.05 Reject the null hypothesis if p-value > 0.05 We reject the null hypotheses. No Yes On the basis of the above findings, we conclude that one of the means is above and two are below the overall mean. all the means are above the overall mean. only one mean is above the overall mean. all the means are below the overall mean. Calculate the mean for each group and the Tukey test statistic Tcalc for each pair. Provide the critical values for both α = .05 and α = .01.(Input the mean values within the input boxes of the first row and input boxes of the first column. Input Tcalc in the appropriate boxes in the table. Round all answers to two decimal places.) Tukey simultaneous comparison t-values (d.f. = 22) Plant B Plant A Plant C Plant B Plant A Plant C Critical values for experimentwise error rate: 0.05 0.01 Based on the hypothesis given below, choose the correct option. H0: σ21 = σ22 = σ23 H1: Not all the variances are equal α = .05 Reject the null hypothesis if H < 6.94 Reject the null hypothesis if H > 6.94 Calculate the value of Hcalc. (Round your answer to 2 decimal places.) Hcalc On the basis of the above-determined Hcalc, we reject the null hypothesis. True False
Inverse Normal Distribution
The method used for finding the corresponding z-critical value in a normal distribution using the known probability is said to be an inverse normal distribution. The inverse normal distribution is a continuous probability distribution with a family of two parameters.
Mean, Median, Mode
It is a descriptive summary of a data set. It can be defined by using some of the measures. The central tendencies do not provide information regarding individual data from the dataset. However, they give a summary of the data set. The central tendency or measure of central tendency is a central or typical value for a probability distribution.
Z-Scores
A z-score is a unit of measurement used in statistics to describe the position of a raw score in terms of its distance from the mean, measured with reference to standard deviation from the mean. Z-scores are useful in statistics because they allow comparison between two scores that belong to different normal distributions.
The results shown below are
| Hourly Productivity of Assemblers in Plants | ||||||||||
| Plant | Finished Units Produced per Hour | |||||||||
| A (9 workers) | 3.6 | 5.1 | 2.8 | 4.6 | 4.7 | 4.1 | 3.4 | 2.9 | 4.5 | |
| B (6 workers) | 2.7 | 3.1 | 5.0 | 1.9 | 2.2 | 3.2 | ||||
| C (10 workers) | 6.8 | 2.5 | 5.4 | 6.7 | 4.6 | 3.9 | 5.4 | 4.9 | 7.1 | 8.4 |
Click here for the Excel Data File
Fill in the missing data. (Round your p-value to 4 decimal places, mean values to 2 decimal places, and other answers to 3 decimal places.)
| Group | Mean | n | Std. Dev | Variance |
| Plant A | ||||
| Plant B | ||||
| Plant C | ||||
| Total | ||||
| One-Factor ANOVA | |||||
| Source | SS | df | MS | F | p-value |
| Treatment | |||||
| Error | |||||
| Total | |||||
Based on the hypotheses given below, choose the correct option.
H0: μ1 = μ2 = μ3
H1: Not all the means are equal
α = .05
-
Reject the null hypothesis if p-value < 0.05
-
Reject the null hypothesis if p-value > 0.05
We reject the null hypotheses.
-
No
-
Yes
On the basis of the above findings, we conclude that
-
one of the means is above and two are below the overall mean.
-
all the means are above the overall mean.
-
only one mean is above the overall mean.
-
all the means are below the overall mean.
Calculate the mean for each group and the Tukey test statistic Tcalc for each pair. Provide the critical values for both α = .05 and α = .01.(Input the mean values within the input boxes of the first row and input boxes of the first column. Input Tcalc in the appropriate boxes in the table. Round all answers to two decimal places.)
| Tukey simultaneous comparison t-values (d.f. = 22) | ||||
| Plant B | Plant A | Plant C | ||
| Plant B | ||||
| Plant A | ||||
| Plant C | ||||
| Critical values for experimentwise error rate: | ||||
| 0.05 | ||||
| 0.01 | ||||
Based on the hypothesis given below, choose the correct option.
H0: σ21 = σ22 = σ23
H1: Not all the variances are equal
α = .05
-
Reject the null hypothesis if H < 6.94
-
Reject the null hypothesis if H > 6.94
Calculate the value of Hcalc. (Round your answer to 2 decimal places.)
Hcalc
On the basis of the above-determined Hcalc, we reject the null hypothesis.
-
True
-
False
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 1 images
