1. You are an office manager for a company that proofreads patent applications,
ID: 3070329 • Letter: 1
Question
1. You are an office manager for a company that proofreads patent applications, and you would like to determine a way to increase your workers' productivity. You have heard that certain types of music can increase productivity in an office, and decide to investigate. Over the next couple of weeks, you collect data on how many patents are proofread in an hour when there is no music playing, when classical music is playing, when jazz is playing, and when Top 40 music is playing. The below are the results No Music Cassical Jazz Mean Standard Deviaton Sar 3.489 4.626 4.984 Sive 167 154 156 163 1.347 1.169 Table 1. Top40 3.247 1.438 Use the information in Table 1 to answer the following questions A) Calculate a 95% Confidence Interval for the difference between the average number of patents proofread when classical music is playing and when no music is playing. B) Calculate a 95% Confidence Interval for the difference between the average number of patents proofread when jazz is playing and when no music is playing. C) Calculate a 95% Confidence Interval for the difference between the average number of patents proofread when Top 40 music is playing and when no music is playing. D) Test the hypothesis that there is no difference in the number of patents proofread when no music is playing versus when classical, jazz, or Top 40 is playing. (i.e you will have three tests) E) Based on your results, would you decide to play music in the office? If so, which type(s)?Explanation / Answer
A.
Assume equal variances
Sample N Mean StDev SE Mean
1 154 4.63 1.17 0.094
2 167 3.49 1.35 0.10
Difference = mu (1) - mu (2)
Estimate for difference: 1.137
95% CI for difference: (0.859, 1.415)
Both use Pooled StDev = 1.2648
When variances are not equal
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 154 4.63 1.17 0.094
2 167 3.49 1.35 0.10
Difference = mu (1) - mu (2)
Estimate for difference: 1.137
95% CI for difference: (0.861, 1.413)
B.
Assume equal variance
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 156 4.98 1.28 0.10
2 167 3.49 1.35 0.10
Difference = mu (1) - mu (2)
Estimate for difference: 1.495
95% CI for difference: (1.207, 1.783)
Both use Pooled StDev = 1.3170
When variances are not equal
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 156 4.98 1.28 0.10
2 167 3.49 1.35 0.10
Difference = mu (1) - mu (2)
Estimate for difference: 1.495
95% CI for difference: (1.207, 1.783)
C.
Assume equal variance
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 163 3.25 1.44 0.11
2 167 3.49 1.35 0.10
Difference = mu (1) - mu (2)
Estimate for difference: -0.242
95% CI for difference: (-0.544, 0.060)
Both use Pooled StDev = 1.3927
When variances are unequal
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 163 3.25 1.44 0.11
2 167 3.49 1.35 0.10
Difference = mu (1) - mu (2)
Estimate for difference: -0.242
95% CI for difference: (-0.544, 0.060)
D.
Assume equal variances
Sample N Mean StDev SE Mean
1 154 4.63 1.17 0.094
2 167 3.49 1.35 0.10
Difference = mu (1) - mu (2)
Estimate for difference: 1.137
T-Test of difference = 0 (vs not =): T-Value = 8.05 P-Value = 0.000 DF = 319
Both use Pooled StDev = 1.2648
When variances are not equal
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 154 4.63 1.17 0.094
2 167 3.49 1.35 0.10
Difference = mu (1) - mu (2)
Estimate for difference: 1.137
T-Test of difference = 0 (vs not =): T-Value = 8.09 P-Value = 0.000 DF = 317
Since p-value<0.05 hence there is significant difference in the proofread when no music is playing versus when classical music is playing.
Assume equal variances
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 156 4.98 1.28 0.10
2 167 3.49 1.35 0.10
Difference = mu (1) - mu (2)
Estimate for difference: 1.495
T-Test of difference = 0 (vs not =): T-Value = 10.20 P-Value = 0.000 DF = 321
Both use Pooled StDev = 1.3170
When variances are not equal
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 156 4.98 1.28 0.10
2 167 3.49 1.35 0.10
Difference = mu (1) - mu (2)
Estimate for difference: 1.495
T-Test of difference = 0 (vs not =): T-Value = 10.21 P-Value = 0.000 DF = 320
Since p-value<0.05 hence there is significant difference in the proofread when no music is playing versus when jazz is playing.
Assume equal variance
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 163 3.25 1.44 0.11
2 167 3.49 1.35 0.10
Difference = mu (1) - mu (2)
Estimate for difference: -0.242
T-Test of difference = 0 (vs not =): T-Value = -1.58 P-Value = 0.115 DF = 328
Both use Pooled StDev = 1.3927
When variances are unequal
Two-Sample T-Test and CI
Sample N Mean StDev SE Mean
1 163 3.25 1.44 0.11
2 167 3.49 1.35 0.10
Difference = mu (1) - mu (2)
Estimate for difference: -0.242
T-Test of difference = 0 (vs not =): T-Value = -1.58 P-Value = 0.116 DF = 325
Since p-value>0.05, hence there is insignificant difference in the proofread when no music is playing versus when Top 40 is playing.
E. Since 95% confidence intervals of Problems A and B contain all positive values and the p-value<0.05 for the differences mentioned in Problems A and B so we decide to play music of the following types:
Classical and Jazz.
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