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2. The Postal Service is attempting to reduce the number of complaints made by the public against its workers. In order to facilitate this task, a staff analyst for the Service regresses the number of complaints lodged against an employee last year (Y) on the hourly wage of the employee for the year (X). She obtains the following results:

Y =10.2 - 1.9X
r2 = .73 sb=.87 n=348 t= slope/ sb

Based on these results, answer the following questions:

a) If wages increased by $1.00, what is the expected effect on the number of complaints received per employee?
b) What percent of the variation in complaints can be explained by wages?
c) At what level of wages can the employee be expected to receive zero (0) complaints? (hint: recalculate the equation with Y=0)
d) Calculate the t-value using the formula above.
e) Can the null hypothesis be rejected at the .05 level of statistical significance that wages have no effect on the number of complaints (Recall the critical values for t-test for a two-tail test is 1.96 and for a one-tail test 1.65)?
f) Discuss the policy implications of the regression analysis.

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Variation complaints for levels of wages are given. The policy implications of the regression analysis is examined.

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2. The Postal Service is attempting to reduce the number of complaints made by the public against its workers. In order to facilitate this task, a staff analyst for the Service regresses the number of complaints lodged against an employee last year (Y) on the hourly wage of the employee for the year (X). She obtains the following results:

Y =10.2 - 1.9X
r2 = .73 sb=.87 n=348 t= slope/ sb

Based on these results, answer the following questions:

a) If ...

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  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
Recent Feedback
  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
  • "excellent work"
  • "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
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