Question

Progressive Products, Inc. wants to know if its six-month effort to reduce the absenteeism behavior of their employees has produced beneficial results. The campaign resulted from an internal study conducted by the Personnel Department that found an average of 5.6 sick days taken annually per employee in the company's main manufacturing plant. A random sample of the number of sick days taken last year (from the records of 12 Progressive employees) is provided below.

7, 2, 5, 14, 6, 3, 4, 3, 4, 7, 2, 9

Based on extensive historical data compiled by Progressive Products, the population standard deviation is known to be 3.45 days. Does this data provide sufficient evidence to support the contention that the campaign has significantly reduced the average number of sick days taken annually at Progressive Products? Use α = .03 to conduct a complete and appropriate hypothesis test.

Step 1 (Hypotheses)

H₀:

H_A:

Step 2 (Decision rule) Using only the appropriate statistical table in your textbook, the critical value for rejecting H₀ is ____. (report your answer to 2 decimal places, using conventional rounding rules)

Step 3 (Test statistic) Using the sample data, the calculated value of the test statistic is ____. (report your answer to 2 decimal places, using conventional rounding rules)

Step 4 (Evaluate the null hypothesis) Should the null hypothesis be rejected? _____

Step 5 (Practical conclusion) Can Progressive Products conclude that the campaign has significantly reduced the average number of sick days taken annually? ____

Using only the appropriate statistical table in your textbook, what is the p-value of this hypothesis test?  

Answer: _____ (Report your answer to 4 decimal places, using conventional rounding rules)

Answer 1

Ho :   µ =   5.6  
Ha :   µ <   5.6   (Left tail test)

critical z value, z* =       -1.88 [Excel formula =NORMSINV(α/no. of tails) ]
          
          
3)Level of Significance ,    α =    0.030  
population std dev ,    σ =    3.4500  
Sample Size ,   n =    12  
Sample Mean,    x̅ = ΣX/n =    5.5000  
          
'   '   '  
          
Standard Error , SE = σ/√n =   3.45/√12=   0.9959  
Z-test statistic= (x̅ - µ )/SE =    (5.5-5.6)/0.9959=   -0.100  
          

4) Decision: Do not reject null hypothesis       

5) No we cannot conclude

p-Value   =   0.4600   [ Excel formula =NORMSDIST(z) ]