Question 662127: You are an HR consultant for REI, a large outdoor gear retailer. REI is considering multiple ‘family-oriented’ policy changes and would like to assess the need for daycare for their employees in the U.S.. Management tells you they estimate that 30% of employees need daycare; however they want you to determine the accuracy of this estimate. As a consultant, you identify a random sample of 400 employees and conduct a telephone interview. You find that 88 of the 400 employees in your sample indicate a need for daycare. (10 pts)
a. Calculate the 99% confidence interval for your parameter of interest based on your sample estimate (i.e., the observed statistic). (3 pts.).
a. p= 88/400= 0.22
se= √(p(1-p)/n)= √(0.22(1-0.22)/400) =0.0207
(99%) p±2.58 (se)= 0.22 ± (2.58)(0.0207)= (0.167, 0.273)
b. Conduct a one-sample hypothesis test for a proportion to determine the accuracy of management’s estimate (i.e. 30%). (3 pts.).
Ho: π=π0
Ha:π≠π0
se=√((π_0 (1-π_0 ))/n)=√((.3(1-.3))/400)=0.023
z=(π ̂-π_0)/〖se〗_0 = (0.22-0.3)/0.023= -3.48
p(z=-3.48)= 0.5-0.4998=0.00015
c. In five sentences or less, tell management your result and conclusion – make sure you explain your results/conclusion in a way that will be useful to management as they move forward with policy change decisions. (4 pts.).
With the p-value much lower than 0.05%, the likelihood that the sample would express the 30% estimate is very unlikely. Reject the null hypothesis. The limit of predictable need for daycare is 22%±5%. There is a .002% error that the general population would show results as high as 30%.
*I wasn't sure if I calculated the se in A correctly by using the right inputs. Is the analyze correctly stated?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Management tells you they estimate that 30% of employees need daycare; however they want you to determine the accuracy of this estimate. As a consultant, you identify a random sample of 400 employees and conduct a telephone interview. You find that 88 of the 400 employees in your sample indicate a need for daycare. (10 pts)
a. Calculate the 99% confidence interval for your parameter of interest based on your sample estimate (i.e., the observed statistic). (3 pts.).
a. p-hat = 88/400= 0.22
se= √(p-hat(1-p-hat)/n)= √(0.22(1-0.22)/400) =0.0207
(99%) p±2.58 (se)= 0.22 ± (2.58)(0.0207)= (0.167, 0.273)
Note: I have changed your notation on (se); but your calculation is correct.
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b. Conduct a one-sample hypothesis test for a proportion to determine the accuracy of management’s estimate (i.e. 30%). (3 pts.).
Ho: π =0.3 (management's claim)
Ha: π ≠ 0.3
se=√((π_0 (1-π_0 ))/n)=sqrt[pq/n) = √((.3(1-.3))/400)=0.023
z=(π ̂-π_0)/〖se〗_0 = (0.22-0.3)/0.023= -3.48
p(z =-3.48)= 0.5-0.4998=0.00015
Note: Your's is a 2-tail test.
The p-value = 2*P(z<-3.48) = 0.00052
c. In five sentences or less, tell management your result and conclusion – make sure you explain your results/conclusion in a way that will be useful to management as they move forward with policy change decisions. (4 pts.).
With the p-value much lower than 0.05%, the likelihood that the sample would express the 30% estimate is very unlikely. Reject the null hypothesis. The limit of predictable need for daycare is 22%±5%. There is a .002% error that the general population would show results as high as 30%.
*I wasn't sure if I calculated the se in A correctly by using the right inputs. Is the analysis correctly stated?
Your statement to management seems overly complicated.
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Cheers,
Stan H.
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