document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=662127>Question 662127</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 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) \r<BR>\n" );
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document.write( "a. Calculate the 99% confidence interval for your parameter of interest based on your sample estimate (i.e., the observed statistic). (3 pts.).<BR>\n" );
document.write( "a. p= 88/400= 0.22<BR>\n" );
document.write( "se= &#8730;(p(1-p)/n)= &#8730;(0.22(1-0.22)/400) =0.0207<BR>\n" );
document.write( "(99%) p±2.58 (se)= 0.22 ± (2.58)(0.0207)=   (0.167, 0.273)\r<BR>\n" );
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document.write( "b. Conduct a one-sample hypothesis test for a proportion to determine the accuracy of management’s estimate (i.e. 30%).  (3 pts.). <BR>\n" );
document.write( "Ho: &#960;=&#960;0<BR>\n" );
document.write( "Ha:&#960;&#8800;&#960;0<BR>\n" );
document.write( "se=&#8730;((&#960;_0 (1-&#960;_0 ))/n)=&#8730;((.3(1-.3))/400)=0.023<BR>\n" );
document.write( "z=(&#960; &#770;-&#960;_0)/&#12310;se&#12311;_0 =  (0.22-0.3)/0.023= -3.48<BR>\n" );
document.write( "p(z=-3.48)= 0.5-0.4998=0.00015\r<BR>\n" );
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document.write( "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.).\r<BR>\n" );
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document.write( "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%.\r<BR>\n" );
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document.write( "*I wasn't sure if I calculated the se in A correctly by using the right inputs. Is the analyze correctly stated? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #412071 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=412071\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 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) \r<BR>\n" );
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document.write( "a. Calculate the 99% confidence interval for your parameter of interest based on your sample estimate (i.e., the observed statistic). (3 pts.).<BR>\n" );
document.write( "a. p-hat = 88/400= 0.22<BR>\n" );
document.write( "se= &#8730;(p-hat(1-p-hat)/n)= &#8730;(0.22(1-0.22)/400) =0.0207<BR>\n" );
document.write( "(99%) p±2.58 (se)= 0.22 ± (2.58)(0.0207)= (0.167, 0.273)<BR>\n" );
document.write( "Note: I have changed your notation on (se); but your calculation is correct.<BR>\n" );
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document.write( "b. Conduct a one-sample hypothesis test for a proportion to determine the accuracy of management’s estimate (i.e. 30%). (3 pts.). <BR>\n" );
document.write( "Ho: &#960; =0.3 (management's claim)<BR>\n" );
document.write( "Ha: &#960; &#8800; 0.3<BR>\n" );
document.write( "se=&#8730;((&#960;_0 (1-&#960;_0 ))/n)=sqrt[pq/n) = &#8730;((.3(1-.3))/400)=0.023<BR>\n" );
document.write( "z=(&#960; &#770;-&#960;_0)/&#12310;se&#12311;_0 = (0.22-0.3)/0.023= -3.48<BR>\n" );
document.write( "p(z =-3.48)= 0.5-0.4998=0.00015<BR>\n" );
document.write( "Note: Your's is a 2-tail test.  <BR>\n" );
document.write( "The p-value = 2*P(z<-3.48) = 0.00052 \r<BR>\n" );
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document.write( "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.). <BR>\n" );
document.write( "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%. \r<BR>\n" );
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document.write( "*I wasn't sure if I calculated the se in A correctly by using the right inputs. Is the analysis correctly stated?<BR>\n" );
document.write( "Your statement to management seems overly complicated.<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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