document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=80062>Question 80062</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Question 1\r<BR>\n" );
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document.write( "A fast food restaurant claims that the mean time is not equal to 3 minutes. A random sample of 45 customers revealed a mean waiting time of 3.3 minutes with a standard deviation of 0.7 minutes. Test the restaurants claim at a, a= 0.05 level of significance. Assume that the population is normally distributed, and the population standard deviation is 0.7. Use the P-value method. \r<BR>\n" );
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document.write( "Question 2<BR>\n" );
document.write( "A researcher at NYU claims that for Mathematics courses the proportion of grades  B or higher was more than 33.0%. In a random sample of 100 grades issued in the year 2006 there were 27 F's, 15 D's, 22 C's, 14 B's, 22 A's. At a=0.10 test the researcher's claim.\r<BR>\n" );
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document.write( "Question 3\r<BR>\n" );
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document.write( "A sociologist wants to determine if the number of individuals per family unit the national average of 2.9. She interviews 31 randomly selected families in a large apartment complex and obtains a mean of 3.1 individuals per family unit with a standard deviation of 0.80. Test the claim that the mean number of individuals per family unit is less than 2.9. Use a significance level of 0.05. Assume that the population has a normal distribution and that the population standard deviation is unknown. Decide to use the t-test or z-test. If you use the z-test use the P-value method. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #57475 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=57475\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A fast food restaurant claims that the mean time is not equal to 3 minutes. A random sample of 45 customers revealed a mean waiting time of 3.3 minutes with a standard deviation of 0.7 minutes. Test the restaurants claim at a, a= 0.05 level of significance. Assume that the population is normally distributed, and the population standard deviation is 0.7. Use the P-value method.<BR>\n" );
document.write( "Ho: mu=3 min<BR>\n" );
document.write( "Ha: mu is not equal to 3 min<BR>\n" );
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document.write( "Critical value : z=+-1.96<BR>\n" );
document.write( "z(3.3)=(3.3-3)/[0.7/sqrt45]<BR>\n" );
document.write( "=0.3/[0.1043]<BR>\n" );
document.write( "=2.87<BR>\n" );
document.write( "p-value=2*0.002=0.004<BR>\n" );
document.write( "Since p<alpha=0.05, Reject Ho.<BR>\n" );
document.write( "Conclusion mu is not equal to 3 min.<BR>\n" );
document.write( "============= <BR>\n" );
document.write( "Question 2<BR>\n" );
document.write( "A researcher at NYU claims that for Mathematics courses the proportion of grades B or higher was more than 33.0%. In a random sample of 100 grades issued in the year 2006 there were 27 F's, 15 D's, 22 C's, 14 B's, 22 A's. At a=0.10 test the researcher's claim. <BR>\n" );
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document.write( "Ho: p=0.33<BR>\n" );
document.write( "Ha: p>0.33<BR>\n" );
document.write( "Critical value for alpha=10%=1.28<BR>\n" );
document.write( "p-hat=(36/100)<BR>\n" );
document.write( "z(0.36)=(0.36-0.33)/sqrt[(0.33*0.67)/100]<BR>\n" );
document.write( "=0.03/0.047<BR>\n" );
document.write( "=0.638<BR>\n" );
document.write( "p-value of 0.638=0.26<BR>\n" );
document.write( "Since p>alpha, Fail to reject Ho<BR>\n" );
document.write( "Conclusion: p=0.33%<BR>\n" );
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document.write( "Question 3 <BR>\n" );
document.write( "A sociologist wants to determine if the number of individuals per family unit the national average of 2.9. She interviews 31 randomly selected families in a large apartment complex and obtains a mean of 3.1 individuals per family unit with a standard deviation of 0.80. Test the claim that the mean number of individuals per family unit is less than 2.9. Use a significance level of 0.05. Assume that the population has a normal distribution and that the population standard deviation is unknown. Decide to use the t-test or z-test. If you use the z-test use the P-value method.<BR>\n" );
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document.write( "Ho: mu=2.9<BR>\n" );
document.write( "Ha: mu < 2.9<BR>\n" );
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document.write( "z(3.1)=(3.1-2.9)/[0.8/sqrt31]=0.0359....<BR>\n" );
document.write( "p-value of 0.0359=0.014<BR>\n" );
document.write( "Since p<alpha=0.05, Reject Ho.<BR>\n" );
document.write( "Conclusion<BR>\n" );
document.write( "mu < 2.9<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.\r<BR>\n" );
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document.write( "0 solutions<BR>\n" );
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