document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1086358>Question 1086358</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Samsung wants to know how long cell phone owners keep their phones before upgrading. A simple random sample of 23 cell phone owners results in a mean of 2.64 years and a standard deviation of 0.71 years. Assume the sample is drawn from a normally distributed population. <BR>\n" );
document.write( "Find the 95% confidence interval of the population mean. \r<BR>\n" );
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document.write( "If you worked for Samsung and decided you wanted to be 99% confident that the sample mean is within 0.25 years of the population mean, how large of a sample would you need to take? Assume that &#963;=0.71 for this calculation. \r<BR>\n" );
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document.write( "Explain why the population parameter may NOT follow a normal distribution. Would you expect the data to show a positive or negative skew? Explain. If the data were not normally distributed, how would this affect the calculations for the confidence interval? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #700549 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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/cgi-bin/show-face.mpl?user=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=700549\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The 95% CI is mean +/-t *s/sqrt(n)<BR>\n" );
document.write( "t df=23, 0.975=2.07<BR>\n" );
document.write( "2.64 +/- 2.07*0.71/4.80; the interval width is 0.31<BR>\n" );
document.write( "(2.33, 2.95) units years<BR>\n" );
document.write( "within 0.25 years<BR>\n" );
document.write( "0.25=z*0.71/ sqrt (n);use z until there is an estimate of sample size.  z(0.995)=2.576<BR>\n" );
document.write( "0.25=1.83/sqrt (n)<BR>\n" );
document.write( "cross multiply and square both sides<BR>\n" );
document.write( "n=(1.83/0.25)^2=53.6<BR>\n" );
document.write( "Need to use t,start with df=54<BR>\n" );
document.write( "2.67*0.71/sqrt(54)=0.2579<BR>\n" );
document.write( "try df=60, and interval is+/-0.25<BR>\n" );
document.write( "NOTE: I'm using a t.<BR>\n" );
document.write( "If sigma is known,then the sample size is 54 as shown above.<BR>\n" );
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document.write( "I wold expect a positive skew, since people will continue to have their phones at 3,4, 5 years.  If the data are not normally distributed, the confidence interval will be wider for a given sample size, but more importantly, another test would be needed, perhaps a non-parametric or distribution-free test. </SPAN>\n" );
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