document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=888626>Question 888626</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I am having issues finding the answer to this statistics problem. Please help me work this out! Thank you!\r<BR>\n" );
document.write( "\n" );
document.write( "According to the College Research Center, the proportion of college students who use only a cellular telephone (no land line) is 76%. Considering this, your instructor surveyed his students to see how they compared. <BR>\n" );
document.write( "From 359 college students surveyed, he found that 262 of them only used a cellular telephone. Consequently, your instructor claims that his college students are different from other college students. \r<BR>\n" );
document.write( "\n" );
document.write( "Test your instructor's claim at a level of significance of 10% using the P-Value Method: \r<BR>\n" );
document.write( "\n" );
document.write( "The following parts break down the four steps of the P-Value Method, which you will use for the hypothesis test in this problem: <BR>\n" );
document.write( "A) State the Hypotheses <BR>\n" );
document.write( "B) Determine the Test Statistic. <BR>\n" );
document.write( "C) Determine the P-Value. <BR>\n" );
document.write( "D) State the conclusion. It should be in terms of the problem (give me more than just Reject Ho or Do Not Reject Ho). \r<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #537582 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=537582\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> note that 262 / 359 = 0.729805014 or 73%<BR>\n" );
document.write( "now 0.76 of 359 = 272.84 or 273 (expected value) vs 262 (observed value)<BR>\n" );
document.write( "now 0.24 0f 359 =  86.16 or  86 (expected value vs 97 (observed value)<BR>\n" );
document.write( "A) The proportion of college students who use only a cellular telephone (no land line) is 76%.<BR>\n" );
document.write( "B) Level of significance is 10% or alpha = 0.10<BR>\n" );
document.write( "C) Degree of freedom is 2 -1 = 1<BR>\n" );
document.write( "x = ((262 - 273)^2 / 273)  + ((86 - 97)^2 / 97)<BR>\n" );
document.write( "x = (-11)^2 / 273  + (-11)^2 / 97 =  0.443223443 + 1.24742268 = 1.690646123 = 1.69<BR>\n" );
document.write( "our chi-square value is 1.69, now we consult the chi-square table to determine our p-value, using our degree of freedom of 1, we scan that row from left to right looking for a chi-square value greater than 1.69 and we find that value of 2.07 with a p-value of 15%<BR>\n" );
document.write( "thus our p-value lies between 15% - 20%<BR>\n" );
document.write( "D) This means  there's a 15-20% chance that the results we observed weren't the result of a change in location (instructor's students, as opposed to the entire nation), but instead just happened by chance. Since we were looking for a chance of less than 10%, we can't say that we're sure our instructor's students are less biased towards using cell phones only - there's a small but statistically significant chance they aren't.\r<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );