document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1193649>Question 1193649</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Matt thinks that he has a special relationship with the number 2. In particular, Matt thinks that he would roll a 2 with a fair 6-sided die more often than you'd expect by chance alone. Suppose p is the true proportion of the time Matt will roll a 2.\r<BR>\n" );
document.write( "\n" );
document.write( "Now suppose Matt makes n = 50 rolls, and a 2 comes up 11 times out of the 50 rolls. Determine the P-value of the test: </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #825687 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3835)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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=825687\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "p = population proportion of the number of times '2' shows up\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "If we assume each of the six sides are equally likely, then p should be p = 1/6 since one side is labeled '2' out of 6 sides total.<BR>\n" );
document.write( "The one-sample proportion test aims to check that claim. \r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Hypotheses:<BR>\n" );
document.write( "H0: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=p+=+1%2F6 ALIGN=MIDDLE  ALT=\"p+=+1%2F6\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "H1: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=p+%3E+1%2F6 ALIGN=MIDDLE  ALT=\"p+%3E+1%2F6\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "The claim is in the alternative hypothesis (H1) because Matt thinks he would roll a '2' more often than expected.<BR>\n" );
document.write( "The inequality sign in the alternative hypothesis determines that we have a right-tailed test.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "x = number of times a '2' shows up = 11<BR>\n" );
document.write( "n = number of rolls = 50\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "phat = sample proportion that estimates p<BR>\n" );
document.write( "phat = x/n<BR>\n" );
document.write( "phat = 11/50<BR>\n" );
document.write( "phat = 0.22\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "SE = standard error<BR>\n" );
document.write( "SE = sqrt(phat*(1-phat)/n)<BR>\n" );
document.write( "SE = sqrt(0.22*(1-0.22)/50)<BR>\n" );
document.write( "SE = 0.058583 which is approximate\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "z = test statistic<BR>\n" );
document.write( "z = (phat - p)/SE<BR>\n" );
document.write( "z = (0.22 - 1/6)/0.058583<BR>\n" );
document.write( "z = 0.91038924830298<BR>\n" );
document.write( "z = 0.910389 which is also approximate\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Now use a calculator such as this one<BR>\n" );
document.write( "<a  rel=nofollow HREF= \"https://onlinestatbook.com/2/calculators/normal_dist.html\">https://onlinestatbook.com/2/calculators/normal_dist.html</a><BR>\n" );
document.write( "to find that P(Z > 0.910389) = 0.1813 approximately<BR>\n" );
document.write( "Recall that we're doing a right-tailed test, so this determines which portion we're shading under the curve. Specifically, we're shading to the right of z = 0.910389 and that area is roughly 0.1813\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "At the common alpha values of things like alpha = 0.05 and alpha = 0.10, the p-value of 0.1813 means we would fail to reject the null. We would conclude that p = 1/6 is indeed the case unless stronger evidence comes along to have us overturn the null. You only reject the null if the p-value is smaller than alpha.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<font color=red>Answer: The p-value is approximately 0.1813</font><BR>\n" );
document.write( "</font> </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );