document.write( "Question 1157127:  please help me \r
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document.write( "A physician measured pulse rates of 26 college men. Based on the sample, he found that a mean pulse rate is 70 pulse beats per minute with a standard deviation of 9.85 beats per minute. Is the mean pulse rate for all college men is different from the current standard of 72 beats per minute, given that the P.value = 0.132?
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document.write( "-	Specify your null and alternative hypotheses.
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document.write( "-	Based on P.value approach, Write your decision and conclusion at 95% confidence level.
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Algebra.Com's Answer #779939 by Boreal(15235)    You can put this solution on YOUR website! Ho: pulse is 72 \n" ); document.write( "Ha: it isn't. \n" ); document.write( "alpha=0.05 P{reject Ho|Ho true} \n" ); document.write( "test stat is a tdf=25, 0.975)=(x bar- mean)/sigma/sqrt (n) \n" ); document.write( "reject if |t|>2.06 \n" ); document.write( "calculation t=(70-72)/9.85/sqrt(26) \n" ); document.write( "=-2*sqrt(26)/9.85 \n" ); document.write( "=-1.035 \n" ); document.write( "I get 1 sided p value of 0.15, or 2 -sided of 0.30\r \n" ); document.write( "\n" ); document.write( "I don't see where the p-value of 0.132 comes from with these data. In that instance and what is shown here, one fails to reject the null hypothesis and says there is insufficient evidence to conclude the pulse rate is different. We could have obtained a result by chance this great or greater with 30% probability.\r \n" ); document.write( "\n" ); document.write( "The 95% CI is t0.975 df=25*s/sqrt(n) for a half interval. That is 3.98 \n" ); document.write( "This is added or subtracted from the mean\r \n" ); document.write( "\n" ); document.write( "(66.02, 73.98) is the interval, and because 72 is in the interval, we can conclude with 95% confidence that the true pulse of the sample of men, which will never be completely known, is not different from 72. \n" ); document.write( " \n" ); document.write( "  |