document.write( "Question 899413: A consumer advocacy group suspects that a local supermarket's 500 grams of sugar actually weigh less than 500 grams. The group took a random sample of 20 such packages, weighed each one and found the mean weight for the sample to be 496 grams with a standard deviation of 8 grams. Using 10% significance level, would you conclude that the mean weight is less than 500 grams? Also, find the 90% confidence interval of the true mean. \r
\n" ); document.write( "\n" ); document.write( "what is the hypothesis?
\n" ); document.write( "H0:
\n" ); document.write( "H1:
\n" ); document.write( "The level of significance is a= ___, df= ___, and the critical region for the value of one sample t-test
\n" ); document.write( "decision rule:
\n" ); document.write( "conclusion:
\n" ); document.write( "compute for the confidence interval:
\n" ); document.write( "
\n" ); document.write( "

Algebra.Com's Answer #545383 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
 
\n" ); document.write( "Hi
\n" ); document.write( "Ho: u ≥ 500
\n" ); document.write( "Ha: u < 500
\n" ); document.write( "------------------
\n" ); document.write( "t(496) = (496-500)/[8/sqrt(20)] = 4/1.7889 = 2.236
\n" ); document.write( "--------
\n" ); document.write( "p-value = P(t> 2.236 when df = 19) = tcdf(2.236,100,19) = .9825
\n" ); document.write( "As .9825 > .10, No, would not conclude mean is <500
\n" ); document.write( "ME = 1.645(8/sqrt(25))
\n" ); document.write( "CI: = 496 - ME < \"mu\" < 496 + ME
\n" ); document.write( " \n" ); document.write( "
\n" );