document.write( "Question 1030980: The average attrition rate of a customer based company is 20% with a standard deviation of 7% (based on the past data). To test this a sample of 100 was taken and 23% average attrition rate was found. What can be concluded from about the attrition rate at 10% significance level?
\n" ); document.write( " The attrition rate is 23% now
\n" ); document.write( " The attrition rate is more than 20% now
\n" ); document.write( " The attrition rate is same as 20%
\n" ); document.write( " The attrition rate is less than 20% now
\n" ); document.write( "I tried this and got the below answer.Can you please check if this is correct.
\n" ); document.write( "H0-No diff in attrition rate
\n" ); document.write( "H1:Increase in attrinium rate
\n" ); document.write( "x>=23=1-(x<=23)
\n" ); document.write( "==1-NORM.DIST(23,20,7/(100)^0.5,TRUE)
\n" ); document.write( "=9.10765E-06 <0.1
\n" ); document.write( " reject null hypothesis
\n" ); document.write( "Is option (a) the correct answer that the attrition rate is 23% now.\r
\n" ); document.write( "\n" ); document.write( "Thanks
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Algebra.Com's Answer #645734 by ewatrrr(24785) $\"\"$ $\"About$

You can put this solution on YOUR website!


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document.write( "Hi,
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document.write( "re: Your reply (Idea being..with a 7% SD ... the 3% difference in the test sample
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document.write( "would be expected to be consistent with the average of 20%) 
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document.write( "I.  Yes, .07/sqrt(100)  
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document.write( "II.  With Excel function: P(z <4.29)) = NORMDIST(4.29)  = .999\r
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document.write( "Ho: p = 0.2
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document.write( "Ha: p > 0.2 (claim)
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document.write( "sample proportion = .23
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document.write( "z(.23) = (.23-.2)/.07/sqrt(100) = .03/.007  = 4.29
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document.write( "p-value = P(z < 4.29) = .999 OR 99.9% (10% significance level)\r
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document.write( "p-value greater than 10%, accept Ho
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document.write( "The attrition rate is same as 20%
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