document.write( "Question 1031434: The average commission charged by a brokerage companies on a sale is $186. However it is seen in the random sample of 133 trades that the average amount of commission paid is $179 with a standard deviation of $61. At 10% significance level can we conclude that the average commission is lower than the industry average?
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document.write( " The average commission is not same as $186
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document.write( " The average commission is same $186
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document.write( " The average commission is lower than $186
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document.write( " The average commission is now $179
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document.write( "Am I correct in assuming that this is an example of T distribution as population standard deviation is not given. Or do we have to you normal distribution.
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document.write( "Thanks.
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Algebra.Com's Answer #646166 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! Yes, your assumption is correct. \n" ); document.write( "Ho=same or more \n" ); document.write( "Ha=less They are asking whether this is lower than the industry average. This is a one-way test, not a two way test, which would ask whether it were the same or different. \n" ); document.write( "alpha=0.10 \n" ); document.write( "critical value about -1.285 \n" ); document.write( "t df=132 = (179-186)/61/sqrt(133) \n" ); document.write( "t=-7(sqrt(133))/61 \n" ); document.write( "=-1.32 \n" ); document.write( "this is significant, because t<-1.285 , and the average commission is less than $186. \n" ); document.write( "We use the t-distribution because the sample sd is used as an estimate for the population sd. The sample size of 30 has been used as a cutoff point for years, for unclear reasons. If the distribution is near normal, a sample size of greater than 10 would be sufficient to use the normal distribution. If the distribution is skewed, even a sample size of 100 might not be sufficient. In any case, the error gets less an less with larger sample sizes, and over 100, the error is about 2%. \n" ); document.write( " |