document.write( "Question 1103717: A hair salon claims that 90% of their customers are satisfied. However, Toni who recently started working there found that of 120 customers last week, only 100 were satisfied. If it were true that 90% of customers are satisfied, what is the chance that among 120 customers 100 or fewer would be satisfied? Is there evidence that the percentage of satisfied customers is lower than 90%? \n" ); document.write( "
Algebra.Com's Answer #718415 by Theo(13342)![]() ![]() You can put this solution on YOUR website! p represents the probability that customers are satisfied, therefore p = .9. \n" ); document.write( "q represents the probability that customers are not satisfied, therefore q = 1 - .9 = .1 \n" ); document.write( "n = 120 which is your sample size.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the mean of this sample is equal to n * p which is equal to 120 *.9 = 108. \n" ); document.write( "the standard deviation of this sample is equal to sqrt(n*p*q) which is equal to sqrt(120*.9*.1) = 3.286335345.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the mean of this sample is equal to 108. \n" ); document.write( "the standard deviation of this sample is equal to 3.286335345.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "you are looking to find the probability of getting a score of less than 100.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the z-score is given by the formula of z = (x-m)/s\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "x is the test score. \n" ); document.write( "m is the mean. \n" ); document.write( "s is the standard error of the sample.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "formula becomes z = (100-108)/3.286335345.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "solve for z to get z = -2.434322478\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the critical z-score for a 99% confidence interval is equal to plus or minus 2.575829303.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the critical z-score for a 95% confidence interval is equal to plus or minus 1.959963986.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "your sample is borderline within limits at 99% confidence interval, and clearly out of limits at 95% confidence interval.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "at 99% confidence interval, with a sample size of 120, the number of satisfied customers should not be less than 108 - 2.575829303 * 3.286335345 = 99.53496112.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "at 95% confidence interval, with a sample size of 120, the number of satisfied customers should not be less than 108 - 1.959963986 * 3.286335345 = 101.5589011.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "either way, the probability is unlikely that getting less than or equal to 100 satisfied customers out of a sample of 120 customers is due to random variations in sample means only.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "accordingly, i believe that this constitutes sufficient evidence that the precentage of satisfied customer is probably less than 90%.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |