SOLUTION: 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 i

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%?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
p represents the probability that customers are satisfied, therefore p = .9.
q represents the probability that customers are not satisfied, therefore q = 1 - .9 = .1
n = 120 which is your sample size.

the mean of this sample is equal to n * p which is equal to 120 *.9 = 108.
the standard deviation of this sample is equal to sqrt(n*p*q) which is equal to sqrt(120*.9*.1) = 3.286335345.

the mean of this sample is equal to 108.
the standard deviation of this sample is equal to 3.286335345.

you are looking to find the probability of getting a score of less than 100.

the z-score is given by the formula of z = (x-m)/s

x is the test score.
m is the mean.
s is the standard error of the sample.

formula becomes z = (100-108)/3.286335345.

solve for z to get z = -2.434322478

the critical z-score for a 99% confidence interval is equal to plus or minus 2.575829303.

the critical z-score for a 95% confidence interval is equal to plus or minus 1.959963986.

your sample is borderline within limits at 99% confidence interval, and clearly out of limits at 95% confidence interval.

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.

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.

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.

accordingly, i believe that this constitutes sufficient evidence that the precentage of satisfied customer is probably less than 90%.

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