Question 1182427
the desired mean is 36 inches.
the sample size is 100.
the sample mean is 36.8 inches.
the sample standard deviation is 1.83 inches.


the z-score formula is:


z = (x - m) / s


z is the z-score
x is the raw score
m is the mean
s is the standard error.


s = standard deviation / square root of sampole size = 1.83 / sqrt(100)  1.83 / 10 = .183.


z-score formula becomes:
z = (36.8 - 36) / .183 = 4.371584699.


at 10% two tail confidence level, the critical z-core is equal to 1.645.
4.37 is way higher than that, so the conclusion is that the manufacturer should change the machine.


since the standard deviation was from the sample rather than the population, the use of the t-score is indicated rather than the z-score.


the calculation of the t-score itself remains the same.
instead of z = (x - m) / s, the formula becomes t = (x - m) / s
you get t = 4.371584699.
the difference is in the critical threshold.
the critical t-score at 99 degrees of freedom for a two tailed normal distribution is equal to 1.66.
the sample t-score is still way above this, therefore the decision stands whether you used t-score or z-score.