SOLUTION: 1. A firm that manufactures tires would like to determine whether the machine that produces these tires is still in good condition. The average diameter of these tires should be 3

Question 1182427: 1. A firm that manufactures tires would like to determine whether the machine that produces these tires is still in good condition. The average diameter of these tires should be 36 inches to fit most of the car models. To check the machine, a sample of 100 tires were measures and showed a mean diameter of 36.8 inches with standard deviation of 1.83 inch. Should the manufacturer change the machine? Use a 10% level of significance
Thanks in advance!
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
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.

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