It is claimed that a random sample of 100 tyres with a mean life of 15269kms is drawn
from a population of tyres which has a mean life of 15200 kms and a standard deviation
of 1248 kms. To test the validity of the claim what is the Z-score. (approximately)
Answers
Answer:
sample size is 100 tires.
mean of sample is 39350
standard deviation of sample is 3260
standard error of the distribution of sample means is 3260 / sqrt(100) = 326.
because your standard deviation is taken from the sample rather than from the population, you would use the t-score rather than the z-score.
because the sample size is large enough, it shouldn't make much difference whether you use the t-score or the z-score.
at 99% confidence limits, the critical z-score would be plus or minus 2.575829303 and the critical t-score with 99 degrees of freedom would be plus or minus 2.62640545.
the basic formula for z-score or t-score is z or t = (x - m) / s
z = s-score
t = t-score
x is the raw score being compared to the mean.
m is the mean
s is the standard error of the distribution of sample means.
x is the mean of the sample = 39350
m is the mean of the population = 40000
s is the standard error of the distribution of sample means = 326.
the z-score / t-score would be equal to (39350 - 40000) / 326 = -1.993865031.
this score is well within the confidence limits of either the z-score or the t-score.
therefore, it's safe to say that this sample could come from a population that had a mean tire life of 40,000.
visually, the results look like this for the z-score.
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- from a population of tyres which has a mean life of 15200 kms and a standard deviation
- from a population of tyres which has a mean life of 15200 kms and a standard deviationof 1248 kms. To test the validity of the claim what is the Z-score. (approximately)