A sample of 100 tyres is taken from a lot. The mean life of tyres is found to be 39350 kilometers with a standard deviation of 3260. Can it be considered as a true random sample from a population with mean life of 40000 kilometers? (use 0.05 level of significance) establish 99% confidence limits within which the mean life of tyres expected to lie. (given that 0.05 = 1.96,0.01 = 2.58)
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Answer:
What to do of this question?
Step-by-step explanation:
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.