A total of 50 repairs are examined and sample mean is found to be $2650. The population standard deviation is assumed to be $425. What is the 95 percent confidence interval for the population mean transmission replacement cost?

The standard error foe the sample mean distribution is: 425/sqrt(50)=60.1

Thus we look at the normal distribution of the normalized variable Z = ( mean- mu) /st. error = (2650-mu)/60.1 where mu is the real (unknown) mean.

The confidence interval end point are calculated from where the normal distribution where the

we solve for the cumulative distribution function such that on each edge we leaves out only 2.5% of all probabilities (for a total of 5%)

from tables of normal distribution function: z= 1.96 (this is well known)

thus we write

0.95 = P(mean-1.96*60.1<mu<mean +1.96*60.1).

So

Lower end of conf. interval = 2650-1.96*60.1=2532

Upper end: =2650-1.96*60.1=2768

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