A local police department attempted to estimate the average rate of speed (mu) o

Question

A local police department attempted to estimate the average rate of speed (mu) of vehicles along a strip of Main Street. With hidden radar, the speed of a random selection of 32 vehicles was measured, which yielded a sample mean of 36 mph. Assume a standard deviation of 4 mph. Find the standard error of the mean. Explain what your obtained value for the standard error of the mean actually means. Find the 95% confidence interval for the population mean. Find the 99% confidence interval for the population mean. What is the difference between the 95% and the 99% confidence interval for the mean? What do you lose and gain by using the 99% confidence interval instead of the 95% CI?

Explanation / Answer

Given,

Sample vehicles(n)=32

Sample mean(x)=36

Standard Deviation(SD)=4

To find,

A.a) Standard error of mean

SE=SD/sqrt(n)

SE=4/sqrt(32)

SE=0.707

Standard error of mean is 0.707

A.b) from Standard error of mean we can interpret that the sample mean is 0.707 mph away from population mean

A.c) we have population SD, hence we can assume normal distribution

critical value(Z)= ±1.96

Confidence interval for population mean

X-Z*SD/sqrt(n) <=µ=< X+Z*SD/sqrt(n)

36-1.96*4/sqrt(32) <=µ=< 36+1.96*4/sqrt(32)

34.614 <=µ=< 37.386

hence, confidence interval for population mean is given as 34.614 <=µ=< 37.386

A.d) critical value(Z)= ±2.33

Confidence interval for population mean

X-Z*SD/sqrt(n) <=µ=< X+Z*SD/sqrt(n)

36-2.33*4/sqrt(32) <=µ=< 36+2.33*4/sqrt(32)

34.352 <=µ=< 37.647

hence, confidence interval for population mean is given as 34.352 <=µ=< 37.647

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