In a sample of 100 boxes of a certain type, the average compressive strength was

Question

In a sample of 100 boxes of a certain type, the average compressive strength was 6230 N, and the standard deviation was 221 N. Find a 95% confidence interval for the mean compressive strength of boxes of this type. Find a 99% confidence interval for the mean compressive strength of boxes of this type. An engineer claims that the mean strength is between 6205 and 6255 N. With what level of confidence can this statement be made? Approximately how many boxes must be sampled so that a 95% confidence interval will specify the mean to within plusminus 25 N?

Explanation / Answer

a. z value for 95% is 1.96 mean=6230 and sd=221

Margin of Error=E=z*sd/sqrt(100)=43.316

CI=mean+/-E=6230+/-43.316=( 6186.684, 6273.316)

b. z value for 99% is 2.58

Margin of Error=E= 57.018

CI=mean+/-E=( 6172.982, 6287.018)

c. We have UCL=mean+E=6255 and LCL=mean-E=6205

So 2E=6255-6205=50

Hence E=50

Now E=sd*z/sqrt(n)

So z=E*sqrt(n)/sd=2.3

We know P(-2.3<z<2.3)=0.98

So level of confidence is 98%

d. E=25, z=1.96 and sd=221

So using formula of E=z*sd/sqrt(n)

n=(z*sd/E)^2

n=300

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