Question

From different tests, average material strength = 70 N/mm^2, and SD=8 N/mm^2. Follows normal distribution. What’s the minimum number of tests required to make a 95% confidence interval on the expected material strength of width <7 N/mm^2 (i.e. 10% of the average measured strength 70 N/mm^2).

Answer 1

It is given that

Average material strength=70 N/mm^2

Standard Deviation=8 N/mm^2

We have to find the minimum number of tests required to make a 95% confidence interval on the expected material strength of width <7 N/mm^2

Applying the below formula for finding the minimum number of tests

Where S.D=standard deviation

Margin of error=7

Value of Z at 95% Confidence interval=1.96

Let us find value of N

margin of error, "E," does not exceed a specified value

Minimum number of test required will be 385