Question

Problem 5 Part 1: A contractor claims the mean value of his compression test is 6500 psi and has a standard deviation of 500 psi. You attempt to check this by taking 8 sample measurements and find your average is 5900 psi. If your level of significance is 0.01, would you believe the means are the same? If using EXCEL or a t-test website, make sure to copy and paste the screenshot. You will need to EXPLAIN the computer results to receive credit. Alternatively, you can show calculations by hand. Part2: Continuing from the past problem. You decided that 8 samples are too small of a data set (so maybe you aren't having enough evidence of your claim). Instead, you measure 18 sample cylinders and find your average is now 5990 psi. For the same a, would you now believe the mean is the same as the contractor?

Answer 1

The contractor has provided the mean value of cubes as 6500psi for the enormous number of cubes he has tested. Thus the provided value is the target mean strength of the concrete design. The characteristic strength of the concrete lies somewhere below the target mean strength exactly at a distance of, $k*\sigma$. Where 'k' is the statistic constant that determines the distance between the target mean strength and the characteristic strength, '$\sigma$' is the Standard Deviation.

As per Indian codes, the statistic constant is 1.65 which means the distance between the target mean strength and the characteristic strength is 1.65 times the standard deviation.

Therefore the Characteristic strength of the concrete as per the claim of the contractor is given by,

fck = fm - 1.65 * o

Where,

fm = Target Mean Strength

fok = Characteristic Strength

fok = 6500 – 1.65 * 500 = 5675psi

Characteristic strength of concrete is 5675psi.

Part 1:

The test done by the reviewer includes testing 8 samples of cubes. He has obtained a mean strength of 5900psi.

This isn't the target mean strength. Rather it is a value obtained below the target mean strength because number of samples considered is very less when compared to the samples supposedly considered by the contractor. For standardization of the value, it is considered to be at the halfway of the distance between the target mean strength and the characteristic strength.

Using this, we can derive the target mean strength of this test.

The characteristic strength as per this test is,

fok = 5900 – 0.825 * 500 = 5487.5psi

The target mean strength is now given as,

fm = 5487.5 + 1.65 * 500 = 6312.5psi

Which is less than the value claimed by the contractor. Thus the value claimed by the contractor is safer.

Part 2:

Similarly, for 18 sample the obtained mean strength is at the halfway of the distance between the target mean strength and the characteristic strength.

The characteristic strength as per this test is,

fok = 5990 – 0.825 * 500 = 5577.5psi

The target mean strength is now given as,

fm = 5577.5 + 1.65 * 500 = 6402.5psi

Thus again for this test the target mean strength obtained is lesser than that claimed by the contractor.

Again after rechecking the value it can be proven that the cubes casted by the contractor are safer.