The following results were obtained from tests carried out on 11 bricks manufact

Question

The following results were obtained from tests carried out on 11 bricks manufactured by Obelisk Works using a new process. The random variable X = Y - 5550.0 where Y represents the compressive strength of the bricks in psi: sigma X = 550.0; sigma ZX^2 = 152,940.0 Use the information given above to find the 99% confidence limits of (i) the true mean and (ii) the true standard deviation of the probability distribution of Y. Assume that Y is a normally distributed random variable. Test the hypothesis that the true mean of the probability distribution of Y is not significantly different from 5,620.0psi. Let alpha = 0.05. Test the hypothesis that the true standard deviation of the probability distribution of Y is not significantly different from 100 psi. Let alpha = 0.05.

Explanation / Answer

Here Mean of X i.e X' = 550/11 = 50 and Var (X) = E(X2 ) - (E(X)) = 152940/11 - (50)2 = 11403.64

So, Standdard Deviation = 106.7878

Now X = Y -5550

So, Y = X + 5550

Mean of Y = Y' = X' + 5550 = 50 +5550 = 6000

Var ( Y) = Var (X + 5550) = Var (X) = 106.7878

If Y is normally distributed, then 99% confidence Intervalis given by Y' - 2.58 SE(Y') Y' + 2.58 SE(Y')

Lower Limit = Y' - 2.59 SD(Y')/n = 6000- 2.58*106.7878/11 = 6000-2.58*32.19774 = 5916.93

Upper Limit = Y' + 2.59 SD(Y')/n = 6000+ 2.58*106.7878/11 = 6000+2.58*32.19774 = 6083.07

(b) Z = x' - X; //n = I5620-6000/106.7878/11I = 11.8021

The value of Z for 5% level of significance is 1.96. Here the value is 11.8021. It is significently different.

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