A process for a certain type of ore is designed to reduce the concentration of impurities to less than 2%. It is known that the standard deviation of impurities for processed ore is 0.5%. Let u(mu) represent the mean impurity level, in percent, for ore specimens treated by this process. The impurity of 75 ore specimens is measured. We will conclude the process is successful if we can demonstrate that the mean impurity is less than 2%.

a) State the appropriate null and alternative hypotheses

b) If the test is made at the 5% level, what is the rejection region?

c) If the sample mean impurity level is 1.82, will the null hypothesis be rejected at the 1% level?

d) If the value 1.9 is a critical point, what is the level of the test?

c) the population standard deviation is known and the sample size is large. if the sample mean purity, x = 1.82, then we would determine the z score by applying the formula

z = (x - µ)/(σ/√n)

Where

x = sample mean

µ = population mean

σ = population standard deviation

n = number of samples

From the information given,

µ = 2

x = 1.82

σ = 0.5

n = 75

z = (1.82 - 2)/(0.5/√75) = - 3.12

Since α = 0.01, the critical value is determined from the normal distribution table.

For the left, α/2 = 0.01/2 = 0.005

The z score for an area to the left of 0.005 is - 2.575

For the right, α/2 = 1 - 0.005 = 0.995

The z score for an area to the right of 0.995 is 2.575

In order to reject the null hypothesis, the test statistic must be smaller than - 2.575 or greater than 2.575. Since - 3.12 < - 2.575 and 3.12 > 2.575, we would to reject the null hypothesis.

d) if critical point is 1.9, we would determine α

The z probability value to 1.9 on the right is 0.971. For the right,

α/2 = 1 - x = 0.971

α/2 = 1 - 0.971 = x

α/2 = 1 - 0.029 = x

α = 2 × 0.029

α = 0.058