Question

As part of the quality-control program for a catalyst manufacturing line, the raw materials (alumina and a binder) are tested for purity. The process requires that the purity of the alumina be greater than 85%. A random sample from a recent shipment of alumina yielded the following results (in percent):

93.2 87.0 92.1 90.1 87.3 93.6
A hypothesis test will be done to determine whethe or not to accept the shipment.

(a) State the appropriate null and alternate hypotheses.
(b) Compute the P-value.
(c) Should the shipment be accepted?Explain.

Answer 1

We accept the alternate hypothesis that the purity of the alumina is greater than 85% and hence, the shipment should be accepted.

Explanation:

We are given the following in the question:

93.2, 87.0, 92.1, 90.1, 87.3, 93.6

Formula:

`\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}`

where
`x_i` are data points,
`\bar{x}` is the mean and n is the number of observations.

`Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}`

`Mean =\displaystyle(543.3)/(6) = 90.55`

Sum of squares of differences = 7.0225 +12.6025 + 2.4025 + 0.2025 + 10.5625 + 9.3025 = 42.095

`SS.D = \sqrt{(42.095)/(5)} = 2.9`

Sample size, n = 6

Alpha, α = 0.05

First, we design the null and the alternate hypothesis

`H_(0): \mu = 85\%\\H_A: \mu > 85\%`

We use One-tailed t test to perform this hypothesis.

Formula:

`t_(stat) = \displaystyle\frac{\bar{x} - \mu}{(\sigma)/(√(n-1)) }` Putting all the values, we have

`t_(stat) = \displaystyle(90.55 - 85)/((2.9)/(√(5)) ) = 4.27`

The p-value is 0.003969.

Since, p-value < 0.05

We reject the null hypothesis and fail to accept it.

We accept the alternate hypothesis that the purity of the alumina is greater than 85% and hence, the shipment should be accepted.