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

Answer:

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

Step-by-step 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.  

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

$Mean =\displaystyle\frac{543.3}{6} = 90.55$

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

$SS.D = \sqrt{\frac{42.095}{5}} = 2.9$

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:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n-1}} }$ Putting all the values, we have

$t_{stat} = \displaystyle\frac{90.55 - 85}{\frac{2.9}{\sqrt{5}} } = 4.27$

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.