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 is to be 85%. A random sample from a recent shipment of alumina yielded the following results (in %):

           93.2               87.0               92.1                      90.1               87.3               93.6

a) Test the requirements of the catalyst manufacturing line using the appropriate hypothesis test (assume a=0.05).

b) Verify your result in part (a) using the appropriate confidence interval.

Answer 1

Part a

One sample t-test

Here, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H0: The purity of the alumina is 85%.

Alternative hypothesis: Ha: The purity of the alumina is not 85%.

H0: µ = 85 versus Ha: µ ≠ 85

This is a two tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 85

Xbar = 90.55

S = 2.901551309

n = 6

df = n – 1 = 5

α = 0.05

Critical value = - 2.5706 and 2.5706

(by using t-table or excel)

t = (Xbar - µ)/[S/sqrt(n)]

t = (90.55 – 85)/[ 2.901551309/sqrt(6)]

t = 4.6853

P-value = 0.0054

(by using t-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is not sufficient evidence to conclude that the purity of the alumina is 85%.

Part b

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

From given data, we have

Xbar = 90.55

S = 2.901551309

n = 6

df = n – 1 = 5

Confidence level = 95%

Critical t value = 2.5706

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 90.55 ± 2.5706*2.901551309/sqrt(6)

Confidence interval = 90.55 ± 3.0450

Lower limit = 90.55 - 3.0450 = 87.5050

Upper limit = 90.55 + 3.0450 = 93.5950

Confidence interval = (87.5050, 93.5950)

The hypothesized value of 85 is not lies within above interval, so we reject the null hypothesis.

There is not sufficient evidence to conclude that the purity of the alumina is 85%.