Question

Two chemical companies can supply a raw metarial. The concentration of a particular element in this material is important. The mean concentration for both suppliers is the same, but we suspect that the variability in concentration may differ between two companies. Ten batches were produced by company 1, resulting in a standard deviation of 4.7 grams per liter. Sixteen batches were produced by company 2, resulting in a standard deviation of 5.8 grams per liter. Is there sufficient evidence to coclude that the two population variances differ? Use α = 0.05.

Answer 1

Here

n1=10

n2=16

S1=4.7

S2=5.8

For checking variance is equal or different we use f test.

F test statistic = S2^2 / S1^2 = 1.2140

Now to calculate f critical value.

Degree of freedom for first sample = n1 - 1= 10 - 1= 9

Degree of freedom for second sample = n2 - 1 = 16 - 1= 15

F critical value = 3.7696

Decision rule if  f-table value is smaller than the calculated value, you can reject the null hypothesis.

Here f table value = 3.7696

F calculate = 1.2140

Therefore 1.2140 < 3.7696

We fail to reject null hypothesis.

There is no sufficient evidence that population variance is differ.