Two chemical companies can supply a raw material. The concentration of a particular element in this material is important. The mean concentration for both suppliers is the same, but you suspect that the variability in concentration may differ for the two companies. The standard deviation of concentration in a random sample of n1 = 10 batches produced by company 1 is s1 = 4.7 grams per liter, and for company 2, a random sample of n2 = 16 batches yields s2 = 5.8 grams per liter. Is there sufficient evidence to conclude that the two population variances differ? Use α = 0.05.

No,  there's no sufficient evidence to make a conclusion that the two population variances differ.

Explanation:

Given that:

For company 1

Sample size = 10

Standard deviation = 4.7

For company 2

Sample size = 16

Standard deviation = 5.8

The null hypothesis and the alternative hypothesis can be computed as:

Since the alternative hypothesis is ≠, it is then a two-tailed test

Using the F test formula to test for the equality of the variations, we have:

F = 0.6567

Hence, the estimated F-value for comparing the standard deviation = 0.6567

The degree of freedom df = ( - 1,  - 1)

df = ( 10- 1, 16 - 1)

df = (9, 15)

The level of significance ∝ = 0.05

The critical value for a two-tailed test when ∝ = 0.05 and df = (9, 15)

by using the Excel function  = FINV (0.025,9,15)

= 3.123

Similarly;

Using the Excel function = FINV(0.975,9,15)

= 0.265

Critical Region: To reject the null hypothesis if f ≥ 3.123 or If f ≤ 0.265.

But since The estimated value for F =  0.6567 which is greater than 0.265, we fail to reject the null hypothesis.

Thus, there's no sufficient evidence to make a conclusion that the two population variances differ.