Answer:
No, there's no sufficient evidence to make a conclusion that the two population variances differ.
Explanation:
Given that:
For company 1
Sample size [tex]n_1[/tex] = 10
Standard deviation [tex]s_1[/tex] = 4.7
For company 2
Sample size [tex]n_2[/tex] = 16
Standard deviation [tex]s_2[/tex] = 5.8
The null hypothesis and the alternative hypothesis can be computed as:
[tex]\mathbf{H_o: \sigma^2_1 = \sigma^2_2}[/tex]
[tex]\mathbf{H_1: \sigma^2_1 \ne \sigma^2_2}[/tex]
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:
[tex]F = \dfrac{s_1^2}{s_2^2}[/tex]
[tex]F = \dfrac{4.7^2}{5.8^2}[/tex]
[tex]F = \dfrac{22.09}{33.64}[/tex]
F = 0.6567
Hence, the estimated F-value for comparing the standard deviation = 0.6567
The degree of freedom df = ([tex]n_1[/tex] - 1, [tex]n_2[/tex] - 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)
[tex]F_{\alpha/2,n_1 -1,n_2-2} = F_{0.05/2, 9,15}[/tex]
[tex]= F_{0.025, 9,15}[/tex]
by using the Excel function = FINV (0.025,9,15)
= 3.123
Similarly;
[tex]F_{1-\alpha/2,n_1 -1,n_2-2} = F_{1-0.05/2, 9,15}[/tex]
[tex]= F_{0.975, 9,15}[/tex]
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.
