Answer:
There is no sufficient evidence to support the claim that men and women differ in repeatability for this assembly task
Step-by-step explanation:
Given
Let subscript 1 represent men and 2 represent women, respectively.
[tex]n_1 = 25[/tex]
[tex]n_2 = 21[/tex]
[tex]s_1 = 0.98[/tex]
[tex]s_2 = 1.02[/tex]
[tex]\alpha = 0.02[/tex]
Required
Determine if here is enough evidence
First, we need to state the hypotheses
[tex]H_o: \sigma_1^2 = \sigma_2^2[/tex]
[tex]H_1: \sigma_1^2 \ne \sigma_2^2[/tex]
Next, calculate the test statistic using:
[tex]F = \frac{s_1^2}{s_2^2}[/tex]
[tex]F = \frac{0.98^2}{1.02^2}[/tex]
[tex]F = 0.923[/tex]
Calculate the rejection region;
But first, calculate the degrees of freedom
[tex]df_1 =n_1 - 1[/tex]
[tex]df_1 =25 - 1[/tex]
[tex]df_1 =24[/tex]
[tex]df_2 = n_2 -1[/tex]
[tex]df_2 = 21 - 1[/tex]
[tex]df_2 = 20[/tex]
Using the F Distribution: table
[tex]c = \frac{\alpha}{2}[/tex]
[tex]c = \frac{0.02}{2}[/tex]
[tex]c = 0.01[/tex]
At 0.01 level (check row 20 and column 24), the critical value is:
[tex]f_{0.01,24,20} = 2.86[/tex] --- the upper bound
At 0.01 level (check row 24 and column 20), the critical value is:
[tex]f_{0.01,20,24} = 2.74[/tex]
Calculate the inverse F distribution.
[tex]f_{0.99,20,24} = \frac{1}{f_{0.01,20,24}} = \frac{1}{2.74} =0.365[/tex] ---- the lower bound
The rejection region is then represented as:
[tex]0.365 < Test\ Statistic < 2.86[/tex]
If the test statistic falls within this region, then the null hypothesis is rejected
[tex]F = 0.923[/tex] --- Test Statistic
[tex]0.365 < 0.923 < 2.86[/tex]
The above inequality is true; so, the null hypothesis is rejected.
This implies that, there is no sufficient evidence.
