To determine if the standard deviation of breaking strength differs between the two composites, we can perform a two-sample F-test for the equality of variances.
First, we need to calculate the sample standard deviations for each composite. Let's denote the sample standard deviation of Composite 1 as s1 and the sample standard deviation of Composite 2 as s2.
For Composite 1:
s1 = √[Σ(xi - x̄)² / (n1 - 1)]
s1 = √[((525.3 - x̄)² + (535 - x̄)² + ... + (524.4 - x̄)²) / (9)]
For Composite 2:
s2 = √[Σ(xi - x̄)² / (n2 - 1)]
s2 = √[((516.2 - x̄)² + (553.9 - x̄)² + ... + (547.7 - x̄)²) / (12)]
Next, we need to calculate the F-test statistic:
F = s1² / s2²
Now, we need to find the critical value for the F distribution with degrees of freedom (n1-1) and (n2-1) at a significance level of 0.05. This critical value can be found using an F-table or a statistical software.
Once we have the critical value, we can compare it to the calculated F-test statistic. If the calculated F-test statistic is greater than the critical value, we reject the null hypothesis that the variances are equal and conclude that the standard deviation of breaking strength differs between the two composites. If the calculated F-test statistic is less than the critical value, we fail to reject the null hypothesis.
The breaking strength of hockey stick shafts made of two different graphite kevlar composites yields the following results . Composite 1: 525.3, 535, 534.7, 528.5, 525.4, 538, 522.2, 512.7, 524.4;
Composite 2: 516.2, 553.9, 577.5, 576.9, 476.1, 509.7, 479.4, 478.9, 501.9, 582.6, 615.5, 523.9, 547.7;
Can you conclude that the standard deviation of breaking strength differs between the two composites. Use the a= 0.05 level of significance. It's a two tailed test. So first find the critical value
3 answers
And what's the critical value
To find the critical value for the F-distribution at a significance level of 0.05, we need to use the F-distribution table or a statistical software. The critical value depends on the degrees of freedom associated with the two samples.
In this case, we have 9 degrees of freedom for Composite 1 (n1-1) and 12 degrees of freedom for Composite 2 (n2-1).
Using an F-distribution table with a significance level of 0.05 and degrees of freedom of 9 and 12, we find the critical value to be approximately 2.83.
Therefore, the critical value for the F-test with degrees of freedom 9 and 12 at a significance level of 0.05 is approximately 2.83.
In this case, we have 9 degrees of freedom for Composite 1 (n1-1) and 12 degrees of freedom for Composite 2 (n2-1).
Using an F-distribution table with a significance level of 0.05 and degrees of freedom of 9 and 12, we find the critical value to be approximately 2.83.
Therefore, the critical value for the F-test with degrees of freedom 9 and 12 at a significance level of 0.05 is approximately 2.83.