The breaking strength of hockey stick shafts made of two different graphite-kevlar composites yields the following results (in Newtons):
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 the breaking strength differs between the two composites? Use the α = 0.05 level of significance.
State the null and alternate hypotheses:
H0: σ1 = σ2
H1: σ1 ≠ σ2
This hypothesis test is a ______ test.
Find the critical value. Round the answer to two decimal places.
The critical value is 3.28
Compute the test statistic. Round the answer to two decimal places.
F = 35.00
Determine whether to reject H0.
Answer (2)
To determine if the standard deviations of the breaking strength of two different graphite-kevlar composites are the same, we need to conduct an F-test. Here's how we proceed step-by-step:
State the Hypotheses
Null Hypothesis (H0): The standard deviations of the two composites are equal. σ 1 = σ 2 .
Alternative Hypothesis (H1): The standard deviations of the two composites are not equal. σ 1 = σ 2 .
Collect Data
Composite 1 breaking strengths: 525.3, 535, 534.7, 528.5, 525.4, 538, 522.2, 512.7, 524.4
Composite 2 breaking strengths: 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
Calculate the Standard Deviation for Each Sample
Use the formula for sample standard deviation to calculate the standard deviations s 1 and s 2 for composite 1 and 2 respectively.
Conduct the F-test
The test statistic F is calculated as F = s 2 2 s 1 2 .
Use the sample with the larger variance as the numerator to ensure F ≥ 1 .
Find the Critical Value
For an α = 0.05 level of significance, check an F-distribution table for 8 and 12 degrees of freedom in the numerator and denominator respectively (since sample sizes are 9 and 13).
The critical value provided is 4.20.
Compute the Test Statistic
The calculated test statistic is F = 33.98 .
Conclusion
Since the calculated F value (33.98) is greater than the critical value (4.20), we reject the null hypothesis.
There is sufficient evidence at the α = 0.05 level of significance to conclude that the standard deviation of breaking strength differs between the two composites.
The F-test was conducted to determine if the standard deviations of two composite hockey stick shafts differ. The null hypothesis was rejected because the calculated test statistic (35.00) exceeded the critical value (3.28). Thus, there is significant evidence indicating that the standard deviation of breaking strengths differs between the two composites.
;
