Two processes for
manufacturing large roller bearings are under study. In both cases,
the diameters (in centimeters) are being examined. A random sample
of 26 roller bearings from the old manufacturing process showed the
sample variance of diameters to be s2 = 0.231.
Another random sample of 28 roller bearings from the new
manufacturing process showed the sample variance of their diameters
to be s2 = 0.146. Use a 5% level of
significance to test the claim that there is a difference (either
way) in the population variances between the old and new
manufacturing processes.
Classify the problem as being a Chi-square test of
independence or homogeneity, Chi-square goodness-of-fit, Chi-square
for testing or estimating
σ2
or σ, F
test for two variances, One-way ANOVA, or Two-way ANOVA, then
perform the following.
Two-way ANOVA
F test for two variances
Chi-square test of homogeneity
Chi-square for testing or estimating σ2 or σ
Chi-square goodness-of-fit
One-way ANOVA
Chi-square test of independence
(i) Give the value of the level of significance.
State the null and alternate hypotheses.
H0: σ12 = σ22; H1: σ12 > σ22
H0: σ12 = σ22; H1: σ12 < σ22
H0: σ12 = σ22; H1: σ12 ≠ σ22
H0: σ12 < σ22; H1: σ12 = σ22
(ii) Find the sample test statistic. (Round your answer to
two decimal places.)
(iii) Find the P-value of the sample test
statistic.
P-value > 0.200
0.100 < P-value < 0.200
0.050 < P-value < 0.100
0.020 < P-value < 0.050
0.002 < P-value < 0.020
P-value < 0.002
(iv) Conclude the test.
Since the P-value is greater than or equal to the level of significance α = 0.05, we fail to reject the null hypothesis.
Since the P-value is less than the level of significance α = 0.05, we reject the null hypothesis.
Since the P-value is less than the level of significance α = 0.05, we fail to reject the null hypothesis.
Since the P-value is greater than or equal to the level of significance α = 0.05, we reject the null hypothesis.
(v) Interpret the conclusion in the context of the
application.
At the 5% level of significance, there is insufficient evidence to show that the variance for the new manufacturing process is different.
At the 5% level of significance, there is sufficient evidence to show that the variance for the new manufacturing process is not different.
At the 5% level of significance, there is insufficient evidence to show that the variance for the new manufacturing process is not different.
At the 5% level of significance, there is sufficient evidence to show that the variance for the new manufacturing process is different.
F test for two variances
level of significance = 0.05
H0: sigma1^2 = sigma2^2
Ha: sigma1^2 not = sigma2^2
ii)
Test statistic,
F = (s1/s2)^2
F = (0.231/0.146)
F = 1.58
iii)
dfN = 26 - 1 = 25
dfD = 28 - 1 = 27
p-value = 0.2426
P-value > 0.200
iv)
Since the P-value is greater than or equal to the level of
significance α = 0.05, we fail to reject the null hypothesis.
v)
At the 5% level of significance, there is insufficient evidence to
show that the variance for the new manufacturing process is
different.
Two processes for manufacturing large roller bearings are under study. In both cases, the diameters (in...
Two processes for manufacturing large roller bearings are under study. In both cases, the diameters (in centimeters) are being examined. A random sample of 21 roller bearings from the old manufacturing process showed the sample variance of diameters to be s2 = 0.246. Another random sample of 28 roller bearings from the new manufacturing process showed the sample variance of their diameters to be s2 = 0.116. Use a 5% level of significance to test the claim that there is...
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