In: Statistics and Probability
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 = 0.214.
Another random sample of 29 roller bearings from the new manufacturing process showed the sample variance of their diameters to be
s2 = 0.115.
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.
One-way ANOVAChi-square test of independence F test for two variancesChi-square goodness-of-fitTwo-way ANOVAChi-square for testing or estimating σ2 or σChi-square test of homogeneity
(i) Give the value of the level of significance.
State the null and alternate hypotheses.
H0: σ12 = σ22; H1: σ12 > σ22H0: σ12 = σ22; H1: σ12 ≠ σ22 H0: σ12 = σ22; H1: σ12 < σ22H0: 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 > 0.2000.100 < P-value < 0.200 0.050 < P-value < 0.1000.020 < P-value < 0.0500.002 < P-value < 0.020P-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.
A random sample of 21 roller bearings from the old manufacturing process showed the sample variance of diameters to be s2 = 0.214.
Another random sample of 29 roller bearings from the new manufacturing process showed the sample variance of their diameters to be s2 = 0.115.
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
.This problem is testing the equality of two population variances.
So it is solved using F test for two variances.
Answer:- F test for two variances
(i) Give the value of the level of significance. = 0.05
test the claim that there is a difference (either way) in the
population variances between the old and new manufacturing
processes.
It is a two-sided test.
The null & alternative hypothesis is :
H0: σ12 = σ22;
Vs
Ha: σ12 ≠ σ22
The decision about the null hypothesis
Since from the sample information we get that FL=0.423<F=1.861<FU=2.232, it is then concluded that the null hypothesis is not rejected.
df1 = n1-1 = 20 ; df2 = n2-1 = 28
P-value = 2* P( Fdf1,df2 > F)
= 2* P( F20,28 > 1.861)
P-value = 2* 0.0641
P-value = 0.1281
Answer:- 0.100 < P-value < 0.20
Since the P-value is greater than or equal to the level of significance α = 0.05, we fail to reject the null hypothesis.
At the 5% level of significance, there is insufficient evidence to show that the variance for the new manufacturing process is different.