Question

(SINCE THE VERY BEGINNING, PLEASE SHOW ALL THE STEPS IN YOUR CALCULATION ONE BY ONE. PLEASE DO NOT JUST PUT THE FORMULA WITH THE FINAL ANSWER.) Thank you for your help.

We want to compare the variation in diameters of an engine part produced by Manufacturer Alpha against those produced by Manufacturer Beta. Sample variance for Manufacturer Alpha, based on n=10 samples, is s₁²= 0.0003. In contrast, the variance of the diameter measurements for a sample of 20 from Manufacturer Beta is s₂² = 0.0001. Do the data provide sufficient information to indicate a smaller variation in diameters for Manufacturer B? Test using α = 0.05 level of significance. State clearly the null hypothesis you are testing.

Answer 1

Solution: Here the given information are

For Alpha

n1=10,  =0.0003

For Beta

n2=20, = 0.0001

Here we want to test the variance between two manufactures.

=variance in the diameters of the Manufacturer Alpha

=variance in the diameters of the Manufacturer Beta

We want to test

vs  

Here,   >, we take test statistic as

Under ,

=  

=

=3 ( calculated value )

= 2.423 ( table value )

Here Calculated value > Table value.

We reject at 5% level of significance.

Conclusion: The variances in the diameters of engine part produced by Manufacturer Alpha and

Manufacturer Beta.