Question

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

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Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis H₀: σ_A²< σ_B²

Alternative hypothesis H_A: σ_A² > σ_B²

Formulate an analysis plan. For this analysis, the significance level is 0.05.

Analyze sample data. Using sample data, the degrees of freedom (DF), and the test statistic (F).

DF₁ = n₁ - 1 = 10 -1

D.F₁ = 9

DF₂ = n₂ - 1 = 20 -1

D.F₂ = 19

Test statistics:-

$F = \frac{s_{1}^{2}}{s_{2}^{2}}$

F = 3.0

Since the first sample had the larger standard deviation, this is a right-tailed test.

p value for the F distribution = 0.021.

Interpret results. Since the P-value (0.021) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test there is sufficient evidence to indicate a smaller variation in diameters for Company B.