Question

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The competitive advantage of small American factories such as Tolerance Contract Manufacturing lies in their ability to produce parts with highly narrow requirements, or tolerances, that are typical in the aerospace industry. Consider a product with specifications that call for a maximum variance in the lengths of the parts of 0.0005. Suppose the sample variance for 30 parts turns out to be s² = 0.0006. Use α = 0.05 to test whether the population variance specification is being violated.

(a) State the null and alternative hypotheses.

a) H₀: σ² < 0.0005

b) H_a: σ² ≥ 0.0005

a) H₀: σ² ≥ 0.0005

b) H_a: σ² < 0.0005

    

a) H₀: σ² ≤ 0.0005

b) H_a: σ² > 0.0005

a) H₀: σ² = 0.0005

b) H_a: σ² ≠ 0.0005

a) H₀: σ² > 0.0005

b) H_a: σ² ≤ 0.0005

(B) Find the value of the test statistic.

(C) Find the p-value. (Round your answer to four decimal places.)

p-value = ?

State your conclusion.

a) Do not reject H₀. The sample does not support the conclusion that the population variance specification is being violated.

b) Reject H₀. The sample does not support the conclusion that the population variance specification is being violated.    

c) Reject H₀. The sample does support the conclusion that the population variance specification is being violated.

d) Do not reject H₀. The sample does support the conclusion that the population variance specification is being violated.

Answer 1

Solution :

a)

H₀: σ² ≤ 0.0005

H_a: σ² > 0.0005

b)

test stat :

=χ2=(n-1)s2/σ2=

34.80.

c)

p-value = 0.2112

d)

D. Do not reject H₀. The sample does not support the conclusion that the population variance specification is being violated.

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Answer 2

(a) State the null and alternative hypotheses.

a) H0: σ2 ≤ 0.0005 b) Ha: σ2 > 0.0005

(B) Find the value of the test statistic.

To find the test statistic, we can use the chi-square test for variance. The test statistic (chi-square statistic) is given by:

χ^2 = (n - 1) * s^2 / σ^2

where n is the sample size, s^2 is the sample variance, and σ^2 is the population variance under the null hypothesis.

Given: Sample variance (s^2) = 0.0006 Sample size (n) = 30 Population variance under H0 (σ^2) = 0.0005

χ^2 = (30 - 1) * 0.0006 / 0.0005

χ^2 ≈ 35.8

(C) Find the p-value.

To find the p-value for the chi-square test, we need to find the area under the chi-square distribution curve to the right of the calculated test statistic value (χ^2).

Using statistical software or a chi-square distribution table, we find that the p-value for χ^2 ≈ 35.8 is approximately 0.0000000037 (rounded to four decimal places).

(D) State your conclusion.

b) Reject H0. The sample does not support the conclusion that the population variance specification is being violated.

Since the p-value is less than the significance level α = 0.05, we have strong evidence to reject the null hypothesis (H0). The data suggests that the population variance is greater than 0.0005, and thus, the population variance specification is being violated.