Question

A pharmaceutical manufacturer purchases a particular material from two different suppliers. The mean level of impurities in the raw material is approximately the same for both suppliers, but the manufacturer is concerned about the variability of the impurities from shipment to shipment. If the level of impurities tends to vary excessively for one source of supply, it could affect the quality of the pharmaceutical product. To compare the variation in percentage impurities for the two suppliers, the manufacturer selects 11 shipments from each of the two suppliers and measures the percentage of impurities in the raw material for each shipment. The sample means and variances are shown in the table.

Supplier A	Supplier B
x1 = 1.88	x2 = 1.82
s12 = 0.274	s22 = 0.094
n1 = 11	n2 = 11

(a) Do the data provide sufficient evidence to indicate a difference in the variability of the shipment impurity levels for the two suppliers? Test using α = 0.01. Based on the results of your test, what recommendation would you make to the pharmaceutical manufacturer? (Round your answers to two decimal places.)

1-2. Null and alternative hypotheses:

H₀: σ₁² ≠ σ₂² versus H_a: σ₁² = σ₂²

H₀: σ₁² = σ₂² versus H_a: σ₁² ≠ σ₂²    

H₀: σ₁² < σ₂² versus H_a: σ₁² > σ₂²

H₀: σ₁² = σ₂² versus H_a: σ₁² > σ₂²

H₀: σ₁² = σ₂² versus H_a: σ₁² < σ₂²

3. Test statistic:  F =  

4. Rejection region:  F >  

5. Conclusion:

H₀ is not rejected. There is sufficient evidence to indicate that the supplier's shipments differ in variability.

H₀ is not rejected. There is insufficient evidence to indicate that the supplier's shipments differ in variability.    

H₀ is rejected. There is insufficient evidence to indicate that the supplier's shipments differ in variability.

H₀ is rejected. There is sufficient evidence to indicate that the supplier's shipments differ in variability.

(b) Find a 99% confidence interval for σ₂². (Round your answers to three decimal places.)

to  

Answer 1

Part a.

1-2. Null and alternative hypotheses:

Null and alternative hypotheses are given as below:

H₀: σ₁² = σ₂² versus H_a: σ₁² ≠ σ₂²    

We are given level of significance = α = 0.01

3. Test statistic:  F =  

Test statistic formula is given as below:

F = S1^2/S2^2 = 0.274/0.094

F = 2.9149

Test statistic = F = 2.91

4. Rejection region:  F >  

n1=11

n2 = 11

df1 = n1 – 1 = 10

df2 = n2 – 1 = 10

α = 0.01

F critical value = 5.846678425

Rejection region: Reject H₀ if F > 5.85

5. Conclusion:

Test statistic F < Critical value

So, we do not reject H₀

H₀ is not rejected. There is insufficient evidence to indicate that the supplier's shipments differ in variability.   

Part b

Here, we have to find 99% confidence interval for population

Confidence interval for population variance

(n – 1)*S² / χ²_{α/2, n – 1} < σ² < (n – 1)*S² / χ²_{1 -α/2, n– 1}

We are given

S² = 0.094

n = 11

df = n – 1 = 10

α =0.01

χ²_{α/2, n – 1} = 25.1882

χ²_{1 -α/2, n– 1} = 2.1559

(n – 1)*S² / χ²_{α/2, n – 1} < σ² < (n – 1)*S² / χ²_{1 -α/2, n– 1}

(11 - 1)*0.094 / 25.1882 < σ² < (11 - 1)*0.094 / 2.1559

0.037319 < σ² < 0.436013

0.037 < σ² < 0.436

[All F critical values are taken from F-table.]