Question

Recently, the RMC manufacturing facility in Tupelo, Mississippi has been experiencing problems with one of their suppliers. This supplier provides a part for which the key quality characteristic is its weight. The problem has not been with the average weight. Instead, parts received at the Tupelo facility have been exhibiting excessive weight variability. Under normal conditions, the standard deviation in weight should be less than or equal to 4 g.

To deal with this problem, a RMC quality engineer has decided to implement a test of hypotheses on each shipment of parts received at the facility. This test is to have a level of significance of 0.05 and can be summarized as follows where  denotes the standard deviation of part weight.

H0: ? ≤ 4g H1: ? > 4g

In designing this test, the engineer wishes to have a Type II error probability of 0.1 when the true standard deviation rises to 10 g.

Part a State any assumptions that are required to perform this test.

Part b Determine the appropriate sample size for this test. You may use this sample size for the remainder of the problem.

Part c What is the power of this test if the true standard deviation of part weight is ? = 12 g?

Suppose a shipment is received, a random sample of parts is measured, and the resulting sample standard deviation is 5.5 g.

Part d Identify the test statistic and critical region for the test. State the conclusion for this test.

Part e Construct an appropriate confidence interval to verify your conclusion from part d.

Part f What is the P-value for this test? State the conclusion for this test. (An interval of probabilities is enough for the P-value)

Answer 1

Solution:

Part a: We assume that weight of a part is normally distributed. Part b: Minitab output: Power and Sample Size Test for One Standard Deviation Testing StDev = null (versus > null) Calculating power for (StDev / null) = ratio Alpha = 0.05 Sample Target Ratio Size Power Actual Power 2.5 70.9 0.918346 Note: P(Type Il error)=1-Power.

Power Curve for One Standard Deviation 1.0 Sample Size 7 0.8 Assumptions Alpha 0.05 Alternative 0.6 Power 0.4 0.2 0.0 + 1.0 1.5 2.0 3.0 3.5 2.5 Ratio Answer: required sample size=7. Part c: Power and Sample Size Test for One Standard Deviation Testing StDev = null (versus > null) Calculating power for (StDev / null) = ratio Alpha = 0.05

Sample Ratio Size Power 370.965915 Power Curve for One Standard Deviation 1.0 Sample Size 7 0.8 Assumptions Alpha 0.05 Alternative 0.6 ower Power 0.4 0.2 0.0 + 1.0 1.5 2.0 3.0 3.5 2.5 Ratio

Answer: Required power=0.965915 Part d: 11.3438 Test statistic = x (n-1)s 42 where, n = 7, s= sample sd = 5.5. Critical value = xã.05,6 = 12.5916 (R code : round(qchisq(1 – 0.05,6), 4)). Since value of test statistic < Critical value, we fail to reject Ho at 5% level of significance. Hence standard deviation of weight is not greater than 4 g. Parte: Minitab output: Method The chi-square method is only for the normal distribution. The Bonett method cannot be calculated with summarized data.

Statistics N StDev Variance 75.50 30.3 95% Confidence Intervals Cl for Cl for Method StDev Variance Chi-Square (3.54, 12.11) (12.6, 146.7) 95% CI for population sd is (3.54, 12.11). f. Minitab output: Test and Cl for One Variance Method Null hypothesis Sigma = 4 Alternative hypothesis Sigma > 4 The chi-square method is only for the normal distribution. The Bonett method cannot be calculated with summarized data.

Statistics N StDev Variance 75.50 30.3 95% One-Sided Confidence Intervals Lower Bound for Lower Bound Method StDev for Variance Chi-Square 3.80 14.4 Tests Test Method Statistic DF P-value Chi-Square 11.34 6 0.078 Answer: P-value=0.078 i.e. 0.05<P-value<0.10.