1 answer

Recently, the RMC manufacturing facility in Tupelo, Mississippi has been experiencing problems with one of their...

Question:

Recently, the RMC manufacturing facility in Tupelo, Mississippi has been experiencing problems with one of their suppliers. T

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 o denotes the standard deviation of part weight Ho: 0 < 4g H: 0 > 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 (5 points): State any assumptions that are required to perform this test. Part b (5 points): Determine the appropriate sample size for this test. You may use this sample size for the remainder of the problem. Part c (4 points): What is the power of this test if the true standard deviation of part weight is o = 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 (6 points): Identify the test statistic and critical region for the test. State the conclusion for this test. Part e (5 points): Construct an appropriate confidence interval to verify your conclusion from part d. Part f (5 points): What is the P-value for this test? State the conclusion for this test. (An interval of probabilities is enough for the P-value)

Answers

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 7 0.9 0.918346

Note: P(Type II error)=1-Power.

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
3 7 0.965915

Answer: Required power=0.965915

Part d:

Test~statistic=\chi^2=\frac{(n-1)s^2}{4^2}=11.3438\\\\ where,~n=7,~s=sample~sd=5.5.\\\\ Critical~value=\chi^2_{0.05,6}=12.5916~(R~code:~round(qchisq(1-0.05,6),4)).\\\\ Since~value~of~test~statistic<Critical~value,~we~fail~to~reject~H_0~at~5\%~level~of\\ significance.

Hence standard deviation of weight is not greater than 4 g.

Part e:

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
7 5.50 30.3


95% Confidence Intervals

CI for CI 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 CI 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
7 5.50 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.

.

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