PSM1_E2021SA 2021 Spring Moed A

Part I

Question 1

Denoting as the process being approved, and given , the probability of having less than two parts being defective (which means the process is approved) follows a binomial distribution:

Question 2

Assertion I is correct and Assertion II is correct.

Question 3

We need to find where is the number of parts that are inspected.

In the modified plan, can take three values:

• : Reject after first sample ( defective in first )
• : Reject after second sample ( defective total in , but in first )
• : Continue to third sample ( defective total in first )

Let and .

Calculating

Calculating
This occurs when but :

For :

Therefore:

Calculating

Calculating expected value

Question 4

Given:

For a Poisson distribution: and

Calculating control limits

Defining false alarm condition
Since is a discrete random variable (count), a false alarm occurs when:

Calculating probability using normal approximation
For large , Poisson approaches normal. Using continuity correction:

Therefore:

Question 5

Given:

• Original process: , LCL =
• Improved process:
• Measurements taken every hours

Finding probability of detection per measurement
Under the improved process, . Detection occurs when (below LCL):

Model as geometric distribution
The number of measurements until first detection follows a geometric distribution with parameter .

Expected number of trials until first success:

Converting to hours
Since measurements are taken every hours:

Question 6

Moving average control chart with last wafers, where each wafer has .

Calculating mean of moving average

Calculating variance of moving average
Since wafers are independent over time:

Therefore:

Calculating Upper Control Limit

Part II

Question 7

Given , then:

Question 8

By the sampling distributions, for the first specification:

By the central limit theorem, for the second specification, we want:

Question 9

We can see from the calculation that the higher the , the larger the absolute value on the right hand side of the inequality. That means that the probability of meeting the second step depends on the numerator, (depends on the exact value of - larger of smaller than ).

Part III

Question 10

The standard error measures the accuracy of the estimate, which is better for larger n, for fixed levels of time.

Question 11

The outer dotted lines in the graph represent the confidence interval, but we want a confidence. Thing is - we need confidence that the parts will have a thickness below - we don’t need a bottom limit - which means we only care about the higher outer dotted line - which comes below at around .

Question 12

This is a classic case for the use of bayes’ rule. We know , that , and that .

That means:

We want to know .

Question 13

According to the central limit theorem:

Part III

Question 14

The width of a confidence interval for the difference between two means depends on:

Where is the pooled standard deviation.

Effect of removing outliers:

1. Sample size effect: decreases from to

   • This tends to increase the standard error slightly
   • Effect: increases marginally

2. Outlier removal effect: Removing extreme values (times min)

   • Substantially reduces sample variability
   • Pooled standard deviation decreases significantly
   • Effect: Much smaller

Dominant effect:
The reduction in from removing outliers has a much larger impact than the small increase in from reduced sample size.

Conclusion:
Since removing outliers dramatically reduces variability while the sample size reduction is minimal ( is only decrease), the overall effect is a reduction in standard error.

Therefore the new confidence interval will be narrower.

Question 15

Assertion I: correct.
Assertion II: incorrect – we can see the difference in the two averages, but not the levels.

Question 16

The boxplots show the errors, meaning is the for a perfect cut. We can see that the median lies at , meaning the the range is about . Which is within degree of the planned cut angle.

Question 17

Given data:

For each measure, we test whether the mean error differs significantly from zero.

1. (no systematic error)
2. (systematic error exists)
4. Critical region: , where

Femoral flexion:

Since , reject . This is significant.

Axial rotation:

Since , do not reject . This is not significant.

Tibia slope:

Since , reject . This is significant.

Therefore:
A significant difference for femoral flexion angle and for tibia slope, but not for femoral axial rotation.

Question 18

The data come from the same knee replacements, so we need a paired test. The average difference is mathematically the same as the difference in averages. Therefore:

An appropriate statistical test is a two-sided paired t-test, and we know the observed average difference will be .

Question 19

We have two groups (RAN and CAN) and want to compare the proportions of patients with different comorbidities (diabetes, hypertension, neither). This creates a contingency table:

	Diabetes	Hypertension	Neither
RAN group	?	?	?
CAN group	?	?	?

We want to test whether group membership (RAN vs CAN) is independent of comorbidity status. This is a test of association between two categorical variables.

The appropriate test is a chi-squared test for association, which tests:

• : Group membership and comorbidity are independent
• : Group membership and comorbidity are associated

Question 20

From the power curve graph with Error Std Dev , Difference in Means = , and :

Assertion I:
The probability of getting a -value less than 0.05 when there truly is a difference is the definition of statistical power. From the graph, at total sample size , the power appears to be approximately 0.80 or slightly above.

Since the true difference in means is (very close to the graph’s ), the power would be very similar. Therefore, the power is not “less than 0.80” but rather around 0.80 or above.

Assertion I is incorrect.

Assertion II Analysis:
Power depends on the effect size, which is the standardized difference:

Original:

New scenario:

Since both effect sizes are identical () and alpha remains , the power curves will indeed be identical.

Assertion II is correct.