ACC544 Decision Support Tools

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07-07-2021

Question:

A company wishes to set control limits for monitoring the direct labour time to produce an important product. Over the past the mean time has been 20 hours with a standard deviation of 10 hours and is believed to be normally distributed. The company proposes to collect random samples of 64 observations to monitor labour time.

1If management wishes to establish x ? control limits covering the 95% confidence interval, calculate the appropriate UCL and LCL.
If management wishes to use smaller samples of 16 observations calculate the control limits covering the 95% confidence interval.
Management is considering three alternative procedures in order to maintain tighter control over labour time:

Sampling more frequently using 16 observations and setting confidence intervals of 90%
Maintaining 95% confidence intervals and increasing sample size to 64 observations
Setting 95% confidence intervals and using sample sizes of 36 observations.

Calculate the control limits for each of the 3 alternatives.

Which procedure will provide the narrowest control limits What are they
(b)
Hypothesis testing
Company A has invested a great deal of time and money in occupational safety training for its employees and claims its occupational sick days are now below the national average. The national average was found to be 1.5 occupational sick days. per 100 employees with a standard deviation of 0.3 days.

Company A randomly selected 100 employees for the last year and found the sample had a mean of 1.3 occupational sick days which Company A believed supported their claim.

Using hypothesis testing with an alpha level of 0.05 and a 1-tail test, show the null and alternative hypotheses, sketch the distribution showing mean and critical region, and determine whether Company A’s belief is supported.

Answer:

The following formulas are used to find the upper control limit UCL, and lower control limit LCL, (Montgomery, 2003)

UCL=mean-Z*StDev/sqrt(n)

LCL=mean+Z*StDev/sqrt(n)

Where mean is the sample mean (20),

StdDev is the population standard deviation (10),

n is the given size of the sample which will be changing for different parts of the problem, and

Z represents the critical value (as obtained from Z table) according to the given level of confidence which will change in different parts of the problem

In what follows, we substitute in the above formula

95% of confidence with Samples of 64 observations :

LCL=20-1.96*10/ sqrt (64)= 17.55

UCL=20+1.96*10/sqrt(64)= 22.45

95% of confidence with 16 observations

LCL= 20-1.96*10/ sqrt (16)= 15.1

UCL= 20+1.96*10/ sqrt (16)= 24.9

In this part we investigate how the level of confidence and sample size will affect the calculation of the confidence interval, more specifically the effect on the interval width.

Increasing the sample size to 16 observations and setting confidence level to 90%

L= 20-1.645*10/ sqrt( 16)= 15.8875

U= 20+1.645*10/ sqrt(16)= 24.1125

Keeping the confidence level at 95% and upraising the sample size to 64 observations

L = 20-1.96*10/ sqrt(64)= 17.55

U=20+1.96*10/ sqrt(64)= 22.45

Holding the confidence level at 95% while reducing the sample sizes of to 36 observations.

L= 20-1.96*10/ sqrt(36)= 16.73333

U= 20+1.96*10/ sqrt(36)= 23.26667

As seen from the result above the second confidence interval has narrower limits (17.55, 22.45)

(b) This is left tail Z test for mean ( Johnson, R.A., Bhattacharyya, G.K., 2014)Null hypothesis: µ ≥ 1 .5

Alternative Hypothesis: µ < 1.5

The test statistic is Z is computed by the formula

z=(mean-1.5)*sqrt(n)/StDev=(1.3-1.5)*sqrt(100)/0.3= -6.6667

-1.6449 is the Critical z is obtained from the Z table corresponding to left tail 0.05

The following graph gives idea about the rejection region of size 0.05 to the left of the critical value (-1.645). Since the observed Z=-6.7 is very far below the critical value, then we reject the null hypothesis and declare the validity of the company claim at 0.05 level of significance.