Statistical Quality Control Analysis

This assignment focuses on applying statistical quality control techniques to real-world scenarios. It explores calculating control limits for labor time using sampling procedures and analyzing occupational sick days with a one-sample t-test. The assignment delves into confidence intervals for different sample sizes and hypothesis testing to determine if company A's occupational sick days are indeed below the national average.

Running Head: DECISION SUPPORT TOOL
Decision Support Tool
Name of the Student
Name of the University
Author Note

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Table of Contents
ANSWER 1.....................................................................................................................................2
ANSWER 2.....................................................................................................................................5
ANSWER 3.....................................................................................................................................6
REFERENCES................................................................................................................................8

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ANSWER 1
a) A random variable is a variable that takes unknown values or functions that
assigns values to every outcome of an experiment.
Discrete random variable and continuous random variables are the two types of
random variables (Miller & Miller, 2015).
A random variable is said to be discrete if it assumes a finite number of values or
a countably infinite number of values. For example, the sum of points obtained in two
throws of a fair die is a discrete random variable.
A random variable is said to be continuous if it assumes uncountably infinite
number of values. For example, the decay time of a radioactive particle is a continuous
random variable.
b) The term expected value of a random variable means the value predicted for the
respective random variable. It is denoted by E(X) and serves as a measure of central
tendency of the probability distribution of X. It is usually defines as the sum of the values
of the random variables multiplied by their respective probabilities. If X assumes the
values x1, x2, …., xk with respective probabilities p1, p2, …., pk, where ∑
i=1
k
xi pi=1, then
E ( X ) =∑
i=1
k
xi pi, provided it is finite.
For example, let a die be rolled once. Let X be the event that the number obtained will be
less than 4. Obtaining a number less than 5 has the following probability distribution:

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Value of X (x) 1 2 3 4
P(X=x) 1/6 1/6 1/6 1/6
Now, E(X) = (1 * 1/6) + (2 * 1/6) + (3 * 1/6) + (4 * 1/6) = 1.67.
c) The expected values of the variables, their respective probabilities and cumulative
probabilities have been calculated and given in the following table.
Sales Units (x) Number of days p(x) Exp Value More than Less than
1 20 0.2 0.2 1 0.2
2 40 0.4 0.8 0.8 0.6
3 20 0.2 0.6 0.4 0.8
4 10 0.1 0.4 0.2 0.9
5 10 0.1 0.5 0.1 1
Total 100 1 2.5
 The probability of selling 1 or 2 units on any one day is (0.2 * 0.4) = 0.08
 The average daily sales is 2.5 units.
 The probability of selling 3 or more units a day is 0.4
 The probability of selling 4 or less units a day is 0.9.
d) The average sales of apples are 5000 with a standard deviation of 500.
 The z score to find out the probability that the sales of apples will be greater than
5500 is Z = ((5500-5000)/500) = 1. The normal probability value corresponding to
this z value obtained from the normal distribution table is 0.8413. The probability that

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the sales will be less than 5500 apples is 0.8413. Thus, the probability that the sales
will be greater than 5500 apples is 1-0.8413 = 0.1587.
 The z score to find out the probability that the sales will be less than 4900 apples is z
= ((4900-5000)/500) = -0.2. The probability corresponding to the z score of -0.2 is
0.4207. Thus, the required probability that the sales will be less than 4900 apples is
0.4207.
 The z score to find out the probability that the sales will be less than 4250 apples is z
= ((4250-5000)/500) = -1.5. The probability corresponding to the z score of -1.5 is
0.0668. Thus, the required probability that the sales will be less than 4250 apples is
0.0668.

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ANSWER 2
Table 2.1: Age and Sex composition of Australia in 2016
Male Female
0-14 2,331,078 2,208,167 4,539,245
15-24 1,622,984 1,540,667 3,163,651
25-54 4,945,735 5,004,821 9,950,556
55-64 1,371,243 1,421,315 2,792,558
65 and over 1,719,932 2,219,507 3,939,439
Total 11,990,972 12,394,477 24,385,449
Age Sex Total
Source: abs.gov.au, 2017
 The probability that any person selected at random from the population is a male is
(11,990,972/24,385,449) = 0.49.
 The probability that any person selected at random from the population is aged between 15
and 24 is (3,163,651/24,385,449) = 0.13
 The joint probability that any person selected at random from the population is a female and
aged between 15 and 24 is ((3,163,651 / 24,385,449)*(12,394,477/24,385,449) = 0.07.
 The conditional probability that any person selected at random from the population is 25 or
over given that the person is a male is ((4,954,735 + 1,371,243 + 1,719,932) / 24,385,449) =
0.33 (De Finetti, 2017).

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ANSWER 3
The mean labour time of a company has been 20 hours with a standard deviation of 10
hours. The labour hours are said to be distributed normally. A random sample of 64 observations
has been collected to monitor the labour hours. The standard error (SE) can be thus given by
standard deviation / √n, where n is the sample size. Therefore, the standard error is SE = 10/8 =
1.25.
1. The management wishes to establish X control limits covering 95% of the confidence
interval. Here X denotes the mean and SD denotes the standard deviation. The UCL is given
by (X + 2 * SE) = 20 + (2 * 1.25) = 22.5 and the LCL is given by ( X – 2 * SE) = 20 - (2 *
1.25) = 17.5 (Miah, 2016).
2. If the sample size is decreased to 16, then the Control limits covering 95% confidence
interval is given by:
UCL = (X + A3 * SD) = (20 + 0.763 * 10) = 27.63.
CL = Mean = 20
LCL = (X - A3 * SD) = (20 - 0.763 * 10) = 12.37.
3. Now the management is considering three alternative procedures in order to maintain tighter
control over labour time.
 The 90% confidence interval for sampling more frequently using 16 observations is
given by ( X ± 1.645 * SD / √n ¿, where n is the sample size. Thus, the 90% confidence
limit is given by (-3.03, 36.45).

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 The 95% confidence interval for a sample size of 64 observations is given by ( X ± 1.96
* SD / √n ¿ where n is the sample size. Thus, the 95% confidence limit is given by
(17.55, 22.45).
 The 95% Confidence interval for a sample size of 36 observations is given by ( X ± 1.96
* SD / √n ¿ where n is the sample size. Thus, the 95% confidence limit is given by
(16.73, 23.27).
The first procedure provides with the narrowest control limits.
b) A lot of money was invested by company A for occupational safety training for its
employees. The company claimed that their occupational sick days are 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.
The null hypothesis for this problem can be given as:
H0: The occupational sick days per 100 person are equal to 1.5 days
The alternate hypothesis can be given as:
HA: The occupational sick days per 100 person is less than 1.5 days
This problem can be tested by using a one sample t-test. The test statistic fir the required test can
be given as:
t= X−μ
σ
√ n
,

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Where X = the population mean, μ = sample mean, σ = population standard deviation and n =
the sample size. Therefore, the value of the test statistic is
t= 1.5−1.3
0.3
√ 100
= 6.67.
The tabulated value of the t statistic at 0.05 level of significance with 100 degrees of
freedom is 1.66. The critical region of the sample will be given by w > 1.66. If the calculated
value of the t statistic falls in the critical region, then the null hypothesis will be rejected. In this
problem the calculated value of the t-statistic is 6.67 which falls in the critical region. Thus, the
null hypothesis is rejected. This test thus supports the belief of company A.

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REFERENCES
3101.0 - Australian Demographic Statistics, Sep 2016. (2017). Abs.gov.au. Retrieved 9 August
2017, from http://www.abs.gov.au/AUSSTATS/abs@.nsf/DetailsPage/3101.0Sep
%202016?OpenDocument
De Finetti, B. (2017). Theory of probability: A critical introductory treatment(Vol. 6). John
Wiley & Sons.
Miah, A. Q. (2016). Statistical Quality Control. In Applied Statistics for Social and Management
Sciences (pp. 353-384). Springer Singapore.
Miller, I., & Miller, M. (2015). John E. Freund's mathematical statistics with applications.
Pearson.