MBA in Management
Statistics for Managers
Final Exam Paper
1. Samples of light bulbs were obtained from two suppliers and tested ‘to destruction’ in the laboratory. The following results were obtained for the length of life:
Length of Life (hours)
700-899
900-1099
1100-1299
1300-1499
Total
Supplier A
14
16
26
12
68
Supplier B
5
36
21
5
67
(a) Which supplier’s bulbs have greater average length of life?
For easier computation, the class intervals were assigned numbers. Interval 700-899 is 1; 900-1099 is 2; 1100-1299 is 3; and 1300-1499 is 4.
SUPPLIER A = 14 (1) + 16 (2) + 26 (3) + 12 (4) / 68
= 2.53 or 3
l 3 or class interval 1100-1299 is the average length of life for Supplier A.
SUPPLIER B = 5 (1) + 36 (2) + 21 (3) + 5 (4) / 67
= 2.39 or 2
l 2 or class interval 900-1099 is the average length of life for Supplier B.
ll Supplier A has bulbs that have a greater average length of life, from 1100 to 1299 hours, compared to Supplier B, whose bulbs have an average length of life is from 900-1099 hours only.
(b) Which supplier’s bulbs are more uniform in quality?
Supplier A mean = 17 Supplier A standard deviation = 5.38
Supplier B mean = 16.75 Supplier B Standard deviation = 12.89
5.38
17
= .3165 or 31.7%
12.89
16.75
= .77 or 77%
l The bulbs of Supplier A are more uniform in quality, as a lower coefficient of variation denotes that the bulbs are more uniform in quality.
(c) Which supplier would you prefer to use?
I would prefer to use Supplier A based on the above computations, because their light bulbs proved to be more cost-efficient, as the bulbs have a greater average length of life and are more uniform in quality than those of Supplier B.
2. At the end of an MBA course in business statistics, the final examination grades have a mean of 70.2 and a standard deviation of 12.3. There were 210 students on the course. Assuming that the distribution of these grades (all whole numbers) in normal, find:
(a) the percentage of grades that should exceed 90
Z = 90 – 70.2
12.3
= 1.61
P (to the right of Z = 1.61)
= .5 – .4463
= .054 or 5.4%
l There is a 5.4% percentage of grades that should exceed 90.
(b) the percentage of grades less than 35
Z = 35 – 70.2
12.3
= -2.86
P (to the left of Z = -2.86)
= .5 – .4979
= .0021 or .21%
l There is a .21% percentage of grades less than 35.
(c) the number of failures (pass = 70 per cent)
69.9 – 70.2
12.3
Z = .024
P (to the right of Z = .024)
P = .5 – .0080
= .492 or 49.2%
49.2 % x 210 = 103.32 or 104
l There are 104 students out of 210 students who failed the final examination.
(d) the lowest distinction mark if at most the highest 5 per cent of grades are to be awarded distinctions.
Z = .45 = 1.645
x – 70.2
12.3
x = (12.3) (1.645) + 70.2
= 90.43
l The lowest distinction mark in the highest 5 percent of grades is 90.43.
3. , researcher for the Colombian Coffee Corporation is interested in determining the rate of coffee usage per household in the United Sates. She believes that yearly consumption per household is normally distributed with an unknown mean and a standard deviation of about 2 pounds.
a) If takes a sample of 16 households and records their consumption of coffee for one year, what is the probability that the sample mean is within one pound of the population mean?
Z = .1915 x 2
= .383 or 38.3%
l The probability that the sample mean is within one pound of the population mean is 38.3%.
b) How large a sample must she take in order to be 98 percent certain that the sample mean is within one pound of the population mean?
38.3 98
16 n
n = 40.94 or 41
l Ms. must take 41 samples in order to be 98 percent certain that the sample mean is within one pound of the population mean.
4. A chain of department stores is moving into a phase of expansion and opening several new stores. As part of the expansion planning process a project team is carrying out an investigation to find out how the sales levels at new stores might be predicted. One approach has been to use regression analysis. The average level of sales per week (y) for each existing store in the chain (fourteen department stores) together with a measure of disposable income per family in each store’s catchment area are given in Table 11.1
Table 11.1
Store number
Average sales per week (₤000s)
Y
Average disposable family income (coded)
x
1
90
301
2
97
267
3
86
397
4
84
227
5
82
273
6
80
253
7
78
203
8
75
263
9
70
190
10
68
212
11
64
157
12
61
141
13
58
119
14
52
133
Mean
74
217
The computer gives the following results after regressing sales against family income:
Variable Coefficient
Income 0.17846
Constant 35.228
Correlations Coefficient = 0.92
Standard error = 4.72
(a) According to the above table, what is the predictable average sales per week with average family disposable income at 221?
y = a + bx
= 35.228 + .17846 (221)
y = 74.67
l The predictable average sale per week assuming disposable income is at 221 is 74.67.
(b) How good is the application of regression analysis in the above data?
The correlations coefficient and the standard error results are suffice evidence that the application of regression analysis in the above data is reliable. The data from the average sales per week is the endogenous and dependent variable in the regression analysis, while the average disposable family income is the exogenous or the independent variable, changes on it that the former reacts to. Correlation coefficient values ranges from negative one to positive one, and a value closer to one (either negative or positive) indicates higher relationship between the two variables. Two closely related variables used in a regression analysis produce more accurate results in showing how changes in the independent variable affects the dependent variable, which, in this case, is the sales per week.
(c) What are the limitations of this method to predict the average sales per week?
For one, there are other factors affecting the computation of sales other than the disposable family income of the consumers. These other factors may come in the form of inflation and interest rates, equity markets, amount of advertising expenditures and number of competitors, which the use of a multiple linear regression would sufficiently accommodate. The other limitation is that the values that can be accurately predicted are restricted to a certain range. Thus, computations for values outside the range will be doubtful because the investigation provided no evidence as to the nature of the statistical relationship outside the existing range.
5. Factory A conducted a sample check of total number of apples per crate and get the following results. The Quality Manager would like to see if the data are out of control.
Sample 1
Sample 2
Sample 3
Sample 4
Sample 5
110
93
99
98
109
103
95
109
95
98
97
110
90
97
100
96
102
105
90
96
105
110
109
93
98
110
91
104
91
101
100
96
104
93
96
93
90
110
109
105
90
105
109
90
108
103
93
93
99
96
97
97
104
103
92
103
100
91
103
105
90
101
96
104
108
97
106
97
105
96
99
94
96
98
90
106
93
104
93
99
90
95
98
109
110
96
96
108
97
103
109
96
91
98
109
90
95
94
107
99
91
101
96
96
109
108
97
101
103
94
96
97
106
96
98
101
107
104
109
104
96
91
96
91
105
(a) Calculate LCL & UCL
UCLx = x + A2 R UCLR = D4 R
= 99.54 + (.58) 19.8 = 2.11 (19.8)
UCLx = 111.02 UCLR = 41.78
LCLx = x – A2 R LCLR = D3 R
= 99.54 – (.58) 19.8 = 0 (19.8)
LCLx = 88.06 LCLR = 0
(b) Calculate Average of Mean ( x )
Sample1x= 99 Sample2x= 98 Sample3x= 101
Sample4x= 98.7 Sample5x= 101
x = 99 + 98 + 101 + 98.7 + 101
x = 99.54
(c) Draw the X – R chart
Credit:ivythesis.typepad.com
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