5. Business Statistics Third Level
Series 2 2002
QUESTION 1
In the year 2000 a random sample of 96 small companies revealed the following profits and losses
distribution:
Profits and Losses (£000) Number of
companies
-50 up to 0 10
0 and up to 10 12
10 and up to 20 21
20 and up to 40 26
40 and up to 80 19
80 and up to 150 8
(a) Calculate the arithmetic mean and the standard deviation of these company profits and losses.
(8 marks)
In the previous year a random sample of 150 small companies showed a mean profit of £28,400 with a
standard deviation of £28,352.
(b) Test to see if there has been a significant increase in the average profit made by small
companies.
(7 marks)
(c) Calculate a 99% confidence interval for the arithmetic mean in the year 2000.
(5 marks)
(Total 20 marks)
3
6. Model Answer to Question 1
(a)
2
f mid pt fx fx
–50 up to 0 10 –25 –250 6,250
Over 0 and up to 10 12 5 60 300
Over 10 and up to 20 21 15 315 4,725
Over 20 and up to 40 26 30 780 23,400
Over 40 and up to 80 19 60 1,140 68,400
Over 80 and up to 150 8 115 920 105,800
96 2,965 208,875
åf å fx å fx 2
x= åx x = 2,965 = £30.89 (000) (£30,885)
åf 96
2 æ ö
2 2
s = å fx – ç å fx ÷
çåf ÷
s= 208,875 æ 2,965 ö
–ç ÷
åf è ø 96 è 96 ø
= 2,175.78 – 953.91 = 1,221.87 = £34.96 (000) (£34,955)
(b) Null hypothesis: There has not been an increase in the average level
of profits
Alternative hypothesis: There has been an increase in the average level
of profits
Critical z for 0.05 significance level I tail test 1.64
30.89 – 28.4 2.49
z = x1 – x2 = = 0.59
s12 + s22 34.962 28.3522 18.09
+
n1 n2 96 150
Alternative answer: 0.58
Conclusions: Accept the null hypothesis, there is insufficient
evidence to claim the average level of profits has
increased.
σ 34.955
(c) ci = x ± 2.58 = 30.89 ± 2.58
n 96
= 30.89 ± 2.58 x 3.57 = 30.89 ± 9.21
= £21.68 to £40.1 (000)
4
7. QUESTION 2
The price of the standard family saloon car and the company market share was recorded for a random
sample of 12 car manufacturers.
Selling 137 138 125 142 168 145 135 145 160 146 136 160
price £00
Market 14 15 10 8 9 7 11 5 3 5 7 2
share %
(a) Plot the data on a scatter diagram and comment.
(4 marks)
(b) Calculate the product-moment correlation coefficient.
(10 marks)
(c) Test to find if the correlation coefficient differs significantly from zero.
(6 marks)
(Total 20 marks)
5
8. Model Answer to Question 2
(a) Price and Market Share
16
14
12
Market Share %
10
Series 1
8
6
4
2
0
12,000 13,000 14,000 15,000 16,000 17,000 18,000
Price £
Comment : Some weak negative relationship.
137 14 18,769 196 1,918
138 15 19,044 225 2,070
125 10 15,625 100 1,250
142 8 20,164 64 1,136
168 9 28,224 81 1,512
145 7 21,025 49 1,015
135 11 18,225 121 1,485
145 5 21,025 25 725
160 3 25,600 9 480
146 5 21,316 25 730
136 7 18,496 49 952
160 2 25,600 4 320
1,737 96 253,113 948 13,593
åx åy å x2 å y2 å xy
(b) r= n å xy – (å x ) (å y )
æ n å x 2 – (å x )2 ö æ n å y 2 – (å y )2 ö
ç ÷ç ÷
è øè ø
12 x 13,593 – 1,737 x 96
r=
(12 x 253,113 – 1,737 2 )(12 x 948 – 96 2 )
163,116 – 166,752
r=
(3,037,356 – 3,017,169 )(11,376 – 9,216 )
–3,636
r= = – 0.5506
(20,187 )(2,160 )
6 CONTINUED ON NEXT PAGE
9. Model Answer to Question 2 continued
(c) Null hypothesis: The correlation coefficient does not differ from zero.
Alternative hypothesis: The correlation coefficient does differ from zero.
Degree of freedom = n – 2 = 12 – 2 = 10
Critical t0.025 = 2.23
r n–2 – 0.55 12 – 2
t= t=
1– r2 1– ( −0.552 )
–1.61
t=
1– 0.3025
–1.61
=
0.835
= –2.086
Conclusions: The calculated value of t is less than the critical
value of t. There is insufficient evidence to reject the null
hypothesis.
The correlation coefficient does not differ from zero.
7
10. QUESTION 3
A company is planning the launch of a new product. It estimates the probability of good market
conditions to be 80%. If market conditions are good the probability of a successful launch is 75%, if
market conditions are poor the probability of a successful launch is 50%.
(a) Find the probability that the launch is successful.
(5 marks)
(b) If the product launch is unsuccessful what is the probability that the market conditions were poor?
(6 marks)
The estimated returns from the new product launch are:
Market conditions are good and the product launch is successful £55 million
Market conditions are good and the product launch is unsuccessful – £13 million
Market conditions are poor and the product launch is successful £37 million
Market conditions are poor and the product launch is unsuccessful – £19 million
(c) What is the expected profit from the new product launch?
(4 marks)
The company sells an established product that has variable levels of weekly sales with arithmetic
mean of £5,000 and standard deviation of £600. You may assume that the sales are normally
distributed.
(d) (i) Find the probability that in one week there are sales of over £6,500
(ii) The sales have to exceed £3,800 in each week for the product to break even, what is the
probability of this happening?
(5 marks)
(Total 20 marks)
8
11. Model Answer to Question 3
(a) Good market conditions and successful = 0.8 x 0.75 = 0.6
Poor market conditions and successful = 0.2 x 0.5 = 0.1
Probability of successful launch = 0.7
Accept decision tree
(b) Good and unsuccessful = 0.8 x 0.25 = 0.2
Poor and unsuccessful = 0.2 x 0.5 = 0.1
Probability unsuccessful = 0.3
Poor and unsuccessf ul 0.1
= = 0.33
Probability unsuccessf ul 0.3
(c) Expected return
Good market conditions and successful launch 0.8 x 0.75 x £55m = £33m
Good market conditions and unsuccessful launch 0.8 x 0.25 x –£13m = -£2.6m
Poor market conditions and successful launch 0.2 x 0.5 x £37m = £3.7m
Poor market conditions and unsuccessful launch 0.2 x 0.5 – £19m = –£1.9m
£32.2m
(d) (i) z = x–µ z= 6,500 – 5,000
=
1,500
sd 600 600
= 2.5, prop = 1 – 0.994 = 0.006
(ii) z = x–µ z= 3,800 – 5,000
=
1,200
= 2
sd 600 600
Proportion = 0.977
9
12. QUESTION 4
(a) What are the benefits of an effective quality control system?
(4 marks)
A company in its quality control procedures sets the warning limit at the 0.025 probability point and the
action limit at the 0.001 probability point. This means for example that the upper action line is set so
that the probability of the mean exceeding the line is 0.001.
The internal diameter of a bored hole is set at 35 mm with a known standard deviation of 0.1 mm.
Random samples of 9 items at a time are taken from the production line to check the accuracy of the
manufacturing process.
(b) (i) Construct a quality control chart to monitor the manufacturing process.
(8 marks)
(ii) The results for 8 samples are given below. Plot these on your quality control chart and
comment on the graph.
(4 marks)
Sample number 1 2 3 4 5 6 7 8
Sample mean (mm) 35.05 34.94 34.89 35.16 35.15 34.95 34.99 35.08
(c) If the process mean changed to 35.02 mm and the standard deviation remained at
0.1 mm calculate the probability that the mean of a random sample of 9 components would be
outside the warning limits.
(4 marks)
(Total 20 marks)
10
13. Model Answer to Question 4
(a) The benefits of a good quality control system are better quality products,
fewer rejects and less waste, better customer relations, more sales
Warning limits = x ± 1.96 σ = 35 ± 1.96
0 .1
= 35 ± 1.96 x 0.033
n 9
upper w lts 35.06 35.07(2)
lower w lts 34.94 34.93(2)
Action limits = x ± 3.09 σ = 35 ± 3.09
0 .1
= 35 ± 3.09 x 0.033
n 9
upper act lts 35.10
lower act lts 34.90
(b) Quality Control Chart
The process goes out of control twice. It seems unstable.
(c) z = x–µ =
35.02 – 35.07
0.1
=
0.05
= 1.5
sd 0.033
9
Probability the mean lies outside the upper warning limit = 0.067
34.93 – 35.02
= 2 .7
0 .1
9
Probability the mean lies outside the lower warning limit = 0.004
Total probability = 0.071 or (0.072 if 0.0333 is used)
11
14. QUESTION 5
(a) In what circumstances is a significance test based on the ‘t’ distribution used in preference to a
significance test based on the normal distribution?
(4 marks)
Data input clerks are sent on an intensive training course to increase their keyboarding speed. The
results of a before and after test for a random sample of 10 clerks gave the following results. Speed is
measured in key depressions per minute (kdp).
Clerk a b c d e f g h i j
Before training
course (kdp) 625 598 685 754 658 690 559 840 758 685
After training
course (kdp) 610 620 690 780 690 702 573 851 744 690
(b) Test whether the training course has increased the clerks’ data input speed.
(12 marks)
(c) What is meant by a Type I and a Type II error?
(4 marks)
(Total 20 marks)
12
15. Model Answer to Question 5
(a) The ‘t’ distribution is used when n the sample size is small <30 and the
standard deviation is estimated from the sample.
(b) Null hypothesis: The course has not increased the speed of the
data input clerks
Alternative hypothesis: The course has increased the speed of the
data input clerks
Degree of freedom n – 1 10 – 1 = 9
Critical t0.05 value = 1.83
625 610 –15 225 615.04
598 620 22 484 148.84
685 690 5 25 23.04
754 780 26 676 262.44
658 690 32 1,024 492.84
690 702 12 144 4.84
559 573 14 196 17.64
840 851 11 121 1.44
758 744 –14 196 566.44
685 690 5 25 23.04
98 3,116 2,155.60
2
åd å d2 å æd – d ö
ç ÷
è ø
æ
å çd – d ÷
ç ÷
ö2
å d2 –
(å d )2
d = åd =
98
= 9 .8 sed = è ø n
n 10 n –1 n –1
982
3,116 –
2,155.8 10
= = 239.53 = 15.48 or
10 – 1 10 – 1
If a two sample means
t = d –0 = 9.8 – 0
15.48
=
9 .8
=2 test is used can score
sed 4 .9
5: nh/ah, conclusions
10
Conclusions: The calculated value of ‘t’ is greater than critical value
of ‘t’ reject the Null Hypothesis the speed of the data input
clerks has increased
(c) A Type I error is the error of rejecting a null hypothesis that is true and
a Type II error is the error of accepting a false null hypothesis which
should be rejected.
13
16. QUESTION 6
(a) (i) What is meant by the standard error of the mean? (4 marks)
(ii) What is the difference between a one and two tail test? (4 marks)
In September 2001, a Travel Agent used a random sample of 36 holiday makers to find out the
average cost per person of a one week holiday in Ruritania. The following information was found:
September 2001
Mean £372.40
Standard deviation £26.10
Sample size 36
In the previous year the average cost of each holiday was £356.20.
(b) Test whether the cost of a one week holiday to Ruritania has increased significantly since the
previous year.
(6 marks)
The company based its views on customer satisfaction from the letters it receives. In the previous
year, 68% of the letters it received were of a positive nature.
(c) The company wishes to adopt a more scientific approach to estimating customer satisfaction.
What sample size would be needed to estimate the proportion of customers’ views to within 2% of
the true figure at the 95% confidence level?
(6 marks)
(Total 20 marks)
14
17. Model Answer to Question 6
(a) (i) The standard error of the mean is the measure of dispersion of the
sample means. It equals
σ
n
(ii) A one tail test tests the direction of the difference between two statistics.
A two tail test tests if there is a difference between the two statistics without
regard to the direction.
(b) Null hypothesis: There is no difference in the holiday cost between
last year and September
Alternative hypothesis: There is an increase in the holiday cost between
last year and September
Critical z value = 1.64
z = x –µ = 372.4 – 356.2
=
16.2
= 3.72
σ/ n 26.1 / 36 4.35
Conclusions: The calculated value of z is greater than the critical value
of z. Reject the null hypothesis. There is evidence to suggest
that the holiday cost has increased significantly.
0.02 0.02
(c) ± 1.96 > ± 1.96 >
p(1 – p) 0.68 x 0.32
n n
2 0.022 0.0004 3.8416 x 0.2176
1.96 > 3.8416 > n> > 2,089.8
0.68 x 0.32 0.2176 0.0004
n n
n = 2,090
15
18. QUESTION 7
(a) In what circumstances is the multiplicative model preferable to the additive model in time series
calculations?
(4 marks)
The table below shows the quarterly sales figures for a company:
Quarter 1 Quarter 2 Quarter 3 Quarter 4
1998 48 56 70 96
1999 64 76 86 132
2000 72 100 108 188
(b) By the method of centred moving averages, calculate the trend values for the time series and plot
the trend on a graph.
(10 marks)
The average seasonal components calculated by the multiplicative method are shown below:
Quarter 1 Quarter 2 Quarter 3 Quarter 4
0.768 0.903 0.980 1.349
(c) Using the average seasonal components provided above and the original sales figures, calculate
the seasonally adjusted sales and plot the results on the graph constructed for part (b). Comment
on your results.
(6 marks)
(Total 20 marks)
16
19. Model Answer to Question 7
(a) The multiplicative model is preferable when the data has a strong trend with the
differences between the trend and the original data values varying proportionally to
the trend values rather than absolutely.
(b)
(c) Qtr Sales Total m avg 1 m avg 2
Trend
1 48
2 56
3 70 270 67.5 69.5
4 96 286 71.5 74
5 64 306 76.5 78.5
6 76 322 80.5 85
7 86 358 89.5 90.5
8 132 366 91.5 94.5
9 72 390 97.5 100.25
10 100 412 103 110
11 108 468 117
12 188
17 CONTINUED ON NEXT PAGE
20. Model Answer to Question 7 continued
Sales Seasonal adjustment Seasonal Adjusted Sales
48 0.768 62.50
56 0.902 62.08
70 0.980 71.43
96 1.349 71.16
64 0.768 83.33
76 0.902 84.26
86 0.980 87.76
132 1.349 97.85
72 0.768 93.75
100 0.902 110.86
108 0.980 110.20
188 1.349 139.36
Comment: The trend and seasonally adjusted data values show little difference.
This implies there is no change in the underlying trend.
18
21. QUESTION 8
(a) When might a χ2 test be used?
(4 marks)
From its records, a company analyses its sales by size and region. The table below shows the results:
Size of order
Region £0 and up to £1,000 and up £3,000 and up £10,000 and
£1,000 to £3,000 to £10,000 over
Northern 40 33 20 15
Midlands 50 45 40 25
Southern 40 32 40 20
(b) Test whether there is a relationship between size of order and region.
(12 marks)
(c) Combining the three regions estimate a 99% confidence interval for the proportion of orders that
are valued at less than £1,000.
(4 marks)
(Total 20 marks)
19
22. Model Answer to Question 8
(a) The Chi-squared test is used to test for association between characteristics
rather than variables, for randomness or changes in proportions.
(b)
Null hypothesis: There is no association between the size of order
and region
Alternative hypothesis: There is association between the size of order
and region
Degrees of freedom = (4 – 1)(3 – 1) = 6
Critical χ2 = 12.59
<£1,000 £1,000<3,000 £3,000<10,000 £10,000+
Region 40 33 20 15
50 45 40 25
40 32 40 20
Total 130 110 100 60
expected freq 35.1 29.7 27.0 16.2
52.0 44.0 40.0 24.0
42.9 36.3 33.0 19.8
contribution 0.684 0.367 1.815 0.089
to chi 0.077 0.023 0.000 0.042
0.196 0.509 1.485 0.002
5.288
2
χ = å (O – E ) = 5.288
E
Conclusions: The calculated χ2 is less than the critical χ2 . There
is insufficient evidence to reject the null hypothesis.
There is no association between the size of order
and region.
(c)
99% confidence z = ± 2.58
p = 130/400 = 0.325, (1 – p) = 0.675
ci = p ± 2.58 p(1 – p ) / n = 0.325 ± 2.58 0.325 x 0.675 / 400
= 0.325 ± 2.58 x 0.0234 = 0.325 ± 0.0604
0.265 to 0.385
20
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