The learning outcomes of this topic are: - Perform a single sample t-test of the mean - Perform a two sample t-test - Interpret significance probabilities - Perform a x2 goodness of fit test This topic will cover: - Hypothesis testing with a sample (confidence intervals, fixed level, significance testing) - Two sample t-test - Significance, errors and power - Frequency data and the x2 test

1. Topic 4:
Inferential Statistics 2

2. This topic will cover:
◦ Hypothesis testing with a sample
 CI, fixed level, significance testing
◦ Two sample t-test
◦ Significance, errors and power
◦ Frequency data and the c2 test

3. By the end of this topic students will be able
to:
◦ Perform a single sample t-test of the mean
◦ Perform a two sample t-test
◦ Interpret significance probabilities
◦ Perform a c2 goodness of fit test

4. ◦ A car manufacturer releases a new car and claims that
its urban cycle fuel efficiency is 18.5 km per litre. A car
enthusiast magazine decides to test this claim.
 Null hypothesis
 H0: m = 18.5
 Alternative hypothesis is
 H1: m ≠ 18.5

6. 𝑥 = 17.79 𝑠 = 0.4782
𝜇−, 𝜇+ = 𝑥 − 𝑡 𝛾
𝑠
𝑛
, 𝑥 + 𝑡 𝛾
𝑠
𝑛
= (17.20, 18.38)
a1
5.00% 2.50% 1.00% 0.50%
a2
10.00% 5.00% 2.00% 1.00%
g 90.00% 95.00% 98.00% 99.00%
n = n - 1
1 6.3138 12.7062 31.8205 63.6567
2 2.9200 4.3027 6.9646 9.9248
3 2.3534 3.1824 4.5407 5.8409
4 2.1318 2.7764 3.7469 4.6041

7. 2.5%2.5%
17.20 18.38
95%

8. ◦ Process:
 State null and alternative hypotheses
 Decide test statistic and its distribution given H0 is true
 Set rejection region (the significance level for the test)
 Calculate test statistic from sample
 Does test statistic fall in rejection region?
◦ Under H0 the statistic T
𝑇 =
𝑥 − 𝜇
𝑠
𝑛
◦ is distributed as T ~ t(n- 1)

9. 2.5%2.5%
-2.7764 2.7764
95%
t(4)

10. H0: m = 18.5 H1: m ≠ 18.5
𝑥 = 17.79 𝑠 = 0.4782
𝑇 =
𝑥 − 𝜇
𝑠
𝑛
=
17.79 − 18.5
0.4782
5
= −3.32
At the 1% significance level the data do not provide
sufficient evidence to reject H0, a mean urban cycle fuel
efficiency of 18.5 km per litre is plausible.
0.5%0.5%
-4.6041 4.6041
99%
t(4)

11. H0: m = 18.5 H1: m< 18.5
𝑥 = 17.79 𝑠 = 0.4782
𝑇 =
𝑥 − 𝜇
𝑠
𝑛
=
17.79 − 18.5
0.4782
5
= −3.32
H0 is not rejected at the 1% significance level, a mean
urban cycle fuel efficiency of 18.5 km per litre is
plausible.
1.0%
-3.7469
99%
t(4)

12. H0: m = 18.5 H1: m≠ 18.5
𝑥 = 17.79 𝑠 = 0.4782
𝑇 =
𝑥 − 𝜇
𝑠
𝑛
=
17.79 − 18.5
0.4782
5
= −3.32
𝑃 𝑇 > 3.32 = 𝑝 = 0.0294
1.47%1.47%
-3.32 3.32
t(4)

13. ◦ Motivation
 To test whether the means of two
populations are equal
◦ Assumptions:
 Both populations essentially normally
distributed
 Independence of populations
 Sample variances differ by less than a factor
of 3

15. Under H0 the statistic T
𝑇 =
𝑥1 − 𝑥2
𝑆 𝑝
1
𝑛1
+
1
𝑛2
is distributed as T ~ t(n1+n2- 2), where
 𝑆 𝑝
2
=
𝑛1−1 𝑠1
2+ 𝑛2−1 𝑠2
2
𝑛1+𝑛2−2



16. ◦ Two machines on a production line are set to produce
the same part. Workers down the line are concerned
that the two machines are not producing parts of the
same size. Test the hypothesis.
◦ H0: m1 – m2 = 0 H1: m1 – m2 ≠ 0
◦ Sample machine 1: 28.54, 28.27, 25.81, 28.41,
29.09, 28.09, 29.72, 23.56
◦ Sample machine 2: 31.85, 29.96, 31.97, 33.68, 23.81,
30.79

17. 2.5%2.5%
-2.1788 2.1788
95%
t(12)
5.0%
-1.7823
95%
t(12)
a1
5.00% 2.50% 1.00% 0.50%
a2
10.00% 5.00% 2.00% 1.00%
g 90.00% 95.00% 98.00% 99.00%
n = n1 + n2 - 2
11 1.7959 2.2010 2.7181 3.1058
12 1.7823 2.1788 2.6810 3.0545
13 1.7709 2.1604 2.6503 3.0123

18. ◦ Sample machine 1: 28.54, 28.27, 25.81, 28.41, 29.09,
28.09, 29.72, 23.56
◦ Sample machine 2: 31.85, 29.96, 31.97, 33.68, 23.81,
30.79
◦ 𝑥1 = 27.68, 𝑠1 = 2.014, 𝑥2 = 30.34, 𝑠2 = 3.44
◦ 𝑆 𝑝
2 =
𝑛1−1 𝑠1
2+ 𝑛2−1 𝑠2
2
𝑛1+𝑛2−2
=
8 − 1 4.1+ 6 − 1 11.8
8+6−2
= 7.3
◦ 𝑇 =
𝑥1−𝑥2
𝑆 𝑝
1
𝑛1
+
1
𝑛2
=
−2.66
7.3
1
8
+
1
6
= −1.82
◦ H0: m1 – m2 = 0 H1: m1 – m2 ≠ 0
◦ H0 cannot be rejected at the 5% significance level

19. Actual
H0 H1
Decision
H0 ☺ Type 2
H1 Type 1 ☺
a = 5%
b
Actual
H0 H1
Decision
H0 1 - a b
H1 a 1 - b
significance
power
Distribution of Sample Means
z = 1.6449

20. ◦ Motivation
 Decide reasonableness of using particular
probability model to describe empirical data
◦ Example uses
 Does Poisson probability distribution fit
empirical data?
 Are two categorical variables independent?

21. Number Frequency
0 10
1 17
2 42
3 34
4 12
5 5
>5 0
120
Average number of customers per
hour = 2.3
𝑃 𝑋 = 𝑥 =
e−𝜆
𝜆 𝑥
𝑥!
=
e−2.3
2.3 𝑥
𝑥!

22. Number Frequency
observed
0 10
1 17
2 42
3 34
4 12
5 5
>5 0
120
𝜒2
=
𝑖=1
𝑘
𝑂𝑖 − 𝐸𝑖
2
𝐸𝑖
≈ 𝜒2
𝑘 − 𝑚 − 1
• k is number of categories
• m is number of estimated
parameters in model
• Only valid if Ei ≥ 5 for all i
Frequency
expected
0.1003 12.0311
0.2306 27.6714
0.2652 31.8222
0.2033 24.3970
0.1169 14.0283
0.0538 6.4530
0.0300 3.5971
1 120

23. Number Frequency
observed
Frequency
expected
0 10 0.1003 12.0311
1 17 0.2306 27.6714
2 42 0.2652 31.8222
3 34 0.2033 24.3970
4 12 0.1169 14.0283
>4 5 0.0838 10.0501
120 1 120
𝜒2
=
𝑖=1
𝑘
𝑂𝑖 − 𝐸𝑖
2
𝐸𝑖
≈ 𝜒2
𝑘 − 𝑚 − 1
• k = 6
• m = 1
• Valid since Ei ≥ 5 for all i

24. significance level
p = 0.05 p = 0.01
aR = 5.00%
aR =
1.00%
n = k - m -1
1 3.841 6.635
2 5.991 9.210
3 7.815 11.345
4 9.488 13.277
5 11.070 15.086
etc.
aR
c2

25. Number Frequency
observed
Frequency
expected
(O – E)2/E
0 10 0.1003 12.0311 0.34
1 17 0.2306 27.6714 4.12
2 42 0.2652 31.8222 3.26
3 34 0.2033 24.3970 3.78
4 12 0.1169 14.0283 0.29
>4 5 0.0838 10.0501 2.54
120 1 120 14.32
𝜒2 =
𝑖=1
𝑘
𝑂𝑖 − 𝐸𝑖
2
𝐸𝑖
= 14.32
H0 is rejected at the 5% significance level.

26. Motivation
• Is there an association between two
categorical variables?
Method of Follow-up
email phone visit
satisfied 15 22 17 54
neither 6 6 3 15
dissatisfied 9 2 10 21
30 30 30 90

27. • Is there an association between two
categorical variables?
• What are expected counts with no
association?
Method of Follow-up
email phone visit
90

28. • Is there an association between two categorical
variables?
• What are expected counts with no association?
Method of Follow-up
satisfied
neither
dissatisfied
90

29. • Is there an association between two categorical
variables?
• What are expected counts with no association?
Method of Follow-up
email phone visit
satisfied
neither
dissatisfied
90

30. • Observed and expected counts.
Method of Follow-up
email phone visit
satisfied 15 (18) 22 (18) 17 (18) 54
neither 6 (5) 6 (5) 3 (5) 15
dissatisfied 9 (7) 2 (7) 10 (7) 21
30 30 30 90
𝜒2
=
𝑖=1
𝑘
𝑂𝑖 − 𝐸𝑖
2
𝐸𝑖
= 8.07 𝜒2
≈ 𝜒2
𝑟 − 1 𝑐 − 1

31. ◦ Simple random sampling
◦ Systematic random sampling
◦ Stratified random sampling
◦ Cluster (area) sampling

32. ◦ Quota sampling
◦ Availability / convenience sampling
◦ Purposive sampling

33. By the end of this topic students will be able
to:
◦ Perform a single sample t-test of the mean
◦ Perform a two sample t-test
◦ Interpret significance probabilities
◦ Perform a c2 goodness of fit test

34. • Buglear, J. Quantitative Methods for Business.
Elsevier Butterworth Heinemann.
• Hinton, PR. Statistics Explained. Routledge.
• Thomas, AB. Research Skills for Management
Studies. Routledge.
• Thomas, AB. Research Concepts for
Management Studies. Routledge.
• Wisniewski, M. Quantitative Methods for
Decision Makers. FT Prentice Hall.

35. Any Questions?