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
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
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)
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)
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
14.
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
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
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.