The document discusses testing hypotheses about the beta coefficients of technology shares. It provides: 1) Hypotheses to test if the average beta is greater than the market average of 1, using a one-tailed test at α=0.10. 2) Calculations to find the test statistic and critical value, showing the test statistic exceeds the critical value, so the null hypothesis is rejected. 3) A 95% confidence interval calculated for the population mean beta coefficient, ranging from 1.0251 to 1.4349. 4) A chi-square test showing the variance of the shares does not differ significantly from 0.15.

1. ECF 6102 - Quantitative
Skills for Business
Tutorial / Research Paper C
Wednesday 13 May
15:30
ML14:114
Andrew Dash

2. What is Beta Coefficient?
• In finance evaluations the “beta coefficient” of a share is often considered a
measure of the stocks volatility (or in financial terms “risk”).
• Shares with beta coefficients greater than 1 generally bear greater risk, and
hence more volatility, than the overall market whereas
• Shares with beta coefficients less than 1 generally are considered less risky
(or less volatile) than the overall market.
• From previous studies it has been established that the “beta coefficient”
has been normally distributed.

3. Research Situation
Imagine we take a random sample of 15 technology shares at the end of 2014.
The mean and standard deviation of the beta coefficients for these data are
calculated as:
𝑿 = 𝟏. 𝟐𝟑 & 𝒔 = 𝟎. 𝟑𝟕
Other Solution Essentials
• 𝑛 = 15 • 𝛼 = 0.10 • 𝜇 = 1
Previous studies have shown that the “beta
coefficient” has been normally distributed.

4. Part A
Set up the appropriate null and alternate
hypotheses to test whether the average
technology shares are riskier than the
market as a whole.

5. One orTwo tailed?
• Keywords in the question are‘…riskier than the market as a whole’.
• In other words, is the average beta coefficient of our sample greater than
the beta coefficient of the market as a whole?
• Therefore, it is a one tailed test.
𝑛 = 15 𝑋 = 1.23 𝑠 = 0.37 𝛼 = 0.10 𝜇 = 1
The Golden Rule
𝑋 > 𝜇 = 𝑈𝑝𝑝𝑒𝑟 𝑇𝑎𝑖𝑙 𝑇𝑒𝑠𝑡
𝑋 < 𝜇 = 𝐿𝑜𝑤𝑒𝑟 𝑇𝑎𝑖𝑙 𝑇𝑒𝑠𝑡
• As 𝑋 (1.23) is greater than 𝜇 (1), it is an upper tailed test.

6. The null & alternate hypotheses
• Setup the alternate hypothesis (𝐻1) first.
• The alternate hypothesis is setup to answer the question posed.
• Therefore:
𝐻0: 𝜇 ≤ 1
𝐻1: 𝜇 > 1
𝑛 = 15 𝑋 = 1.23 𝑠 = 0.37 𝛼 = 0.10 𝜇 = 1

7. Part B
Using your sampling decision tree,
establish the appropriate test statistic
and rejection region for the test using
alpha = 0.10.

8. Sampling DecisionTree
𝑛 = 15 𝑋 = 1.23 𝑠 = 0.37 𝛼 = 0.10 𝜇 = 1
(M Waring , personal communication, March 2015)

9. CriticalValues of t
𝑛 = 15 𝑋 = 1.23 𝑠 = 0.37 𝛼 = 0.10 𝜇 = 1 𝑡 𝑐𝑣 = +1.3450
(M Waring , personal communication, March 2015)

10. Part C
Test the hypothesis in Part A

11. The Decision Rule
• Our hypothesis: 𝐻0: 𝜇 ≤ 1
𝐻1: 𝜇 > 1
• Decision Rule: Reject 𝐻0 if 𝑡test > +1.3450, otherwise do not reject 𝐻0.
Do not reject H0
0
Reject H0
t
α=0.10
𝑡 𝑐𝑣 = +1.3450
𝑛 = 15 𝑋 = 1.23 𝑠 = 0.37 𝛼 = 0.10 𝜇 = 1 𝑡 𝑐𝑣 = +1.3450
Image adapted from example provided by M Waring (personal communication, 30 March 2015)

12. Testing our hypothesis
• Our test statistic: 𝑡𝑡𝑒𝑠𝑡 =
𝑋−𝜇
𝑠 𝑋
=
1.23−1
0.09553
= +2.4075
Do not reject H0
0
Reject H0
t
𝑡 𝑐𝑣 = +1.3450
𝑡𝑡𝑒𝑠𝑡 = +2.4075
𝑡𝑡𝑒𝑠𝑡 = +2.4075𝑛 = 15 𝑋 = 1.23 𝑠 = 0.37 𝛼 = 0.10 𝜇 = 1 𝑡 𝑐𝑣 = +1.3450
Image adapted from example provided by M Waring (personal communication, 30 March 2015)

13. Our Conclusion
• Since 𝑡𝑡𝑒𝑠𝑡 2.4075 > 𝑡 𝑐𝑣(1.3450), we reject 𝐻0 as there is enough
evidence to conclude that the average technology shares are riskier than the
market as a whole.
Null Hypothesis m= 1
Level of Significance 0.1
Sample Size 15
Sample Mean 1.23
Sample Standard Deviation 0.37
Standard Error of the Mean 0.0955
Degrees of Freedom 14
t Test Statistic 2.4075
Upper-Tail Test Calculations Area
Upper Critical Value 1.3450
p-Value 0.0152
Reject the null hypothesis
Data
Intermediate Calculations
t Test for Hypothesis of the Mean
𝑡𝑡𝑒𝑠𝑡 = +2.4075𝑛 = 15 𝑋 = 1.23 𝑠 = 0.37 𝛼 = 0.10 𝜇 = 1 𝑡 𝑐𝑣 = +1.3450

14. Part D
Use PhStat to determine the approximate p-
value associated with this test.
What does the p-value theoretically mean?

15. PhStat and the p-Value
• “The p-value is the probability of getting a test statistic equal to or more
extreme than the sample result, given that the null hypothesis, 𝐻0 is
true.The p-value is also known as the observed level of significance”
(Berenson, Levine, & Krehbiel, 2012).
𝑡𝑡𝑒𝑠𝑡 = +2.4075𝑛 = 15 𝑋 = 1.23 𝑠 = 0.37 𝛼 = 0.10 𝜇 = 1 𝑡 𝑐𝑣 = +1.3450

16. PhStat and the p-Value
• The strength of the decision concerning H0 is found by
comparing the p-value to the alpha () level.
• If p-Value is low, 𝐻0 must go…
𝑡𝑡𝑒𝑠𝑡 = +2.4075𝑛 = 15 𝑋 = 1.23 𝑠 = 0.37 𝛼 = 0.10 𝜇 = 1 𝑡 𝑐𝑣 = +1.3450
Null Hypothesis m= 1
Level of Significance 0.1
Sample Size 15
Sample Mean 1.23
Sample Standard Deviation 0.37
Standard Error of the Mean 0.0955
Degrees of Freedom 14
t Test Statistic 2.4075
Upper-Tail Test
Upper Critical Value 1.345030374
p-Value 0.015213385
Reject the null hypothesis
Data
Intermediate Calculations
t Test for Hypothesis of the Mean
Our Data
• p-Value (0.0152) < 𝛼 (0.1).
• Therefore we reject 𝐻0.

17. PhStat and the p-Value
• The p-Value is an important result because it measures the amount of
statistical evidence that supports the alternative hypothesis.
• A small p-Value indicates that there is ample evidence to support the
alternative hypothesis.
• A large p-Value indicates that there is little evidence to support the
alternative hypothesis.
𝑡𝑡𝑒𝑠𝑡 = +2.4075𝑛 = 15 𝑋 = 1.23 𝑠 = 0.37 𝛼 = 0.10 𝜇 = 1 𝑡 𝑐𝑣 = +1.3450

18. Part E
If we had tested this same hypothesis
using the Z distribution at alpha = 0.10
would you have drawn the same
conclusion?

19. Finding the value of 𝑍𝑡𝑒𝑠𝑡 & 𝑍 𝑐𝑣
𝑡𝑡𝑒𝑠𝑡 = +2.4075𝑛 = 15 𝑋 = 1.23 𝑠 = 0.37 𝛼 = 0.10 𝜇 = 1 𝑡 𝑐𝑣 = +1.3450
𝑍 𝑐𝑣 = +1.28
𝑍𝑡𝑒𝑠𝑡 =
𝑋 − 𝜇
𝑠 𝑋
=
1.23 − 1
0.09553
= +2.4075
(M Waring , personal communication, March 2015)

20. Testing our hypothesis
• Our decision rule: Reject 𝐻0 if 𝑍𝑡𝑒𝑠𝑡 +2.4075 > 𝑍 𝑐𝑣(+1.28), otherwise
do not reject 𝐻0
Do not reject H0
0
Reject H0
t
𝑍 𝑐𝑣 = +1.28
𝑍𝑡𝑒𝑠𝑡 = +2.4075
𝑍𝑡𝑒𝑠𝑡 = +2.4075𝑛 = 15 𝑋 = 1.23 𝑠 = 0.37 𝛼 = 0.10 𝜇 = 1 𝑍 𝑐𝑣 = +1.28
Image adapted from example provided by M Waring (personal communication, 30 March 2015)

21. Our Conclusion
• Since 𝑍𝑡𝑒𝑠𝑡 2.4075 > 𝑍 𝑐𝑣(1.28), therefore we reject 𝐻0 as there is enough
evidence to conclude that the average technology shares are riskier than the
market as a whole.
𝑍𝑡𝑒𝑠𝑡 = +2.4075𝑛 = 15 𝑋 = 1.23 𝑠 = 0.37 𝛼 = 0.10 𝜇 = 1 𝑍 𝑐𝑣 = +1.28
Null Hypothesis m= 1
Level of Significance 0.1
Population Standard Deviation 0.37
Sample Size 15
Sample Mean 1.23
Standard Error of the Mean 0.0955
Z Test Statistic 2.4075
Upper-Tail Test
Upper Critical Value 1.2816
p-Value 0.0080
Reject the null hypothesis
Data
Intermediate Calculations
Z Test of Hypothesis for the Mean

22. Part F
Develop a 95% confidence interval for the
“beta coefficient”.

23. Solution Essentials
𝑛 = 15 𝑥 = 1.23 𝑠 = 0.37 1 − 𝛼 = 95%
(M Waring , personal communication, March 2015)

24. CriticalValues of t
Therefore, 𝑡 𝛼/2, 𝑑. 𝑓. = 𝑡0.025; 14 =  2.1448
𝑛 = 15 𝑥 = 1.23 𝑠 = 0.37 1 − 𝛼 = 95% 𝑡0.025; 14 = ±2.1448
(M Waring , personal communication, March 2015)

25. • The formula for the confidence interval estimation of the mean is:
𝑋 − 𝑡 𝛼 2
𝑠
𝑛
≤ 𝜇 ≤ 𝑋 + 𝑡 𝛼 2
𝑠
𝑛
The Calculations
.5000 .5000
0.95
(1-)
.4750 .4750
/2= 0.025 /2=0.025
s x
mm -2.1448 m +2.1448 s x
t-2.1448 0 2.1448
𝑛 = 15 𝑥 = 1.23 𝑠 = 0.37 1 − 𝛼 = 95% 𝑡0.025; 14 = ±2.1448
Image adapted from example provided by M Waring (personal communication, 30 March 2015)

26. Therefore, we can be 95% confident that the mean beta coefficient lies between 1.0251
and 1.4349
The Calculations
1.23 − 2.1448
0.37
15
≤ 𝜇 ≤ 1.23 + 2.1448
0.37
15
𝑋 − 𝑡 𝛼 2
𝑠
𝑛
≤ 𝜇 ≤ 𝑋 + 𝑡 𝛼 2
𝑠
𝑛
1.0251 ≤ 𝜇 ≤ 1.4349
1.23 − 0.2049 ≤ 𝜇 ≤ 1.23 + 0.2049
𝑛 = 15 𝑥 = 1.23 𝑠 = 0.37 1 − 𝛼 = 95% 𝑡0.025; 14 = ±2.1448

27. Confidence Interval Estimate for the Mean
Data
Sample Standard Deviation 0.37
Sample Mean 1.23
Sample Size 15
Confidence Level 95%
Standard Error of the Mean 0.095533589
Degrees of Freedom 14
t Value 2.1448
Interval Half Width 0.2049
Interval Lower Limit 1.0251
Interval Upper Limit 1.4349
Intermediate Calculations
Confidence Interval
Just in case…
1.0251 ≤ 𝜇 ≤ 1.4349
𝑛 = 15 𝑥 = 1.23 𝑠 = 0.37 1 − 𝛼 = 95% 𝑡0.025; 14 = ±2.1448

28. Part G
Use PhStat software to conduct a test to
determine if the variance of the shares beta
value differs from 0.15 at alpha = 0.05.
Show this printout

29. Chi-SquareTest ofVariance
Data
Null Hypothesis s^2= 0.15
Level of Significance 0.05
Sample Size 15
Sample Standard Deviation 0.37
Degrees of Freedom 14
Half Area 0.025
Chi-Square Statistic 12.7773
Two-Tail Test
Lower Critical Value 5.6287
Upper Critical Value 26.1189
p-Value 0.4559
Do not reject the null hypothesis
Intermediate Calculations
Chi-Square Test of Variance

30. Part H
Use PhStat output to confirm your answer
to a) – d) above.

31. The Confirmation
Since 𝑡𝑡𝑒𝑠𝑡 2.4075 > 𝑡 𝑐𝑣(1.3450),
we reject 𝐻0 as there is enough
evidence to conclude that the
average technology shares are
riskier than the market as a whole.
Null Hypothesis m= 1
Level of Significance 0.1
Sample Size 15
Sample Mean 1.23
Sample Standard Deviation 0.37
Standard Error of the Mean 0.0955
Degrees of Freedom 14
t Test Statistic 2.4075
Upper-Tail Test Calculations Area
Upper Critical Value 1.3450
p-Value 0.0152
Reject the null hypothesis
Data
Intermediate Calculations
t Test for Hypothesis of the Mean
𝑡𝑡𝑒𝑠𝑡 = +2.4075𝑛 = 15 𝑋 = 1.23 𝑠 = 0.37 𝛼 = 0.10 𝜇 = 1 𝑡 𝑐𝑣 = +1.3450

32. References
• Mark L. Berenson,T. C. K. D. M. L. (2012). Basic Business Statistics:
Concepts and Applications (E. Svendsen Ed. Vol. 12): Pearson
Education, Inc.