An economist wonders if corporate productivity in some countries
is more volatile than in other countries. One measure of a
company's productivity is annual percentage yield based on total
company assets.
A random sample of leading companies in France gave the following
percentage yields based on assets.
4.6 | 5.1 | 3.1 | 3.8 | 2.9 | 3.5 | 2.8 | 4.4 | 5.7 | 3.4 | 4.1 |
6.8 | 2.9 | 3.2 | 7.2 | 6.5 | 5.0 | 3.3 | 2.8 | 2.5 | 4.5 |
Use a calculator to verify that the sample variance is
s^{2} ≈ 1.977 for this sample of French
companies.
Another random sample of leading companies in Germany gave the
following percentage yields based on assets.
2.9 | 3.1 | 3.4 | 4.9 | 5.4 | 5.5 | 5.0 | 5.4 | 3.2 |
3.5 | 3.7 | 2.6 | 2.8 | 3.0 | 3.0 | 2.2 | 4.7 | 3.2 |
Use a calculator to verify that s^{2} ≈ 1.176
for this sample of German companies.
Test the claim that there is a difference (either way) in the
population variance of percentage yields for leading companies in
France and Germany. Use a 5% level of significance. How could your
test conclusion relate to the economist's question regarding
volatility (data spread) of corporate productivity of
large companies in France compared with companies in Germany?
(a) What is the level of significance?
State the null and alternate hypotheses.
H_{o}: σ_{1}^{2} = σ_{2}^{2}; H_{1}: σ_{1}^{2} > σ_{2}^{2}
H_{o}: σ_{1}^{2} > σ_{2}^{2}; H_{1}: σ_{1}^{2} = σ_{2}^{2}
H_{o}: σ_{2}^{2} = σ_{1}^{2}; H_{1}: σ_{2}^{2} > σ_{1}^{2}
H_{o}: σ_{1}^{2} = σ_{2}^{2}; H_{1}: σ_{1}^{2} ≠ σ_{2}^{2}
(b) Find the value of the sample F statistic. (Use 2
decimal places.)
What are the degrees of freedom?
df_{N} | |
df_{D} |
What assumptions are you making about the original distribution?
The populations follow independent chi-square distributions. We have random samples from each population.
The populations follow independent normal distributions.
The populations follow dependent normal distributions. We have random samples from each population.
The populations follow independent normal distributions. We have random samples from each population.
(c) Find or estimate the P-value of the sample test
statistic. (Use 4 decimal places.)
p-value > 0.200
0.100 < p-value < 0.200
0.050 < p-value < 0.100
0.020 < p-value < 0.050
0.002 < p-value < 0.020
p-value < 0.002
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis?
At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.
At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.
(e) Interpret your conclusion in the context of the
application.
Fail to reject the null hypothesis, there is sufficient evidence that the variance in percentage yields on assets is greater in the French companies.
Reject the null hypothesis, there is insufficient evidence that the variance in percentage yields on assets is greater in the French companies.
Reject the null hypothesis, there is sufficient evidence that the variance in percentage yields on assets is different in both companies.
Fail to reject the null hypothesis, there is insufficient evidence that the variance in percentage yields on assets is different in both companies.
a)
significance level = 0.05
Ho: σ1^2 = σ2^2;
H1: σ1^2 ≠ σ2^2
Option D
b)
Test statistic,
F = (s1/s2)^2
F = (1.977/1.176)
F = 1.68
dfN = 21 - 1 = 20
dfD = 18 - 1 = 17
The populations follow independent normal distributions. We have random samples from each population.
c)
0.020 < p-value < 0.050
d)
At the α = 0.05 level, we reject the null hypothesis and conclude
the data are statistically significant.
e)
Reject the null hypothesis, there is sufficient evidence that the
variance in percentage yields on assets is different in both
companies.
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