Question

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.

4.5 5.9 3.5 3.3 2.2 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 s2 ≈ 2.190 for this sample of French companies.

Another random sample of leading companies in Germany gave the following percentage yields based on assets.

2.5 3.5 3.2 4.1 5.9 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 s2 ≈ 1.255 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.

Ho: σ12 = σ22; H1: σ12 > σ22

Ho: σ12 > σ22; H1: σ12 = σ22

Ho: σ22 = σ12; H1: σ22 > σ12

Ho: σ12 = σ22; H1: σ12 ≠ σ22

(b) Find the value of the sample F statistic. (Use 2 decimal places.)

What are the degrees of freedom?

dfN

dfD

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 dependent normal distributions. We have random samples from each population. The populations follow independent normal distributions. 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.

Answer 1

a) H₀:

    H₁:

b) The test statistic F =

                                  = 2.190/1.255

                                  = 1.75

df_N = 21 - 1 = 20

df_D = 18 - 1 = 17

The populations follow independent normal distributions. We have random samples from each population.

c) P-value = 2 * P(F > 1.75)

                = 2 * 0.1241

                = 0.2482

P-value > 0.200

d) Since P-value > (0.2482 > 0.05), so we should not reject the null hypothesis.

At the = 0.05 level, we fail to reject the null hypothesis and conclude that the data are not statistically significant.

e) Fail to reject the null hypothesis, there is insufficient evidence that the variance in percentage yields on assets is different in both companies.