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.

4.7	5.1	3.1	3.7	2.5	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 ≈ 2.046 for this sample of French companies.

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

3.0	3.8	3.2	4.1	5.2	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 ≈ 1.044 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: σ₁² = σ₂²; H₁: σ₁² > σ₂²H_o: σ₁² > σ₂²; H₁: σ₁² = σ₂²     H_o: σ₂² = σ₁²; H₁: σ₂² > σ₁²H_o: 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 independent normal 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.

(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

x=france company

s²=sum((x-mean)²/(n-1)=40.9181/(21-1)=2.046

mean=sum(x)/n=87.7/21=4.1762

following information has been generated using ms-excel

	x	x-mean	(x-mean)2
	4.7	0.5238	0.274366
	5.1	0.9238	0.853406
	3.1	-1.0762	1.158206
	3.7	-0.4762	0.226766
	2.5	-1.6762	2.809646
	3.5	-0.6762	0.457246
	2.8	-1.3762	1.893926
	4.4	0.2238	0.050086
	5.7	1.5238	2.321966
	3.4	-0.7762	0.602486
	4.1	-0.0762	0.005806
	6.8	2.6238	6.884326
	2.9	-1.2762	1.628686
	3.2	-0.9762	0.952966
	7.2	3.0238	9.143366
	6.5	2.3238	5.400046
	5	0.8238	0.678646
	3.3	-0.8762	0.767726
	2.8	-1.3762	1.893926
	2.5	-1.6762	2.809646
	4.5	0.3238	0.104846
n=	21	21	21
sum=	87.7	-0.0002	40.9181

  second part

y=german company

s²=sum((y-mean)²/(n-1)=17.7561/(18-1)=1.044

mean=sum(x)/n=67.1/18=4.1762

	y	y-mean	(y-mean)2
	3	-0.7278	0.529693
	3.8	0.0722	0.005213
	3.2	-0.5278	0.278573
	4.1	0.3722	0.138533
	5.2	1.4722	2.167373
	5.5	1.7722	3.140693
	5	1.2722	1.618493
	5.4	1.6722	2.796253
	3.2	-0.5278	0.278573
	3.5	-0.2278	0.051893
	3.7	-0.0278	0.000773
	2.6	-1.1278	1.271933
	2.8	-0.9278	0.860813
	3	-0.7278	0.529693
	3	-0.7278	0.529693
	2.2	-1.5278	2.334173
	4.7	0.9722	0.945173
	3.2	-0.5278	0.278573
n=	18	18	18
sum=	67.1	-0.0004	17.75611

(a) What is the level of significance?

answer is 0.05=5%

null hypothesis Ho: σ12= σ22;

alternate hypothesis H1: σ12 ≠ σ22

(b) F=s1²/s2²=2.046/1.044=1.960

What are the degrees of freedom?

answer is

dfN=20
dfD=17

What assumptions are you making about the original distribution?

answer is The populations follow independent normal distributions.

(c)Find or estimate the P-value of the sample test statistic. (Use 4 decimal places.)

anser is 0.050 < p-value < 0.100

p-value=0.0830 ( using ms-excel command=FDIST(1.96,20,17))

(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 fail to reject the null hypothesis and conclude the data are not statistically significant.

since p-value =0.083 is greater than level of significance alpha=0.05,so we fail to reject null hypothesis

(e)Interpret your conclusion in the context of the application.

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