Question

1. Consider a multinomial experiment with n = 307 and k = 4. The null hypothesis to be tested is H₀: p₁ = p₂ = p₃ = p₄ = 0.25. The observed frequencies resulting from the experiment are: (You may find it useful to reference the appropriate table: chi-square table or F table)

Category	1	2	3	4
Frequency	85	58	89	75

a. Choose the appropriate alternative hypothesis.

• Not all population proportions are equal to 0.25.

• All population proportions differ from 0.25.

b-1. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.)

b-2. Find the p-value.

• p-value   0.10
• 0.01  p-value < 0.025

• p-value < 0.01

• 0.025  p-value < 0.05
• 0.05  p-value < 0.10

c. At the 10% significance level, what is the conclusion to the hypothesis test?

• Reject H₀ since the p-value is less than the significance level.

• Reject H₀ since the p-value is greater than the significance level.

• Do not reject H₀ since the p-value is greater than the significance level.

• Do not reject H₀ since the p-value is less than the significance level.

2. An analyst is trying to determine whether the prices of certain stocks on the NASDAQ are independent of the industry to which they belong. She examines four industries and, classifies the stock prices in these industries into one of three categories (high-priced, average-priced, low-priced).

	Industry
Stock Price	I	II	III	IV
High	18	12	24	24
Average	19	18	20	25
Low	8	8	9	12

a. Choose the competing hypotheses to determine whether stock price depends on the industry.

• H₀: Stock price is independent of the industry.; H_A: Stock price is dependent on the industry.

• H₀: Stock price is dependent on the industry.; H_A: Stock price is independent on the industry.

b-1. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.)

b-2. Find the p-value.

• p-value < 0.01

• 0.01 p-value < 0.025

• 0.025  p-value < 0.05
• 0.05 p-value < 0.10
• p-value  0.10

c. At a 1% significance level, what can the analyst conclude?

• Do not reject H₀; there is not enough evidence to support the claim that the stock price is dependent on the industry.

• Reject H₀; there is enough evidence to support the claim that the stock price is dependent on the industry.

• Reject H₀; there is not enough evidence to support the claim that the stock price is dependent on the industry.

• Do not reject H₀; there is enough evidence to support the claim that the stock price is dependent on the industry.

3. Using 20 observations, the multiple regression model y = β₀ + β₁x₁ + β₂x₂ + ε was estimated. A portion of the regression results is shown in the accompanying table:

	df	SS	MS	F	Significance F
Regression	2	2.10E+12	1.12E+12	63.503	1.30E-08
Residual	17	3.10E+11	1.77E+10
Total	19	2.43E+12

	Coefficients		Standard Error		t Stat		p-value		Lower 95%	Upper 95%
Intercept	−988,484		130,933		−7.550		0.000		−1,264,728	−712,240
x1	28,503		32,372		0.880		0.391		−39,796	96,802
x2	29,494		33,046		0.893		0.385		−40,227	99,215

a. At the 5% significance level, are the explanatory variables jointly significant?

• No, since the p-value of the appropriate test is less than 0.05.

• Yes, since the p-value of the appropriate test is more than 0.05.

• Yes, since the p-value of the appropriate test is less than 0.05

• No, since the p-value of the appropriate test is more than 0.05.

b. At the 5% significance level, is each explanatory variable individually significant?

• Yes, since both p-values of the appropriate test are less than 0.05.

• Yes, since both p-values of the appropriate test are more than 0.05.

• No, since both p-values of the appropriate test are not less than 0.05.

• No, since both p-values of the appropriate test are not more than 0.05.

c. What is the likely problem with this model?

• Multicollinearity since the standard errors are biased.

• Multicollinearity since the explanatory variables are individually and jointly significant.

• Multicollinearity since the explanatory variables are individually significant but jointly insignificant.

• Multicollinearity since the explanatory variables are individually insignificant but jointly significant.

4. The following table lists a portion of Major League Baseball’s (MLB’s) leading pitchers, each pitcher’s salary (In $ millions), and earned run average (ERA) for 2008.

	Salary			ERA
J. Santana		17.0			2.31
C. Lee		3.0			2.39
⋮		⋮			⋮
C. Hamels		0.2			3.00

	Salary	ERA
J. Santana	17.0	2.31
C. Lee	3.0	2.39
T. Lincecum	0.3	2.42
C. Sabathia	10.0	2.20
R. Halladay	10.0	2.39
J. Peavy	5.4	2.15
D. Matsuzaka	7.8	2.43
R. Dempster	7.1	2.32
B. Sheets	11.7	3.04
C. Hamels	0.2	3.00

a-1. Estimate the model: Salaryˆ=Salary^= β₀ + β₁ERA + ε. (Negative values should be indicated by a minus sign. Enter your answers, in millions, rounded to 2 decimal places.)

a-2. Interpret the coefficient of ERA.

• A one-unit increase in ERA, predicted salary decreases by $2.89 million.

• A one-unit increase in ERA, predicted salary increases by $2.89 million.

• A one-unit increase in ERA, predicted salary decreases by $11.48 million.

• A one-unit increase in ERA, predicted salary increases by $11.48 million.

b. Use the estimated model to predict salary for each player, given his ERA. For example, use the sample regression equation to predict the salary for J. Santana with ERA = 2.31. (Round coefficient estimates to at least 4 decimal places and final answers, in millions, to 2 decimal places.)

c. Derive the corresponding residuals. (Negative values should be indicated by a minus sign. Round coefficient estimates to at least 4 decimal places and final answers, in millions, to 2 decimal places.)

Answer 1

1.

Category	Observed Frequency (O)	Proportion, p	Expected Frequency (E)	(O-E)²/E
1	85	0.25	307 * 0.25 = 76.75	(85 - 76.75)²/76.75 = 0.8868
2	58	0.25	307 * 0.25 = 76.75	(58 - 76.75)²/76.75 = 4.5806
3	89	0.25	307 * 0.25 = 76.75	(89 - 76.75)²/76.75 = 1.9552
4	75	0.25	307 * 0.25 = 76.75	(75 - 76.75)²/76.75 = 0.0399
Total	307	1.00	307	7.4625

a. Appropriate alternative hypothesis:

Not all population proportions are equal to 0.25.

b-1.

Test statistic:

χ² = ∑ ((O-E)²/E) = 7.463

b-2. df = n-1 = 3

p-value = CHISQ.DIST.RT(7.4625, 3) = 0.0585

0.05  p-value < 0.10

c. Conclusion to the hypothesis test:

Reject H0 since the p-value is less than the significance level.

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2.

Observed Frequencies
Stock Price	I	II	III	IV	Total
High	18	12	24	24	78
Average	19	18	20	25	82
Low	8	8	9	12	37
Total	45	38	53	61	197
Expected Frequencies
	I	II	III	IV	Total
High	45 * 78 / 197 = 17.8173	38 * 78 / 197 = 15.0457	53 * 78 / 197 = 20.9848	61 * 78 / 197 = 24.1523	78
Average	45 * 82 / 197 = 18.731	38 * 82 / 197 = 15.8173	53 * 82 / 197 = 22.0609	61 * 82 / 197 = 25.3909	82
Low	45 * 37 / 197 = 8.4518	38 * 37 / 197 = 7.1371	53 * 37 / 197 = 9.9543	61 * 37 / 197 = 11.4569	37
Total	45	38	53	61	197
	(fo-fe)²/fe
High	(18 - 17.8173)²/17.8173 = 0.0019	(12 - 15.0457)²/15.0457 = 0.6165	(24 - 20.9848)²/20.9848 = 0.4332	(24 - 24.1523)²/24.1523 = 0.001
Average	(19 - 18.731)²/18.731 = 0.0039	(18 - 15.8173)²/15.8173 = 0.3012	(20 - 22.0609)²/22.0609 = 0.1925	(25 - 25.3909)²/25.3909 = 0.006
Low	(8 - 8.4518)²/8.4518 = 0.0241	(8 - 7.1371)²/7.1371 = 0.1043	(9 - 9.9543)²/9.9543 = 0.0915	(12 - 11.4569)²/11.4569 = 0.0257

a)

H0: Stock price is independent of the industry.;

HA: Stock price is dependent on the industry.

Test statistic:

χ² = ∑ ((fo-fe)²/fe) = 1.8020

df = (r-1)(c-1) = 6

p-value = CHISQ.DIST.RT(1.802, 6) = 0.937

p-value >0.10

conclusion:

Do not reject H0; there is not enough evidence to support the claim that the stock price is dependent on the industry.

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