COURSE TITLE: STATISTICS

CHAPTER 1: WHAT IS STATISTICS?

Exercise 1: The age is a ratio scale level. A 40-year-old is twice as old as someone 20 years old.

Exercise 2: Nominal scale. We could arrange the states in any order.

CHAPTER 2: DESCRIBING DATA - Frequency Distributions and Graphic Presentation

Exercise 1:

1. How many classes would you use?

. There are 30 observations. So must > 30.

24 = 16 <30. 25 = 32>30.

So 5 classes are recommended.

2. What class interval would you suggest?

The highest value is 30,872 USD. The lowest value is 17,357 USD. The number of classes (k) is 5.

The interval width should be at least . We round up 2,703 to 3,000

Therefore, class interval of 3,000 is reasonable.

Exercise 2:

Selling price (unit: thousand USD) Frequency Relative Frequency

15 up to 18 8 8/80 = 0.1000 = 10%

18 up to 21 23 23/80 = 0.2875 = 28.75%

21 up to 24 17 17/80 = 0.2125 = 21.25%

24 up to 27 18 18/80 = 0.2250 = 22.50%

27 up to 30 8 8/80 = 0.1000 = 10%

30 up to 33 4 4/80 = 0.0500 = 5%

33 up to 36 2 2/80 = 0.0250 = 2.5%

Total 80 1.000

1. How many vehicles were sold for 18,000 USD up to 21,000 USD?

23 vehicles

2. What percent of the vehicles sold for a price between 18,000 USD and 21,000 USD?

28.75%

3. What percent of the vehicles sold for 30,000 USD or more?

5% + 2.5% = 7.5%

CHAPTER 3: DESCRIBING DATA - Numerical Measures

Exercise 1: What was the weighted mean price of a computer?

= 237 USD

Exercise 2: What is the geometric mean percent increase?

 The number 1.0491 represents 4.91% increase in sales in 2005

 The number 1.0575 represents 5.75% increase in sales in 2006

 The number 1.0812 represents 8.12% increase in sales in 2007

 The number 1.2160 represents 21.60% increase in sales in 2008

 There are 4 years (2005, 2006, 2007 and 2008). So n = 4

The geometric mean percent increase is

=

Then 1.099 - 1 = 0.99 = 0.099 = 9.9%

Exercise 3: What is the geometric mean annual percent increase?

 There are 20 years (from 1984 to 2004). So n = 20

 The value at start of period (1984) is 23,000 trucks

 The value at end of period (2004) is 120,520 trucks

The geometric mean annual percent increase is = 8.63%

Exercise 4:

1. What is the range of the weights?

Range = largest value - smallest value = 112 --90 = 22 tons.

2. Compute the arithmetic mean weight.

3. Compute the mean deviation of the weight.

X

95 103 -8 8

103 103 0 0

105 103 2 2

110 103 7 7

90 103 -13 13

104 103 1 1

105 103 2 2

112 103 9 9

= 42

Exercise 5: The monthly salaries of 5 employees at Lotus Department Store are as follows:

3,536 USD 3,173 USD 3,448 USD 3,121 USD 3,622 USD

1. Compute the population variance.

N = 5

= 39,490.8

2. Compute the population standard deviation.

Exercise 6: The weight of the contents of several small aspirin bottles are as follows:

4 grams 5 grams 5 grams 6 grams

2 grams 4 grams 2 grams

1. Compute the sample variance.

n = 7

X

X-

4 4 0 0

5 4 1 1

2 4 -2 4

4 4 0 0

5 4 1 1

2 4 -2 4

6 4 2 4

Total = 14

= = 2.33

2. Determine the sample standard deviation.

=1.53

CHAPTER 5: A SURVEY OF PROBABILITY CONCEPTS

Exercise 1: Suppose you are going to survey about a new pension plan. You are going to take a sample of 2,000 employees. The employees are classified as follows:

Classification Event Number of Employees

Supervisors A 120

Maintenance B 50

Production C 1,460

Management D 302

Secretarial E 68

1. What is the probability that the first person selected is: either in maintenance or a secretary?

P(maintenance or secretary) = P(B or E) = P(B) + P(E)

P(B) =

P(D) =

So P(maintenance or secretary) = P(B) + P(E) = + = 0.059

2. What is the probability that the first person selected is not in management?

P(management) = P(D) =

P(not management) = P(not D) = P(A) + P(B) + P(C) + P(E) = 1 - P(D) = 1 - = 0.849

Exercise 2: There are 12 rolls of film in a box, 8 of which are defective. 4 rolls are to be selected, one after the other, without being returned to the box. What is the probability all 4 rolls of film are defective?

 The first roll of film selected from the box being found defective is even D1, P(D1) =

 The second roll of film selected from the box being found defective is event D2, P(D2|D1). It means that the probability of selecting the second roll being defective, given that the first roll was defective. The first roll is selected with being returned to the box. So there are now only 11 rolls of film in the box. The first roll is defective. So there are now only 7 defective rolls of film in the box. Therefore, P(D2|D1) =

 The third roll of film selected from the box being found defective is event D3, P(D3|D1 and D2) =

 The fourth roll of film selected from the box being found defective is event D4, P(D4|D1 and D2 and D3) =

P(D1 and D2 and D3 and D4) = P(D1) x P(D2|D1) x P(D3|D1 and D2) x P(D4|D1 and D2 and D3) =

CHAPTER 8: SAMPLING METHODS AND THE CENTRAL LIMIT THEOREM

Exercise 1: The Quality Assurance Department for Cola, Inc. maintains the records regarding the amount of cola in the bottle. Their records indicate that the amount of cola follows the normal probability distribution. The mean amount per bottle is 31.2 ounces and the population standard deviation is 0.4 ounces. At 8:00 AM today the quality technician randomly selected 16 bottles from the filling line. The mean amount of cola contained in the bottles is 31.08 ounces. What can you conclude about the filling process?

In the Appendix D, locate the probability corresponding to a z value of 1.20. It is 0.3849.

The probability that z is greater than -1.20 is 0.3849 + 0.5 = 0.8849 (Recall that area under the normal probability distribution is 1. Half of this area on the left side is 0.5, and half of this area on the right side is 0.5. On the left, the probability is 0.3948. On the right, the probability is 0.5. So the probability that z is greater than -1.20 = 0.3849 + 0.5). There is less than a 12 percent chance of this event happening (1- 0.8849 = 0.1151 = 11.51% < 12%).

Exercise 2: The mean hourly wage for plumbers in the Atlanta region is 28 USD. What is the likelihood that we could select a sample of 50 plumbers with a mean wage of 28.5 USD or more? The standard deviation of the sample is 2 USD per hour.

In the Appendix D, locate the probability corresponding to a z value of 1.77. It is 0.4616.

The probability that z is greater than 1.77 is 0.5 - 0.4616 = 0.384.

CHAPTER 9: ESTIMATION AND CONFIDENCE INTERVAL

Exercise 1: The mean daily sales at the Bun-and-Run (a fast food restaurant) is 20,000 USD for a sample of 40 days. The standard deviation of the sample is 3,000 USD.

1. What is the estimated mean daily sale of the population (the mean daily sales for all days)?

The estimated mean daily sale of the population is 20,000 USD.

What is this estimate called?

This is called the point estimate.

2. What is the 99 percent confidence interval?

In this example, the sample size is large (40 days). So we use z distribution.



 99% = 0.99. The area under the normal distribution is divided into 2 parts (left side and right side). So 0.99/2 = 0.495. In the Appendix D, find the value of 0.4950 (There are 2 values: 0.4949 and 0.4951). As convention, we select higher value (0.4951). The value of z corresponding to 0.4951 is 2.58. So z = 2.58

 s = $3,000

 n = 40 days

3. Interpret your findings.

The endpoints of the confidence interval are 18,776 (found by 20,000 - 1224) and 21,224 (found by 20,000 + 1224). About 99 percent of the intervals similarly constructed would include the population mean.

Exercise 2: The Dottie Restaurant bakes and sells cookies at 50 different locations in the Philadelphia area. The manager of this restaurant is concerned about absenteeism among her workers. The information below reports the number of days absent for a sample of 10 workers during the last two-week day period.

4 1 2 2 1 2 2 1 0 3

1. What is the population mean? What is the best estimate of that value?

The population mean is not known. So the best estimate is the sample mean

2. Develop a 95 percent confidence interval for the population mean.

In this example, there population standard deviation is not known, and the sample size is small. So we use t distribution. The formula is



 To find the value of t we use Appendix F. In this case, we want 95 percent level of confidence, so we move to the column headed "95%". The degree of freedom is n-1 = 10-1 = 9. So we select the row with 9 degrees of freedom. The value is 2.262.



 n = 10 (workers)

Exercise 3: The owner of the West Gas Station wished to determine the proportion of customers who use a credit card to pay at the pump. He surveys 1,400 customers and finds that 420 used credit cards to pay at the pump.

1. Estimate the value of the population proportion.

2. Compute the standard error of the proportion.

3. Develop a 99 percent confidence interval for the population proportion.

 For the proportion we use z distribution. 99% = 0.99. The area under the normal distribution is divided into 2 parts (left side and right side). So 0.99/2 = 0.495. In the Appendix D, find the value of 0.4950 (There are 2 values: 0.4949 and 0.4951). As convention, we select higher value (0.4951). The value of z corresponding to 0.4951 is 2.58. So z = 2.58

4. Interpret your findings.

The interval is between 0.27 (found by 0.3 - 0.03) and 0.33 (found by 0.3 + 0.03). About 99 percent of the similarly constructed intervals would include the population mean.

CHAPTER 10: ONE-SAMPLE TESTS OF HYPOTHESIS

Exercise 1: The following information is available

H0:

H1:

The sample mean is 49, and the sample size is 36. The population standard deviation is 5. Use the 0.05 significance level.

1. Is this a one-tailed or two-tailed test?

This is a two-tailed test.

2. What is the decision rule?

 Because this is a two-tailed test, 0.05 significance level must be divided by 2. So .

 Appendix D is based on half of the area under the curve (0.5). Then 0.5 - 0.025 = 0.475

 In the Appendix D, locate 0.475. It is 1.96.

 Therefore, the decision rule is to reject H0 and accept H1 when computed z does not fall in the region from -1.96 and +1.96.

3. What is the value of the test statistic?

 Look at H0 and H1, we know that the population mean is 50.

 = -1.2

4. What is your decision regarding H0?

Because -1.2 is in between -1.96 and +1.96 we can not reject H0 ( is not different from 50).

Exercise 2: The following information is available

H0:

H1:

A sample of 36 observations is selected from a normal population. The sample mean is 21, and the standard deviation is 5. Use the 0.05 significance level.

1. Is this a one-tailed or two-tailed test?

This is a one-tailed test.

2. What is the decision rule?

 Because this is a one-tailed test, we have to put all the rejection region in one tail, in this case it is in the right tail (see the direction of H1). 0.05 (we do not need to divide the significance level into 2).

 0.5 - 0.05 = 0.45

 In the Appendix D, locate 0.45. There is not 0.45 (there are 2 values: 0.4495 and 0.4505). As the convention, we select higher value (0.4505). It is 1.65 (the z value is 1.65).

 Therefore, the decision rule is to reject H0 and accept H1 when computed z > 1.65.

3. What is the value of the test statistic?

 Look at H0 and H1, we know that the population mean is 20.

 = 1.2

4. What is your decision regarding H0?

Because 1.2 is smaller than 1.65 we fail to reject H0 at 0.05 significance level.

Exercise 3: The following hypotheses are given

H0:

H1:

A sample of 100 observations revealed that p = 0.75. At the 0.05 significance level, can the null hypothesis (H0) be rejected?

1. State the decision rule.

 This is the test concerning proportion. So z is the test statistic.

 This is a one-tailed test. So we do not need to divide the significance level into 2. Therefore, 0.5 - 0.05 = 0.450. This probability is in the right tail (see the direction of H1).

 In the Appendix D, locate 0.45. There is not 0.45 (there are 2 values: 0.4495 and 0.4505). As the convention, we select higher value (0.4505). It is 1.65 (the z value is 1.65).

 The decision rule is: H0 is rejected if the computed

2. Compute the value of the test statistic

 Looking at H0 and H1 we know that



3. What is your decision regarding the null hypothesis (H0)?

 H0 is not rejected.

Exercise 4: Given the following hypothesis:

H0:

H1:

For a random sample of 10 observations, the sample mean was 12 and the sample standard deviation is 3. Using the 0.05 significance level.

1. State the decision rule.

 This is the test for a population mean (small sample and the population standard deviation is unknown). So t is the test statistic.

 This is a one-tailed test. In the Appendix F, locate "Level of Significance for One-Tailed Test", 0.05. The degree of freedom = n - 1 = 10 - 1 = 9. The t value is 1.833.

 The decision rule is: H0 is rejected if the computed

2. Compute the value of the test statistic

 Looking at H0 and H1 we know that



3. What is your decision regarding the null hypothesis (H0)?

 Reject H0. The mean is greater than 10.

CHAPTER 11: TWO-SAMPLE TESTS OF HYPOTHESIS

Exercise 1: A sample of 40 observations is selected from one population. The sample mean is 102 and the sample standard deviation is 5. A sample of 50 observations is selected from a second population. The sample mean is 99 and the sample standard deviation is 6. Conduct the following test of hypothesis using the 0.04 significance level.

H0:

H1:

1. Is this a one-tailed or a two-tailed test?

 This is a two-tailed test.

2. State the decision rule.

 Because this is a two-tailed test, 0.04 significance level must be divided by 2. So .

 Appendix D is based on half of the area under the curve (0.5). Then 0.5 - 0.02 = 0.48

 In the Appendix D, locate 0.48. There is not 0.48 (there are 2 values: 0.4798 and 0.4803). 0.4798 is the closest value to 0.48. So we use 0.4798. It is 2.05 (the z value is 2.05).

 Reject H0 if the computed or the computed

3. Compute the value of the test statistic



4. What is your decision regarding the null hypothesis (H0)?

 Because z is greater than 2.05, we reject H0 and accept H1.

Exercise 2: The null and alternate hypotheses are:

H0:

H1:

A sample of 100 observations from the first population indicated that X1 is 70. A sample of 150 observations from the second population revealed X2 to be 90. Use the 0.05 significance level to test the hypothesis.

1. State the decision rule.

 This is a two-sample test about proportion. So z is the test statistic.

 This is a one-tailed test. So we do not need to divide the significance level into 2. Therefore, 0.5 - 0.05 = 0.450.

 In the Appendix D, locate 0.45. There is not 0.45 (there are 2 values: 0.4495 and 0.4505). As the convention, we select higher value (0.4505). It is 1.65 (the z value is 1.65).

 H0 is rejected if z 1.65

2. Compute the pooled proportion.



3. Compute the value of the test statistic.





4. What is your decision regarding the null hypothesis (H0)?

 Because the computed z is smaller than 1.65, H0 is not rejected.

Exercise 3: A sample of scores on an examination of the course "Advanced English" are:

Men 72 69 98 66 85 76 79 80 77

Women 81 67 90 78 81 80 76

At the 0.01 significance level, is the mean grade of the women higher than the mean grade of the men?

H0: (f = female; m = male)

H1:

 This is the test comparing population means with small samples. So t is the test statistic.

 This is a one-tailed test. In the Appendix F, locate "Level of Significance for One-Tailed Test" and 0.01. Degree of freedom = 7 + 9 - 2 = 14. It is 2.624.

 The decision rule is to reject H0 if the computed t 2.624

 is the variance of the first sample. It is 6.88 (use the MegaStat).

 is the variance of the second sample. It is 9.49 (use the MegaStat).





 Because the computed t is smaller than 2.624, we do not reject H0. It means that there is no difference between the mean grade of women and the mean grade of men.

Exercise 4: The null and alternate hypotheses are:

H0:

H1:

The following sample information shows the number of defective units produced on the day shift and the afternoon shift for a sample of 4 days last month.

Day 1 Day 2 Day 3 Day 4

Day shift 10 12 15 19

Afternoon shift 8 9 12 15

At the 0.05 significance level, can we conclude there are more defects produced on the afternoon shift?

 This is a two-sample test of hypothesis, dependent samples. So t is the test statistic.

 This is a one-tailed test. In the Appendix F, locate "Level of Significance for One-Tailed Test", 0.05. The degree of freedom = 4 - 1 = 3. The t value is 2.353.

 The decision rule is: Reject H0 if the computed t 2.353

Day 1 Day 2 Day 3 Day 4

Day shift 10 12 15 19

Afternoon shift 8 9 12 15

d 10-8=2 12-9=3 15-12=3 19-15=4

Day 1 Day 2 Day 3 Day 4

Day shift 10 12 15 19

Afternoon shift 8 9 12 15

d 10 - 8 = 2 12 - 9 = 3 15 - 12 = 3 19 - 15 = 4

d-

2 - 3 = -1 3 - 3 = 0 3 - 3 = 0 4 - 3 = 1

1 0 0 1

= 1 + 1 = 2

Because the computed t is greater than 2.353, we reject H0. There are more defective parts produced on the day shift.

CHAPTER 12: ANALYSIS OF VARIANCE

Exercise 1:

 In this example, we are comparing 2 population variances, so the test statistic is F.

 This is a two-tailed test. So 0.02 must be divided into 2 ( )

 The critical value is obtained from Appendix G.

 The standard deviation of the first population (10) is larger than the standard deviation of the second population (7). So this sample will be chosen for the degree of freedom in the numerator. The degrees of freedom in the numerator = n - 1 = 8 - 1 = 7

 The standard deviation of the second population (7) is smaller than the standard deviation of the first population (10). So this sample will be chosen for the degree of freedom in the denominator. The degrees of freedom in the denominator = n - 1 = 6 - 1 = 5

 In the Appendix G (page 724), the critical value is 10.5

 The decision rule is: reject H0 if the computed F > 10.5



 Because the computed F is smaller than 10.5, we do not reject H0. There is no difference in the variations of the two populations.

Exercise 2: The following is sample information. Test the hypothesis that the treatment means are equal. Use the 0.05 significance level.

Treatment 1 Treatment 2 Treatment 3

8 3 3

6 2 4

10 4 5

9 3 4

1. State the null hypothesis and the alternate hypothesis.

H0:

H1: Treatment means are not all the same.

2. What is the decision rule?

 The critical value is obtained from Appendix G (page 723, 5 percent level of significance).

o Degrees of freedom in the numerator = k - 1 = 3 - 1 = 2

o Degrees of freedom in the denominator = n - k = 12 - 3 = 9

o The critical value is 4.26

 The decision rule is: Reject H0 if the computed F > 4.26

3. Complete the following ANOVA table.

Source of Variation Sum of Squares

(SS) Degrees of Freedom

(df) Mean Square

(MS) F

Treatment 62.17 k - 1 = 3 - 1 = 2

Error 74.92 - 62.17 = 12.75 n - k = 12 - 3 = 9

Total 74.92

4. State your decision regarding the null hypothesis.

 Because the computed F (21.94) is greater than 4.26, we reject H0. The treatment means are not all the same.

Exercise 2: The following data are given for a two-factor ANOVA

Block Treatment

1 2 3

A 31 25 35

B 33 26 33

C 28 24 30

D 30 29 28

E 28 26 27

Using 0.05 significance level, conduct a test of hypothesis to determine whether the block or the treatment means differ.

1. State null hypothesis and alternate hypothesis for treatments.

H0:

H1: Not all treatment means are equal.

2. State the decision rule for treatments

 The critical value is obtained from Appendix G (page 723, 5 percent level of significance).

o Because we are conducting the test of hypothesis concerning the treatment means, the degree of freedom in the numerator = k-1 = 3-1 = 2.

o The degree of freedom in the denominator = (k-1)(b-1) = (3-1)(5-1) = 8.

o The critical value is 4.46 (See Appendix G, page 723).

 The decision rule: Reject H0 if the computed F > 4.46

3. State null hypothesis and alternate hypothesis for blocks.

H0:

H1: Not all block means are equal.

4. State the decision rule for blocks.

 The critical value is obtained from Appendix G (page 723, 5 percent level of significance).

o Because we are conducting the test of hypothesis concerning the block means, the degree of freedom in the numerator = b-1 = 5-1 = 4.

o The degree of freedom in the denominator = (k-1)(b-1) = (3-1)(5-1) = 8.

o The critical value is 3.84 (See Appendix G, page 723).

 The decision rule: Reject H0 if the computed F > 3.84

5. Complete the following ANOVA table.

Source of Variation Sum of Squares

(SS) Degrees of Freedom

(df) Mean Square

(MS) F

Treatments 62.53 k-1 = 3-1 = 2

Blocks 139.73 - 62.53 -43.47 = 33.73 b-1 = 5-1 = 4

Error 43.47 (k-1)(b-1) = (2)(4)=8 =

Total 139.73

6. Give your decision regarding the two sets of hypotheses.

 For the treatments: Because the computed F (5.75) is greater than the critical value (4.46), we reject H0. There is a difference in the treatments.

 For the blocks: Because the computed F (1.55) is smaller than the critical value (3.84), we do not reject H0. There is no difference among blocks.

CHAPTER 13: LINEAR REGRESSION AND CORRELATION

Exercise 1: The following sample observations were randomly selected.

X: 4 5 3 6 10

Y: 4 6 5 7 7

Given that the standard deviation of Y is 1.3038, the standard deviation of X is 2.7019, and coefficient of correlation is 0.7522.

1. Determine the regression equation.

: Standard deviation of Y (the dependent variable) = 1.3038

: Standard deviation of X (the independent variable) = 2.7019

r: The coefficient of determination = 0.7522

: The mean of Y (the dependent variable) =

: The mean of X (the independent variable) =

Therefore, the regression equation is: Y' = a + b.X = 3.7671 + 0.363.X

2. Determine the value of Y' when X is 7.

Y' = a + b.X = 3.7671 + 0.363.X = 3.7671 + 0.363*(7) = 6.3081

Exercise 2: Given the following ANOVA table

SOURCE DF SS MS F

Regression 1 1000 1000 26.00

Error 13 500 38.46

Total 14 1500

1. Determine the coefficient of determination.

SSR = SS regression = 1000

SST = SS Total = 1500

2. What is the coefficient of correlation?

3. Determine the standard error of estimate.

SSE = SS Error = 500

SS Total = n - 1 = 14. Therefore, n = 14+1 = 15

CHAPTER 14: MULTIPLE REGRESSION AND CORRELATION ANALYSIS

Exercise 1: The director of marketing at a computer company is studying monthly sales. Three variables are thought to relate to the monthly sale (Y). These three variables are: regional population (X1), per-capita income (X2), and regional unemployment rate (X3). The regression equation was computed to be (in USD): Y' = 64,100 + 0.394X1 + 9.6X2 - 11,600X3

1. What is the full name of the equation?

Multiple regression equation

2. Interpret the number 64,100

The Y-intercept.

3. What are the estimated monthly sales for a particular region with a population of 796,000 people, per-capita income of 6,940 USD, and an unemployment rate of 6 percent.

Y' = 64,100 + 0.394X1 + 9.6X2 - 11,600X3 =

Y' = 64,100 + 0.394*(796,000) + 9.6 * (6,940) - 11,600 * (6) = 374,748 USD.

4. The following output was obtained

SOURCE SS DF MS F

Regression 3.8479 1 3.8479 3.91

Residual 2.9521 3 0.9840

Total 6.8000 4

Compute and interpret

56.6% of the variation in monthly sales can be explained by regional population, per-capita income, and regional unemployment rate.

Exercise 2: Refer to the following ANOVA table

SOURCE DF SS MS F

Regression 3 21 7.0 2.33

Error 15 45 3.0

Total 18 66

NOTE: YOU DO NOT NEED TO DISPLAY THE FOLLOWING TABLE

SOURCE DF SS MS F

Regression k SSR MSR=

Error n-(k+1) SSE MSE=

Total n-1 SS Total

1. How large was the sample?

n - 1 = 18. Therefore, n = 18+1 = 19.

2. How many independent variables are there?

In the above table, k = 3. So there are 3 independent variables.

3. Compute the coefficient of multiple determination.

4. Compute the multiple standard error of estimate.

Exercise 3: The following output was obtained

SOURCE DF SS MS F

Regression 5 100 20

Error 20 40 2

Total 25 140

Predictor Coef StDev t-ratio

Constant 3.00 1.50 2.00

X1 4.00 3.00 1.33

X2 3.00 0.20 15.00

X3 0.20 0.05 4.00

X4 -2.50 1.00 -2.50

X5 3.00 4.00 0.75

1. What is the sample size?

n = 25+1 =26

2. Compute the value of r2?

3. Conduct a global test of hypothesis to determine whether any of the regression coefficients are significant. Use the 0.05 significance level.

H0:

H1: Not all the are 0

 To conduct the global test, we use the F.

 The critical value is obtained from the Appendix G (page 723, 5 percent level of significance).

The degrees of freedom in the numerator = k = 5

The degrees of freedom in the denominator = n-(k+1) = 26-(5+1) = 20

 The critical value is 2.71

 The decision rule: reject H0 if the computed F > 2.71

 The computed F = .

 Because the computed F (10) is greater than 2.71, we reject H0. At least one regression coefficient is not zero.

4. Test the regression coefficients individually. Would you consider omitting any variables? If so, which one? Use the 0.05 significance level.

 To test the regression coefficients individually, we use the t-test.

 The critical value for t is in Appendix F (page 722).

o Locate the Level of Significance for Two-tailed Test. Go to 0.05 significance level.

o The degrees of freedom is n-(k+1)=26-(5+1)=20.

 The critical value is 2.086.

 The decision rule: H0 is rejected in each case if or . Looking at the above table, the coefficients for X1(t=1.33) and X5 (t=0.75) are not statistically significant. X1 and X5 should be dropped (omitted).