t-test_z_test

Test Statistics

Z- test Using One-Sample Mean Problem: An electrical Company claims that lives of the light bulbs if   manufactures are normally distributed with a mean of 1000 hours and a standard deviation of 150 hours. What can you say about this claim if a random sample of 100 bulbs produced by this company has a mean life of 980 hours? Use 0.05 level of significance.

Solution: Step 1.State the null and alternative hypothesis. Ho: The average life of all the bulbs produced by the firm is 1000 hours (μ = 1000). Ha: The average life of all the bulbs produced by the firm is not equal to 1000 hours (μ ≠ 1000). Step 2. Specify the level of significance. Use α = 0.05 Step 3. Determine the test statistic to be used. Z-test using one sample mean should be used since n=100 (n≥30) at α = 0.05, Zt = ± 1.96

Step 4. Compute the value. Ẍ - µ     √n Zc= s 980 – 1000 √ 100 = ----------------------150 -20 √ 100 -200 = -----------= ---------------------150 150

Zc = -1.33

Step 5. The /Zc= -1.33/ < / Zt = ± 1.96. Since the /Zc value/ is less than /Zt value/, accept the null hypothesis and reject the alternative hypothesis. Step 6. Therefore, there is a sufficient evidence to support the claim of the manufacturer that the average life of the bulbs it produces is 1000 hours.

Z- Test Using Two-Sample Means Problem: A bank is opening a new branch in one of two

neighborhoods. One of the factors considered by the bank as whether the average monthly family income (in thousand pesos) in the two neighborhoods differed. From census records, the bank drew two random samples of 100 families each and obtained the following information: Neighborhood Sample A

Sample B

X₁ = 10,800

X₂ = 10,300

S₁ = 300

S₂ = 400

n₁ = 100

n₂ = 100

The bank wishes to test the null hypothesis that the two neighborhoods have the same mean income. What should the bank conclude? Test the hypothesis using α = 0.05.

Solution: Step 1.State the null and alternative hypothesis. Ho: The average monthly family income of the two neighborhoods, A and B are equal.

is

Ha: The average monthly family income of neighborhood A not equal to neighborhood B.

Step 2. Specify the level of significance. Use α = 0.05 Step 3. Determine the test statistic to be used. Z-test using two sample means should be used since n=100 (n≥30) at α = 0.05, Zt = ± 1.96.

Step 4. Compute the value. X₁ -- X₂ Zc=

S₁² n₁

S₂² n₂

10,800 - 10,300 = 300² 100

400² 100 500

500

=

= 90,000 100

160,000 100

500

2,500 = 500 50

= 900

1,600

Zc= 10

Step 5. Decision: The /Zc= 10/ > / Zt = ± 1.96. Since the /Zc value/ is greater than /Zt value/, reject the null hypothesis and accept the alternative hypothesis. Step 6. Conclusion The average family income of neighborhood A is higher than that of neighborhood B at α = 0.05.

T- test Using Two-sample Means Problem: A taxi company is trying to determine whether the use of radial tires and belted tires provide the same fuel. Twelve cars were driven twice over a prescribed test course; and for each test, a car used a different type of tire ( radial or belted) in random order. The data obtained were recorded as follows: Type of tire

X

S

Radial

5.75

1.10

Belted

5.61

1.30

At 0.05 level of significance, can we conclude that cars equipped with radial tires and belted tires provide the same fuel consumption? Assume that data are normally distributed.

Step 1.State the null and alternative hypothesis. Ho: There is no significant difference on the fuel consumption of cars using radial and belted tires. Ha: There is a significant difference…. Step 2. Specify the level of significance. 0.05 Step 3. Determine the test statistic to be used. t-test; df = 12+12-2 = 22 tabular value: ±2.074

two-tailed test

Step 4. Compute the value. X₁ - X₂ tc = [S₁² (n₁ - 1) + S₂² (n₂ - 1)] (n₁ + n₂) – 2

1

1

n₁

n₂

5.75 – 5.61 =

[ (1.10)² (12 – 1) + (1.30)² (12 – 1) ] (⅟₁₂ + ⅟₁₂) (12 + 12) -2

0.14 = 1.21 (11) + 1.69 (11) 22

(⅙)

0.14 = 13.31 + 18.59 ( ⅙) 22 0.14 = 31.9 22

( 1/6 )

0.14 = 0.241666666 0.14 =

tc = 0.285 0.49159604

Step 5. Decision: The /tc= 0.285/ < / Tt = ± 2.074. Since the /tc value/ is less than /Tt value/, accept the null hypothesis and reject the alternative hypothesis. Step 6. Conclusion Therefore, we can say that there is no significant difference on the fuel consumption of cars using radial and belted tires at α = 0.05, with df = 22.

Analysis of Variance Example 1. In a Statistics class, 10 students are randomly assigned to each of four groups. Each group is asked to perform a set of tasks after exposure to the experimental treatment. Do the groups differ in task performance? Use the raw score method to test this problem at 5% level of significance.

Analysis of Variance Group 1

Group 2

Group 3

Group 4

20

19

18

16

18

18

18

16

17

18

15

15

17

17

14

15

15

15

13

14

14

14

12

12

12

13

12

12

11

13

11

10

10

10

10

8

10

8

9

5

Anova: Single Factor

SUMMARY Groups

Count

Sum

Average

Variance

Column 1

10

144

14.4

12.71111

Column 2

10

145

14.5

13.16667

Column 3

10

132

13.2

9.511111

Column 4

10

123

12.3

13.56667

ANOVA Source of Variation

SS

df

MS

F

P-value

Between Groups

33

3

11

0.898774

0.451281

Within Groups

440.6

36

12.23889

Total

473.6

39

F crit  

2.866266

 Analysis of Variance

Assignment: 1. Three different milling machines were being considered for purchase by a clothing manufacturing company. The company would be purchasing hundreds of these machines. To make sure it made the best decision, the company borrowed five of each machines and were assigned to 15 technicians. Each machine was put through a series of tasks and was rated using a standard test. The following results were obtained:

1. State the hypothesis: Ho: There is no significant difference among the performance using a standardized test of the three machines. Ha: There is a significant difference among the performance using a standardized test of the three machines. 2. Level of significance: α = 0.01

3. Test statistic: F – test

Analysis of Variance

Machine 1

Machine 2

Machine 3

24.5

28.4

26.1

23.5

34.2

28.3

26.4

29.5

24.3

27.1

32.2

26.2

29.9

30.1

23.8

SUMMARY Groups

Count

Sum

Average Variance

Column 1

5

131.4

26.28

6.172

Column 2

5

154.4

30.88

5.357

Column 3

5

128.7

25.74

3.183

ANOVA Source of Variation

SS

Between 79.78533 Groups

df

2

Within Groups

58.848

12

Total

138.6333

14

MS

F

39.89267 8.13472

4.904

P-value

0.00585

F crit  

6.9266

5. Decision: Since the computed value is greater than the

tabular value, reject the null hypothesis and accept the alternative hypothesis. 6. Conclusion: Thus, there is a significant difference among the performance using a standardized test of the three machines at α = 0.01.

Pearson’s Product Moment Correlation Coefficient Assignment

1. To determine the relationship between years of education and salary potential, 10 persons who have been employed for 5 years were interviewed. The results obtained on their number of years of higher education (college degree and higher) and their monthly salaries are shown in the following table:

Assignment Below are data from an experiment on the effects of sleep deprivation on the time it takes to solve mathematical problems. Subjects went without sleep for either 16, 24, 32 or 40 hours. Are increased levels of sleep loss associated with longer problem solving time? Do a test statistic using 5% level of   significance.

Problem Solving Time (in hours) Hours without Sleep

16

24

32

40

4

4

5

5

4

6

6

6

5

6

8

7

6

7

8

10

7

8

9

11

7

8

9

11

1. Ho: There is no significant difference on the time solving two mathematical problems when categorized according to the hours without sleep.

Ha: There is a significant difference on … 2. 0.05 level of significance 3. Analysis of Variance using F-test Ft (3, 21) = 3.07

4. Compute: ANOVA: Single Factor

SUMMARY Groups

Count

Sum

Average Variance

16 Hours

6

33

5.5

1.9

24 Hours

6

39

6.5

2.3

32 Hours

6

45

7.5

2.7

40 Hours

6

50

8.333333 7.066667

ANOVA Source of Variation

SS

df

MS

F

P-value

F crit

Between Groups

27.125

3

9.041667 2.589499 0.08139 3.098391

Within Groups

69.83333

20

3.491667

Total

96.95833

23

5. Since the computed value (Fc = 2.60) is less than the tabular value (Ft = 3.07), accept the null hypothesis and reject the alternative hypothesis.

6. Thus, there is no significant difference on the time solving two mathematical problems when categorized according to the hours without sleep at α = 0.05.

Chi-Square