Ch8 ppt

Section 8-2Steps in Hypothesis Testing – Traditional Method

Chapter 8Hypothesis Testing

Section 8-2

Exercise #12b


Using the z table find the critical value (or values). = 0.01

– 2.33

– 2.33 0

Left - tailed test.

0.010.49

Section 8-2

Exercise # 12g


Using the z table find the critical value (or values). = 0.05

1.65 0

0.45

0.05

Right - tailed test.

1.65

Section 8-2

Exercise #12h


Using the z table find the critical value (or values).

= 0.01Two - tailed test.

2.58, – 2.58

– 2.58 2.58 0

0.495

0.005

Section 8-3z Test for a Mean


Section 8-3

Exercise #5


A report in USA TODAY stated that the average age of commercial jets in the United States is 14 years. An executive of a large airline company selects a sample of 36 planes and finds the averageage of the planes is 11.8 years. The standard deviation of the sample is 2.7 years. At = 0.01, can it be concluded that the average age of the planes in his Company is less than the national average?

C.V . = – 2.33

z = X – s

n

= 11.8 – 142.7

36

= – 4.89

– 2.33 0

– 4.89

0H : 14 H1 : < 14 (claim)

Reject the null hypothesis. There is enough evidence to support the claim that the average age of the planes in the executive’s airline is less than national average.

C.V . = – 2.33

0H : 14 H1 : < 14 (claim)

z = X – s

n

= 11.8 – 142.7

36

= – 4.89

Section 8-3

Exercise #7


The average one-year-old (both sexes) is 29 inches tall. A random sample of 30 one-year-olds in a large day care franchise resulted in the following heights. At = 0.05, can it be concluded that the average height differs from 29 inches?

25 32 35 25 30 26.5 26 25.5 29.5 32

30 28.5 30 32 28 31.5 29 29.5 30 34

29 32 27 28 33 28 27 32 29 29.5

– 1.96 0 1.96

C.V . = ± 1.96

0.944

z = X –

n

= 29.45 – 292.61

30

s = 2.61 X = 29.45

= 0.944

H0 : = 29 H : 29 ( )1 claim

Do not reject the null hypothesis. There is not enough evidence to support the claim that the average height differs from 29 inches.

C.V . = ± 1.96

z = X –

n

= 29.45 – 292.61

30

s = 2.61 X = 29.45

= 0.944

H0 : = 29 H : 29 ( )1 claim

Section 8-3

Exercise #13


To see if young men ages 8 through 17 years spend more or less than the national average of $24.44 per shopping trip to a local mall, the managersurveyed 33 young men and found the average amount spent per visit was $22.97.The standard deviation of the sample was $3.70. At = 0.02, can it be concluded that the average amount spent at a local mall is not equal to the national average of $24.44.

= – 2.28

= 22.97– 24.44

3.7033

C.V. = ± 2.33

– 2.33 2.330

H0 : = $24.44 H : $24.44 (claim)1

n = 33 X = $22.97

s = $3.70 = 0.02

z = X –

n

– 2.33 2.330

Do not reject the null hypothesis.

– 2.28

C.V. = ± 2.33

H0 : = $24.44 H : $24.44 (claim)1

n = 33 X = $22.97

s = $3.70 = 0.02

There is not enough evidence to support the claim that the amount spent at a local mall is not equal to the national average.

– 2.28

C.V. = ± 2.33

H0 : = $24.44 H : $24.44 (claim)1

n = 33 X = $22.97

s = $3.70 = 0.02

Section 8-3

Exercise #17


A study found that the average stopping distance of a school bus traveling 50 miles per hour was 264 feet (Snapshot, USA TODAY, March12, 1992). A group of automotive engineers decided to conduct a study of itsschool buses and found that for 20 buses, the average stopping distance of buses traveling 50 miles per hour was 262.3 feet.The standard deviation of the populationwas 3 feet. Test the claim that the average stopping distance of the company’s buses is actually less than 264 feet. Find the P-value. On the basis of the P-value, should the null hypothesis be rejected at = 0.01? Assume that the variable isnormally distributed.

The area corresponding to z = 2.53 is 0.4943. The P-value is 0.5 – 0.4943 = 0.0057. The decision is to reject the null hypothesis since 0.0057 < 0.01. There is enough evidence to support the claim that the average stopping distance is less than 264 feet.

= – 2.53

z = X –

n

= 262.3 – 264

320

H : 2640 H1 : < 264 (claim)

Section 8-4t Test for a Mean


Section 8-4

Exercise #3a


Find the critical value (or values) for the t test for each.

Right - tailed = 0.05

d.f . = 9 C.V . = + 1.833

n = 10

Section 8-4

Exercise #3b


Two - tailed

n = 18

= 0.10


d.f . = 17

C.V = ± 1.740

Section 8-4

Exercise #3c


Left - tailed

n = 6

= 0.01


d.f .= 5

C.V .= – 3.365

Section 8-4

Exercise #7


The average salary of graduates entering the actuarial field is reported to be $40,000. To test this, a statistics professor surveys 20 graduates and finds their average salary to be $43,228with a standard deviation of $4,000. Using = 0.05, has he shown the reported salary incorrect?

= 43,228 – 40,0004000

20

= 3.61

t = X –

n

d.f . = 19 C.V . = ± 2.093

H0 : = $40,000 1H : $40,000 (claim)

X = $43,228 = $40,000 n = 20

= 0.05

s = $4,000

3.61

2.093

2.0930

Reject the null hypothesis. There is enough evidence to support the claim that the average salary is not $40,000.

d.f . = 19 C.V . = ± 2.093 X = $43,228 = $40,000 n = 20

= 0.05

s = $4,000

H0 : = $40,000 1H : $40,000 (claim)

Section 8-4

Exercise #9


A researcher estimates that the average height of the buildings of 30 or more stories in a large city is at least 700 feet. A random sample of 10 buildings is selected, and the heights in feet are shown:

485 511 841 725 615520 535 635 616 582At = 0.025, is there enough evidence to reject the claim?

= 700 ft. n = 10 buildings

X= 606.5 s = 109.1 = 0.025

0 – 2.262

C.V. = – 2.262 d.f. = 9

= 606.5 – 700109.1

10

= – 2.71

0 1H : 700 (claim) H : < 700

t = X – S

n

– 2.71

Reject the null hypothesis. There is enough evidence to reject the claim that the average height of the building is at least 700 feet

= 700 ft. n = 10 buildings

X= 606.5 s = 109.1 = 0.025

C.V. = – 2.262 d.f. = 9

0 1H : 700 (claim) H : < 700

Section 8-4

Exercise #13


Last year the average cost of making a movie was $54.8 million. This year, a random sample of 15 recent action movies had an average production cost of $62.3 million with a variance of $90.25 million.At the 0.05 level of significance, can it be concluded that it costs more than average to produce an action movie?

= $ 54.8 million n = 15 movies

X= $ 62.3 million s 2 = $ 90.5 million

= 0.05

0 1.761

d.f . = 14

= $62.3 – $54.89.5

15

= 3.06

C.V. = 1.761

t = X – S

n

3.06

H : $54.8 H : > $54.8 (claim)0 1

Reject the null hypothesis. There is enough evidence to support the claim that the cost to produce and action movie is more than $54.8 million.

= $ 54.8 million n = 15 movies

X= $ 62.3 million s 2 = $ 90.5 million

= 0.05 d.f . = 14 C.V. = 1.761

H : $54.8 H : > $54.8 (claim)0 1

Section 8-4

Exercise #17


A report by the Gallup Poll stated that on average a woman visits her physician 5.8 times a year. A researcher randomly selected 20 women and the following data was obtained.

At = 0.05 can it be concluded that the average is still 5.8? Use the P - value method.

3 2 1 3 7 2 9 4 6 6

8 0 5 6 4 2 1 3 4 1

0 1: = 5.8 (claim) : 5.8H H

= 5.8 X = 3.85 n = 20 women

s = 2.52 = 0.05 d.f. = 19

= 3.85 – 5.8

2.52

20

= – 3.46

0 1: = 5.8 (claim) : 5.8H H

t = X – S

n

P - value < 0.05

Since the P - value is less than 0.05, reject the null hypothesis. There is not enough evidence to support the claim that the average number of visits is the same as in the Gallup Poll.

= 5.8 X = 3.85 n = 20 women

s = 2.52 = 0.05 d.f. = 19

0 1: = 5.8 (claim) : 5.8H H P - value < 0.05

Section 8-5z Test for a Proportion


Section 8-5

Exercise #7


It has been reported that 40% of the adult population participates in computer hobbies during their leisure time. A random sample of 180 adults found that 65 engaged in computer hobbies. At = 0.01, is there sufficient evidenceto conclude that the proportion differsfrom 40%?

0 1H = 0.40 H 0.40 ( ): p : p claim

– 2.58 0 2.58

– 1.07

z = p̂ – p

pqn

= 0.361– 0.40

(0.40)(0.60)

180

= – 1.07

p̂ = 65180

= 0.361 p = 0.40 q = 0.60

C.V . = ± 2.58

0 1H = 0.40 H 0.40 ( ): p : p claim

Do not reject the null hypothesis. There is not enough evidence to support the claim that the proportion is different from 0.40 (40%).

p̂ = 65180

= 0.361 p = 0.40 q = 0.60

C.V . = ± 2.58

0 1H = 0.40 H 0.40 ( ): p : p claim

Section 8-5

Exercise #9


An item in USA TODAY reported that 63% of Americans owned an answering machine. A survey of 143 employees at a large school showed that 85 owned an answering machine. At a = 0.05, test the claim that the percentage is the same as stated in USA TODAY .

0 1H = 0.63 ( ) H 0.63: p claim : p

z = p̂ – p

pqn

= 0.5944– 0.63

(0.63)(0.37)143

= – 0.88

p̂ = 85143

= 0.5944 p = 0.63 q = 0.37

C.V . = ± 1.96

– 1.96 0 1.96

– 0.88

0 1H = 0.63 ( ) H 0.63: p claim : p

Do not reject the null hypothesis. There is not enough evidence to reject the claim that the percentage is the same.

z = p̂ – p

pqn

= 0.5944– 0.63

(0.63)(0.37)143

= – 0.88

0 1H = 0.63 ( ) H 0.63: p claim : p

p̂ = 85143

= 0.5944 p = 0.63 q = 0.37

C.V . = ± 1.96

Section 8-5

Exercise #15


Researchers suspect that 18% of all high school students smoke at least one pack of cigarettes a day. At Wilson High School, with an enrollment of 300 students, a study found that 50 studentssmoked at least one pack of cigarettes a day. At = 0.05, test the claim that 18% of all high school students smoke at least one pack of cigarettes a day. Use the P - value method.

z =p̂ – p

pqn

=0.1667– 0.18

(0.18)(0.82)300

=– 0.60

Area = 0.2257

p̂ =

50300 = 0.1667 q = 0.82 p = 0.18

P - value = 2 (0.5 – 0.2257) = 0.5486

H0 :p = 0.18 (claim) 1H : 0.18p

Since P - value > 0.05, do not reject the null hypothesis. There is not enough evidence to reject the claim that 18% of all high school students smoke at least a pack of cigarettes a day.

p̂ =

50300 = 0.1667

z =p̂ – p

pqn

=0.1667– 0.18

(0.18)(0.82)300

=– 0.60

q = 0.82 p = 0.18

H0 :p = 0.18 (claim) 1H : 0.18p

Section 8-5

Exercise #19


A report by the NCAA states that 57.6% of football injuries occur during practices. A head trainer claims that this is too high for his conference, so he randomly selects 36 injuries and finds that 17 occurred during practices. Is his claim correct, using = 0.05 ?

0 1H 0 576 H : < 0.576 (claim)p p : .

= 0.472 – 0.576

(0.576)(0.424)36

= – 1.26

= 0.05 C.V . = –1.65

– 1.65 0

0.05

z = p̂ – p

pqn

p̂ = 1736

= 0.472 , p = 0.576 q = 0.424 n = 36

0 1H 0 576 H : < 0.576 (claim)p p : .

Do not reject the null hypothesis. There is not enough evidence to support the claim that the percentage of injuries during practice is below 57.6%.

– 1.26

= 0.05 C.V . = –1.65

– 1.65 0

0.05

0 1H 0 576 H : < 0.576 (claim)p p : .

Section 8-6c2 Test for a Variance or Standard Deviation


Section 8-6

Exercise #5


Test the claim that the standard deviation of the number of aircraft stolen each year in the United States is less than 15 if a sample of 12 years had a standard deviation of 13.6. Use = 0.05.

H1 : < 15 (claim) 0H : 15

0 4.575

0.025

9 04

. Do not reject the null hypothesis.

There is not enough evidence to support the claim that the standard deviation is less than 15.

C.V . = 4.575 11. .d f

=

(12 – 1)(13.6)2

152 = 9.04

22

– ( 1)= 2 n s

= 0.05

9.04

H1 : < 15 (claim) 0H : 15

Section 8-6

Exercise #7


The manager of a large company claims that the standard deviation of the time (in minutes) that it takes a telephone call to be transferred to the correct office in her company is 1.2 minutes or less. A sample of 15 calls is selected, and the calls are timed. The standard deviation of the sample is 1.8 minutes. At = 0.01, test the claim that the standard deviation is less than or equal to 1.2 minutes. Use the P-value method.

0H : 1.2 ( )claim H1 : > 1.2

0 31.319

0.005

P - value < 0.005 (0.0047)

=

(15 – 1)(1.8)2

(1.2)2 = 31.5

31.5

= 0.01 s = 1.8 n = 15 d.f. = 14

22

– ( 1)= 2 n s

0H : 1.2 ( )claim H1 : > 1.2

Reject the null hypothesis. There is not enough evidence to support the claim that the standard deviation is less than or equal to 1.2 minutes.

P - value < 0.005 (0.0047)

31.5

0 31.319

0.005

Section 8-6

Exercise #9


A random sample of 20 different kinds of doughnuts had the following calorie contents. At = 0.01, is there sufficient evidence to conclude that the standard deviation is greater than 20 calories?

290 320 260 220 300 310 310 270 250 230

270 260 310 200 250 250 270 210 260 300

d.f . = 19 = 0.01 C.V . = 36.191 s = 35.11

= (20 – 1)(35.11)2

202 = 58.55

0 36.191

H1 : > 20 (claim) 0H : 20

22

– ( 1)= 2 n s

58.55

Reject the null hypothesis. There is enough evidence to support the claim that the standard deviation is greater than 20.

d.f . = 19 = 0.01 C.V . = 36.191 s = 35.11

= (20 – 1)(35.11)2

202 = 58.55

22

– ( 1)= 2 n s

H1 : > 20 (claim) 0H : 20

Section 8-6

Exercise #13


A random sample of home run totals for National League Home Run Champions from 1938 to 2001 is shown. At the 0.05 level of significance, is there sufficient evidence to conclude that the variance is greater than 25?

34 47 43 23 36 50 42

44 43 40 39 41 47 45

14 – 1 45 38=

25

.

2

2– ( 1)=

2 n s

0 22.362

s2 = 45.38 = 0.05 d.f . = 13

C.V. = 22.362

= 23.622

23.6

2 20 1H : 25 H : > 25( ) claim

Reject the null hypothesis. There is evidence to support the claim that the variance is greater than 25.

0 22.362

23.6

s2 = 45.38 = 0.05 d.f . = 13

C.V. = 22.362

2 20 1H : 25 H : > 25( ) claim

2 23.622

Section 8-7Additional Topics Regarding Hypothesis Testing


Section 8-7

Exercise #1


A ski shop manager claims that the average of the sales for her shop is $1800 a day during the winter months. Ten winter days are selected at random, and the mean of the sales is $1830. The standard deviation of the population is $200. Can one reject the claim at = 0.05?Find the 95% confidence interval of themean. Does the confidence interval Interpretation agree with the hypothesis test results? Explain. Assume that the variable is normally distributed.

H : 18000 1H : = 1800 (claim)

z = X – n

C.V . = ± 1.96

=1830 – 1800

200

10

= 0.47

Do not reject the null hypothesis. There is not evidence to reject the claim that the average of the sales in $1800.

– 1.96 0 1.96 0.47

H : 18000 1H : = 1800 (claim)

The 95% confidence interval of the mean is:

1830 – 1.96(

200

10) < < 1830 + 1.96(

200

10)

1706.04 < < 1953.96

The hypothesized mean is within the interval, thus we can be 95% confident that the average sales will be between $1,706.94 and $1,953.96.

X – z

2

n < < X + z

2

n

Section 8-7

Exercise #5


From past studies the average time college freshmen spend studying is 22 hours per week. The standard deviation is 4 hours. This year, 60 students were surveyed, and the average time thatthey spent studying was 20.8 hours. Test the claim that the time students spend studying has changed. Use = 0.01. It is believed that the standard deviation is unchanged. Find the 99% confidence interval of the mean. Do the results agree? Explain.

H0 : = 22 1H : 22 ( )claim

C.V . = ± 2.58

= 20.8 – 224

60

= – 2.32

z = X – n

– 2.58 0 2.58 – 2.32

H0 : = 22 1H : 22 ( )claim

Do not reject the null hypothesis. There is not enough evidence to support the claim that the average studying time has changed.

C.V . = ± 2.58

– 2.58 0 2.58 – 2.32

H0 : = 22 1H : 22 ( )claim

The 99% confidence interval of the mean is:

20.8 – 2.58(

4

60) < < 20.8 – 2.58(

4

60)

19.47 < < 22.13

The 99% confidence interval contains the hypothesized mean of 22. There is not enough evidence to support the claim that the average studying time has changed.

X – z

2

n < < X + z

2

n

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Ch8 ppt