chapter 17

Oxford Fajar Sdn. Bhd. (008974-T) 2014

CHAPTER 13 CHAPTER 13

Focus on Exam 17

1 (a) Type I error means rejecting the null hypothesis (H0) when it is in fact true.

(b) H0: m = 3 kg

H1: m > 3 kg

Under H0 and at a = 0.01, accept H

1 if z > 2.326.

Let X be the mean mass of the watermelon.

ZX

n

X

=

=

ms

~ ( , )

.

N

0 1

30 65

30

To use the new organic fertilizer, z > 2.326

x 30.65

30

2.326

>

x

x

> +

>

=

3 0 276

3 276

3 276

.

.

.k 2 (a) The hypotheses: H

0: m = 10 kg

H1: m > 10 kg

(b) Let X be the mean weight gain of a calf.

If H0 is true, then Z

X

n

N(0, 1).

=

Using a right-tailed test at 5% significance level, reject H0 if z > 1.645.

Test statistic: z11.5 10

2.5

16 = 2.4

=

CHAPTER 17


ACE AHEAD Mathematics (T) Third Term2

Since zobserved

> 1.645, reject H0 and conclude that there is sufficient evidence to show that

the new feed increases the weight gain of a calf at the 5% significance level. 3 H

0: m = 140 cm

H1: m 140 cm

If H0 is true, Z

X

n=

ms /

~ ( , ).N 0 1

For a two-tailed test and at 1% level of significance, reject H0 of z > 2.576.

Test statistic: z =

=

126 5 14031 3

402 73

..

.

Since zobserved

< 2.576, therefore reject H0 and conclude that the mean is different from 140 cm

at the 1% level of significance.

4 H0: m RM2400 (or m = RM2400)

H1: m > RM2400

If H0 is true,

=

ZX

n/~ N(0, 1).

For a right-tailed test, at the 2.5% significance level, reject H0 if z > 1.96.

Test statistic: z =2520 2400

450

50

1 89 = .

Since zobserved

< 1.96, do not reject H0 at the 2.5% level of significance and conclude that the

insurance company should not be concerned.

5 Let X be a random variable such that X ~ N( m, s 2). H

0: m = m

0

H1: m m

0

Then ZX

n

~ N(0, 1).0

=

At 5% level of significance, H0 is rejected if z >1 96. .

Hence H0 is rejected if

x

n

1.96.0

>

xn

xn

< > +m s m s0 01 96 1 96. . or


Fully Worked Solutions 3

H0: m = m

0

H1: m < m

0

Then ZX

n

~ N(0, 1).0

=

At 1% level of significance, H0 is rejected if z < 2.326.

Hence H0 is rejected if

x

n

2.326.0

<

> +x

n2.326

0

6 (a) Significance level is the probability that test statistic lies in the critical region. It is also a fixed probablility of rejecting H

0 when it is in fact true.

(b) The hypotheses are H0: m = 12 hours

H1: m < 12 hours

If H0 is true, Z

X

n

~ N(0, 1).

=

At the 5% significance level, reject H0 if z 1.645.<

Calculation: .

.

s 2 25049

3 24

10 712

=

=

Test statistic: z =

=

11 3 12

10 712

50

1 51

.

.

.

Since zobserved

> 1.645, do not reject H0 and conclude that the doctors claim is justified.

7 (a) The hypotheses are H0: m = 25 g

H1: m > 25 g

(b) Using a right-tailed test and at the 1% significance level, reject H0 if z > 2.326.

If H0 is true, then Z

X

n

~ N(0, 1).

=

Calculation: .

.

s 2 6059

11 9

12 102

=

=

Test statistic: z =

=

26 2 25

12 102

60

2 67

.

.

.



Since zobserved

> 2.326, reject H0 and conclude that mean intake of fat by women in the city

exceeds 25 g per day at the 1% significance level. (c) The women chosen are assumed to be random and independent (not related) to each other.

8 H0: m = 30 minutes

H1: m < 30 minutes

Using a one-tailed test and at the 5% level of significance, reject H0 if z < 1.645.

Test Statistic: ZX

n

=

x

x

n= =

=

433515028 9. minutes

s 2 =

1

149137 301

4335

150

2

= 80.668

s = 8.98 minutes

z28.9 30

8.98

1501.500

=

=

Since zobserved

> 1.645, do not reject H0 and conclude that the mean time spent shopping by

customers is 30 minutes at the 5% level of significance. 9 Let X be the mass of paint in a tin.

H0: m =1.005 kg

H1: m > 1.005 kg

At the 5% significance level, reject H0 if z > 1.645.

Given x( 1) 2.09i

= and n = 160.

x

nx

x

= +

=

=

=

2 09

1601

1 013

1

11 005

1

1591 2

2 2

2

.

.

( . )

( ) (

s

00 005 1 160 0 005

1

1590 6788 0 0209 0 004

2. ) ( ) ( . )

. . .

x + = + [[ ]=

=

0 004163

0 064522.

.s



Test statistic: zX

n=

=

=

1 005

1 013 1 0050 06452

160

1 57

.

/

. ..

.

s

Since zobserved

< 1.645, do not reject H0 and we conclude that the mean mass of paint in a tin do

not exceed 1.005 kg at the 5% level of significance.

10 (a) Let m be the population mean.

Xii

( ) ==

2 1 291

100

.

E Xii

( ) == 2 1 291

100

.

E EX Xii

ii= =

=

=1

100

1

100

100 2 1 29 201 29. .

= ( ) =

=

=

=

m E EX

Xii 1

100

100

201 29

1002 0129

.

. kg

s 22

1

100

2

1

100

2 0129

99

2 0 0129

9

= ( ) =( )

=

( )

=

=

Var X

X

X

ii

ii

.

.

99

2

99

2 0 0129 2

99

100 0 01292

1

100

1

100

2

=

( )+

( ) ( )+

( )= =

X Xii

ii

..

1100

3 9

99

0 0258 1 29

990 0129

0 0392

2

2

=

( )+

=

. . ..

. kg

(b) H0: m = 2 kg

H1: m > 2 kg

Under H0 , the value of the test statistic,



zobserved =

=

2 0129 2

0 0392

100

0 652

.

.

.

At 10% level of significance, reject H0 if z

observed > 1.282.

Hence, do not reject H0 and conclude that the manufacturer is understating the average

mean of the contents of salt of a bag.

11 Given x 291.2kg =

Unbiased estimate of xx

n291.2

1002.912kg

= =

=

=

Given x 915.8kg2 2 =

Unbiased estimate of 22

2 2

2

1

100 915.82.912

99 100

n x x

n n ns s

s

= =

= = 0.6851= 0.828

H0: m = 3 kg

H1: m < 3 kg

Test statistic: ZX

n

=

Since s is unknown, the unbiased estimate s is used, therefore Z N 0,1( ) approximately. At 10% significance level, critical region is z < 1.282.

Under the null hypothesis, z2.912 30.8277

1001.063

=

=

Since zobserved

> 1.282, therefore there is no sufficient evidence to reject H0.

Let 100a % be the largest level of significance at which the test would result in rejection of the null hypothesis.

For H0 to be rejected, z1.063 a <

z 1.063 >

From tables, P[z > 1.063] = 0.144

Therefore the largest level of significance level at which the test would reject H0 is 14.4%.



12 Let X denotes the random variable of mass of ducks.

E (X ) = m, Var(X ) = 2, n = 200

By central limit theorem, X is approximately normal.

X N ,

200

2

and standard deviation of X is 200 .

Given x x382.8 and 824.22 = =

The unbiased estimate of Xn

x1

382.8

2001.914kg

= =

=

=

The unbiased estimate of ( )

= =

n x

x

n 1

12 2 2

2

=

=

1199

824.2382.8200

0.46

22

H0: m = 1.8 kg

H1: m > 1.8 kg

Using one-tail test and at the 5% level of significance, reject H0 if z

observed > 1.645.

Test statistics: ZX

n

=

=

=

ms

1 914 1 81 08

200

1 49

. ..

.

Since zobserved

< 1.645, do not reject H0 and conclude that, at the 5% level of significance, the

mean mass of a duck is equal to 1.8 kg.

13 Let X be the mean lifespan of the sample of bulbs X N ,150

15.

2

H

0: m = 5500 hours

H1: m > 5500 hours

At the 5% significance level, reject H0 if z >1.645

Calculation: x x

n

82690

155512.67=

= =

Test statistic: zx

n

=

=

=

ms

5512 67 5500150

15

0 327

.

.



Since zobserved

< 1.645, do not reject H0 and conclude that there is no evidence to support the

suppliers claim.

14 H0: m = 32.5 cm

H1: m > 32.5 cm

Given x > 35 cm, n = 5 and a = 10 cm.

z > 35 32.5

10

50

z > 1.768

Therefore H0 is rejected if z > 1.768

Type I error = a = P[z > 1.768]

= 0.0385

15 Let X be the mean lifespan of the sample of tyres.

(a) (i) Lifespan of tyres has normal distribution.

Sample mean lifespan also has a normal distribution.

X N ,

121where

121

1202500

22 2

=

X ~ N(m, 2510.3952)

(ii) n = 121 (large sample)

By central limit theorem, X is a approximately normal.

X ~ N

121

2 where

2 = 121

120 25002

X ~ N(m, 2510.3952)

(b) The hypotheses are H0: m = 50 000 km

H1: m < 50 000 km

If H0 is true, then Z =

X

n

ms

~ N(0, 1)

where 2

= 121120

25002

= 2510.395

For left-tailed test and at 1% level of significance, reject H0 if z < 2.326.

Test statistic: z = 49 200 50 0002510 395

121

.

= 3.51



Since z < 2.326, reject H0 and conclude that there is no sufficient evidence to justify the

manufacturers claim at the 1% significance level. 16 Let p be the proportion of pets that like Petsuka and X be the number of pets that like

Petsuka.

H0: p = 0.8

H1: p < 0.8

If H0 is true, X ~ B(300, 0.8)

X ~ N(240, 48)

Reject H0 if z

observed < 1.645.

Test statistic: z =

=

227 5 240

48

1 81

.

.

Since zobserved

< 1.465, H0 is rejected at the 5% level of significance and conclude that the

manufacturers claim cannot be accepted.

17 H0: p = 0.3

H1: p < 0.3

If H0 is true, then p = 0.3

X ~ B(120, 0.3)

X ~ N(36, 25.2)

At 1% level of significance, reject H0 if z < 2.326

z =

=

21 5 36

25 2

2 89

.

.

.

Since zobserved

< 2.326, therefore we reject H0 and conclude that p < 0.3 at the 1% level of

significance.

18 Let X be the number of replies that were in favour of the proposal.

H0: p = 0.7

H1: p > 0.7

If H0 is true, then X ~ B(360, 0.7)

X ~ N(252, 75.6)

Using the right-tailed test and at the 1% level, reject H0 if z > 2.326.

Test statistic: z =

=

271 5 252

75 6

2 24

.

.

.

Since zobserved

< 2.326, do not reject H0 and we conclude that the population proportion in favour

of the proposal is 0.7 at the 1% level of significance.



19 Let X be the number of voters who intended to support the party.

H0: p = 0.42

H1: p < 0.42

If H0 is true, then X ~ B(700, 0.42)

X ~ N(294, 170.52)

Using the left-tailed test and at the 5% level, reject H0 if z < 1.645.

Test statistic: z =

=

271 5 294

170 52

1 72

.

.

.

Since zobserved

< 1.645, we reject H0 and conclude that the support has decreased since the last

election at the 5% level of significance.

20 H

H

0

1

1

61

6

:

:

p

p

=

>

Let X be the number of a five occurring.

If H0 is true, then X

B 15

1

6, .

Using a one-tailed test at 10% significance level, reject H0 if P(X c) < 0.10

where c is the critical value.

P PX X

C C

5 1 4

1

5

6

1

6

5

6

1

615

0

15 0

151

14

[ ] = [ ]

=

+

+

+

+

152

13 2

153

12 3

15

5

6

1

6

5

6

1

6C C

CC4

11 45

6

1

6

1 0 9104

0 0898

=

=

.

.

Since P(X 5) < 10%, the observed value of x = 5 is in the critical region. Hence, reject H0

and conclude that the die is biased in favour of a five.

21 H0: p = 0.5

H1: p 0.5

Let X represent the number of tails obtained.

If H0 is true, then X ~ B(20, 0.5).

At = 0.05 and using a two-tailed test, the two required critical values are c1 and c2 such that

P(X c1) 0.025 and P (X c

2) 0.025.

From tables, P(X 6) = 0.0577

P(X 5) = 0.0207

c1 = 5



P(X 14) = 1 P(X 13)

= 1 0.9423

= 0.0577

P(X 15) = 1 P(X 14)

= 1 0.9793

= 0.0207

c2 = 15

Therefore the critical region for X is X 5 or X 15.

As 6 lies between 5 and 15, do not reject H0 and conclude that the coin is not biased.

Type I error occurs when H0 is rejected when it is in fact true. This occurs when X 5 or X 15.

P[type I error] = P(X 5|p = 0.5) + P(X 15|p = 0.5)

= 0.0207 + 0.0207

= 0.042

22 Let X be the number of school children who own graphic calculators.

H0: p = 0.4

H1: p > 0.4

If H0 is true, p = 0.4.

So, X ~ B(200, 0.4)

X ~ N(80, 48)

Reject H0 if z

observed > 1.645

Test statistic: z =

=

92 5 80

48

1 80

.

.

Since zobserved

> 1.645, H0 is rejected and conclude that more than 40% of the school students

own graphic calculators.

23 H0: p = 0.95

H1: p < 0.95

X ~ B(120, 0.95)

X ~ N(114, 5.7)

Reject H0 if z < 1.96

z =

=

110 5 114

5 7

1 47

.

.

.

Do not reject H0 since z > 1.96 and we conclude that the wholesalers claim is justified.



24 H0: p = 0.65

H1: p > 0.65

X ~ B(80, 0.65)

X ~ N(52, 18.2)

Reject H0 if z > 1.282

z =

=

55 5 52

18 2

0 820

.

.

.

Do not reject H0 since the value of z is less than the critical value. There is no evidence to

justify the claim made by the council at 10% level of significance.

chapter 17

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