Solutions 8 - Hypothesis Testing1 8 - Hypothesis... · Section 8 Hypothesis Testing Solutions...

Section 8 Hypothesis Testing Solutions Teaching Team Frances A. Pfab

Section 8: Solutions

Case Study 1 – Packaging Flour

1. • Means

• One sample v. standard

• Not functioning correctly ⇒ Two-sided

HO: µ = 1 kg (i.e. standard weight) HA: µ ≠ 1 kg

• t-distribution since n < 30

Sample values:

n = 25 x = 1.12 kg s = 0.2

Test statistic:

t = n/s

x µ−

= 25/2.0

0.112.1 − = 3.0 with 24 d.f.


• Reject HO at the 1% level of significance.

• There is a significant difference between the average weight of the sample and the weight stated on the packet at the 1% level of significance. This implies that the machine is not functioning correctly.

2. As above:

s = 0.7 t = 0.86

• Reject HA.

• No significant difference.

3. (a)(1) As above:

n = 250 s = 0.2 z = 9.49

• Reject HO at 1% level of significance.

• There is a significant difference at the 1% level of significance.

Note: Use critical values for z, since n ≥ 30.


3. (a)(2) n = 250 s = 0.7 z = 2.71



3. (b)(1) n = 150 s = 0.2 z = 7.35



3. (b)(2) n = 150 s = 0.7 z = 2.10


• There is a significant difference at the 5% level of significance


4. n = 50 x = 0.92kg s = 0.02kg

One-sided test – i.e. underfilled.

HO: µ ≥ 1 HA: µ < 1

zcalc = –28.28

• Reject HO.

• There is evidence that average weight of the bags is significantly less than the stated weight at the 1% level of significance.


Case Study 2 – Food Purchasing

Edinburgh:

16.44£=µ

Barnton:

n = 120 x = £53.43 s2 = £49.12 s = 7.01

1. HO: µ = 44.16 HA: µ ≠ 44.16

Two-sided test:

z = ns

x µ−

= 12001.7

16.4443.53 −

= 95.1001.7

27.9

= 640.027.9

= 14.48



• The average amount spent in the Barnton area is significantly different from the average amount spent last year in Edinburgh at the 1% level of significance.

2. Now a one-sided test:

HO: µ ≤ 44.16 HA: µ > 44.16

z = 14.48



• The average amount spend in the Barnton area is significantly higher than the average amount spent last year in Edinburgh at the 1% level of significance.


Case Study 3 – Gun Powder

• Means

• One sample v population

• Difference ⇒ Two-sided

• t-test since n < 30

HO: µ = 3,000 ft/sec HA: µ ≠ 3,000 ft/sec

x )xx( − )xx( − 2

3005

2925

2945

2965

2995

3005

2935

2905

45

–35

–15

5

35

45

–25

–55

2025

1225

225

25

1225

2025

625

3025

∑ x = 23680 ∑ − )xx( 2 = 10400

n = 8

823680

nx

x == ∑ = 2960

∑ − 2)xx( = 10400


s2 = 7

10400)1n()xx( 2

=−−∑ = 1485.71

s = 38.54

t = 54.3800.40

854.3830002960

ns/x

=−

=µ− × 2.828

= –2.935

tcalc = –2.935 with 7 degrees of freedom

From tables:

t2.5% = –2.365 at the 5% level of significance

t0.5% = –3.499 at the 1% level of significance


• The average velocity of the 8 shells tested was significantly different from the manufacturer's claim of 3,000 ft/sec at the 5% level of significance.


Case Study 4 – Wages Policy

Firm: 12 managers:

n = 12 average salary = x = £26,750 s = £3,100

• One sample against population (rival firm)

• Means

• Difference ⇒ Two-sided test

• t-test since n < 30

HO: µ = £28,500 HA: µ ≠ £28,500

t = ns x µ−

t = 100,3750,1

123100500,28750,26

−=− × 3.46

tcalc = –1.953 with 11 degrees of freedom

From tables:

t2.5% = 2.201 with 11 degrees of freedom


• Reject HA, since t2.5% < tcalc < 0.

• There is no significant difference between the average salary of the managers in the first firm and the average of £28,500 in the second firm.


Case Study 5 – Silicon Chips

Use F-test to test variability

HO: 2Aσ ≤

H

2Bσ

A: 2Aσ > 2

Bσ

B

A

VarVar=F with (9,9) degrees of freedom

2

2

)7.95()4.173(

=

calcF = 3.283 with (9,9) degrees of freedom

From tables:

5%F = 3.18 with (9,9) d.f. at 5% = 5.35 with (9,9) d.f. at 1% 1%F

• Reject HO at 5% level of significance

• The variability for supplier A is significantly greater than the variability for supplier B at the 5% level of significance.


Case Study 6 – Bulb Production

Use F-test.

HO: 2Aσ ≤

H

2Bσ

A: 2Aσ > 2

Bσ

Fcalc = 000,37000,92

VarVar

B

A = = 2.486 with (49,49) d.f.

From tables: F5% = 1.60 at 5% level of significance, using (50,50) d.f.

F1% = 1.94 at 1% level of significance.


• The variability of bulbs produced on line A is significantly greater than the variability of those produced on line B at the 1% level of significance.


Case Study 7 – Stick-on Soles

Use paired t-test: Pair 1 2 3 4 5 6 7 8

Old 18 17 14 11 10 7 5 6

New 31 20 18 17 9 8 10 7

Difference (x) 13 3 4 6 –1 1 5 1

Difference = x = New – Old

This could be analysed as a two-tailed test.

• HO: The old and new materials show the same amount of wear: µ = 0.

• HA: The materials show different amounts of wear: µ ≠ 0.

x= 4.0 s2 = 18.57 s = 4.31

tcalc = ns x µ−

= 831.400.4 −

= 31.40.4 × 2.83

= 2.626 with 7 degrees of freedom


Since t2.5% < tcalc < t0.5%:


• The average thickness of the new material is significantly different from that of the old material at the 5% level of significance.

This would be better if it were treated as a one-tailed test.

• HO: The new materials do not show less wear than the old.

• HA: The new materials show less wear than the old.

HO: µ ≤ 0 HA: µ > 0


tcalc = 2.626 with 7 degrees of freedom

Since t5% < tcalc < t1%:


• The average thickness with the new material is significantly greater than with the old material at the 5% level of significance, the new material shows significantly less wear.


Case Study 8 – Personal Computers

Ax = £45.60 sA = £5.33 nA = 40

Bx = £39.65 sB = £5.64 nB = 40

First check that the variances are approximately equal.

HO: ≤ 2Bσ

2Aσ

HA: > 2Bσ

2Aσ

Fcalc = 2

2

2A

2B

5.335.64

ss

=

= 1.12 with 39,39 degrees of freedom

F5% = 1.69

Since Fcalc < F5%:

• Reject HO.

• Variance of model B is not significantly greater than variance of model A.

• Testing the means:

HO: µA = µB (Two-tailed test) HA: µA ≠ µB

Pooled variance:


s2 = 2)n(n

39(5.64)39(5.33)BA

22

−++

= 78

1240.571107.95 + = 78

2348.52

s2 = 30.11

s = 5.49

Test statistic:

z =

21

21

n1

n1s

xx

+

−

=

401

4015.49

39.6545.60

+

−

= 0.0250.0255.49

5.95+×

= 0.2245.49

5.95×

= 4.84


Since zcalc > z0.5%:


• There is a significant difference in the average cost of maintenance and repair for the two models at the 1% level of significance.

It may also be useful to use a one-sided test to test whether the annual average cost for Model B is significantly lower than that for Model A.

HO: µA ≤ µB HA: µA > µB

zcalc= 4.85

Since zcalc > z1%:


• The annual average cost for Model B is significantly lower than the annual average cost for Model A at the 1% level of significance.


Case Study 9 – Smoking

From the data:

π = 0.75 p = 0.60 n = 64

HO: π ≥ 0.75 HA: π < 0.75

Test statistic:

z =

n)(1

pπ−ππ− =

640.75)0.75(1

0.750.60−− =

640.250.75

0.15×

−

= ⎟⎠⎞

⎜⎝⎛−

80.433

0.15 = 0.0541

0.15− = –2.77


• The proportion of employees found in the sample who are in favour of a no-smoking policy is significantly lower than the management’s initial hypothesis, at the 1% level of significance.


Case Study 10 – Tax Returns

From the data:

π = 0.20 p = 0.288 n = 250

HO: π ≤ 0.2 HA: π > 0.2

Test statistic:

z =

n)(1

pπ−ππ− =

2500.20)0.20(10.200.288−− =

2500.800.20

0.088×

⎟⎠⎞

⎜⎝⎛15.810.4

0.088 = 0.0250.088 = 3.52 =


• The proportion of errors observed in the firm’s sample is significantly higher than the reported national rate, at the 1% level of significance.


Case Study 11 – Telephone Services

From the data:

π = 0.936p = 0.872 n = 100

6 6

HO: π = 0.93HA: π ≠ 0.93

Test statistic:

n)(1

pπ−ππ−

1000.0640.936

0.064×

−

1000.936)0.936(1

0.9360.872−− = = z =

⎟⎠⎞

⎜⎝⎛−

100.2450.064 = =

0.0240.064− = –2.61

5


• The percentage of orders in the sample that were

t completed by the promised date is significantlydifferent from the percentage reported by Oftel athe 1% level of significance.


Case Study 12 – Sugar Bags

Formulate the hypotheses:

the process mean is not 1 kg.

Use a two-tailed test as we are testing for a differe

Since variance is known, we use the

calc

HO: µ = µ0: the process mean is still 1 kg. HA: µ ≠ µ0:

nce.

n ≥ 30, and thez-test:

nxσµ− z =

x = 1.03 1.00

σ = 0.01 σ = 0.1 n =100

µ =2

n = 100 =10

Test statistic:

z = n

xσ

µ−

= 101.0

00.103.1 −

= 01.003.0 = 3


At the 0.01 level of significance:

z = 2.58

calc > z0.5%:

• Reject H at the 1% level of significance.

• The process mean is no longer 1.0 kg.

0.5%

Since z

O


Case Study 13 – Maths Tests


HO: µ = µ0: this year's scores are typical. HA: µ ≠ µ0: this year's scores are not typical.

esting for a

Although n < 30, the z-test is used as the historical he scores follow a normal

The test is two-tailed, as we are tdifference.

data indicate that tdistribution.

nxσ

µ− z =

From the question:

µ = 75 σ = 6 x = 82 n =16 n = 16 = 4

Test statistic:

z = n

xσ

µ−

= 467582 −

= 5.1

7 = 4.67


1.64

•

• We conclude that this group of students is not typi

question:

ous years?’

This requires a one-tailed test.

• HO thi ar's rage score is equal to the average score of previous years.

• HA: µ µ0: this year's average score is greater than previous years.

From the previous calculation:

calc = 4.67

From tables:

z2.5% =

Since zcalc > z2.5%:

Reject HO at the 5% level of significance.

cal.

We can now ask the

‘Are this year's scores higher than the average of previ

: µ ≤ µ0: s ye aveor less than

>

z


At the 10% level of significance

ter vious years.

At the

z = 1.65

Since z > z :

•

er

z10% = 1.28

Since zcalc > z10%:


• The average score of this year's scores is greathan pre

1% level of significance

1%

calc 1%

Reject HO at the 1% level of significance.

• The average score of this year's scores is greatthan previous years.


Case Study 14 – Light Bulbs

As the question asks if the mean has changed, we use a two-tailed test.


is still 1500 hours

• HA ean lifetime has changed and is ng 50 hours.

n standard deviation and distribution are unknown, we

• HO: µ = µ0: the mean lifetime

: µ ≠ µ0: the mno lo er 1 0

Si ce the sample size < 30, and the population

use the t-test.

s = 90

x µ−t = with (n – 1) degrees of freedom

freedom = n – 1 = 9

with 9 d.f.:

ns

Degrees of

From the t tables,

t2.5% = 2.26

t0.05% = 3.25


nsx µ− t =

where:

µ = 1500 s = 90 x = 1410 n =10 n = 10 = 3.16

Test statistic:

tcalc = 16.3/90

15001410 −

= 5.28

.16

Since t0.5% < tcalc < t2.5%:


• There is evidence to suggest that the mean lifetime of bulbs has changed.

A one-tailed test would be required to test for a decrease.

90− = –3


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