Section 8 Hypothesis Testing Solutions Teaching Team Frances A. Pfab
Section 8: Solutions Page 1
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.
Section 8: Solutions Page 2
• 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.
Section 8: Solutions Page 3
3. (a)(2) n = 250 s = 0.7 z = 2.71
• Reject HO at 1% level of significance.
• There is a significant difference at the 1% level of significance.
3. (b)(1) n = 150 s = 0.2 z = 7.35
• Reject HO at 1% level of significance.
• There is a significant difference at the 1% level of significance.
3. (b)(2) n = 150 s = 0.7 z = 2.10
• Reject HO at 5% level of significance.
• There is a significant difference at the 5% level of significance
Section 8: Solutions Page 4
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.
Section 8: Solutions Page 5
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
Section 8: Solutions Page 6
• Reject HO at the 1% level of significance.
• 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
Section 8: Solutions Page 7
• Reject HO at the 1% level of significance.
• 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.
Section 8: Solutions Page 8
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
Section 8: Solutions Page 9
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
• Reject HO at 5% 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.
Section 8: Solutions Page 10
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
Section 8: Solutions Page 11
• 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.
Section 8: Solutions Page 12
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.
Section 8: Solutions Page 13
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.
• Reject HO 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.
Section 8: Solutions Page 14
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
Section 8: Solutions Page 15
Since t2.5% < tcalc < t0.5%:
• Reject HO at the 5% level of significance.
• 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
Section 8: Solutions Page 16
tcalc = 2.626 with 7 degrees of freedom
Since t5% < tcalc < t1%:
• Reject HO at the 5% level of significance.
• 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.
Section 8: Solutions Page 17
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:
Section 8: Solutions Page 18
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
Section 8: Solutions Page 19
Since zcalc > z0.5%:
• Reject HO at the 1% level of significance.
• 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%:
• Reject HO at the 1% level of significance.
• The annual average cost for Model B is significantly lower than the annual average cost for Model A at the 1% level of significance.
Section 8: Solutions Page 20
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
• Reject HO at the 1% level of significance.
• 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.
Section 8: Solutions Page 21
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 =
• Reject HO at the 1% level of significance.
• The proportion of errors observed in the firm’s sample is significantly higher than the reported national rate, at the 1% level of significance.
Section 8: Solutions Page 22
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
• Reject HO at the 1% level of significance.
• 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.
Section 8: Solutions Page 23
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
Section 8: Solutions Page 24
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
Section 8: Solutions Page 25
Case Study 13 – Maths Tests
Formulate the hypotheses:
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
Section 8: Solutions Page 26
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
Section 8: Solutions Page 27
At the 10% level of significance
ter vious years.
At the
z = 1.65
Since z > z :
•
er
z10% = 1.28
Since zcalc > z10%:
• Reject HO at the 10% level of significance.
• 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.
Section 8: Solutions Page 28
Case Study 14 – Light Bulbs
As the question asks if the mean has changed, we use a two-tailed test.
Formulate the hypotheses:
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
Section 8: Solutions Page 29
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%:
• Reject HO at the 5% level of significance.
• 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
Section 8: Solutions Page 30