Hypothesis Testing Scenarios

This assignment presents several hypothesis testing problems across different statistical concepts. Students are tasked with formulating hypotheses, calculating test statistics (z-values), determining p-values, and making conclusions based on varying significance levels (5% and 1%). Topics covered include testing proportions, comparing means between two samples, and analyzing data related to family bills and policy holders' proportions.

BUSN1009 – Quantitative Methods
Assessment 4: Major Assignment
Student id and name
[Pick the date]

Question 1
 For 95% confidence
Z value for 95% confidence = 1.96
Error E = 100
As per six-sigma estimate the standard deviation can be computed as shown below:
6 S= ( 2500−600 )
S=316.67
Now,
Sample ¿ ≥ ( ZS
E )2
n ≥ ( 1.96∗316.67
100 )
2
≥ 38.52
Therefore, it can be said that sample size 39 would be taken into account.
 For 90% confidence
Z value for 90% confidence = 1.645
Error E = 100
As per six-sigma estimate the standard deviation can be computed as shown below:
6 S= ( 2500−600 )
S=316.67
Now,
1

Sample ¿ ≥ ( ZS
E )
2
n ≥ ( 1.645∗316.67
100 )2
≥ 27.13
Therefore, it can be said that sample size app. 28 would be taken into account.
It can be seen that sample size has been reduced when the Chinese investors want to be 90%
confident in place of 95% confidence.
Question 2
 For 95% confidence
Z value for 95% confidence = 1.96
Error E = $2
Standard deviation S = $12.50
Now,
Sample ¿ ≥ ( ZS
E )2
n ≥ ( 1.96∗12.50
2 )
2
≥ 150.06
Therefore, it can be said that sample size 151 would be taken into account.
 For 90% confidence
2

Z value for 90% confidence = 1.645
Error E = $2
Standard deviation S = $12.50
Now,
Sample ¿ ≥ ( ZS
E )
2
n ≥ ( 1.645∗12.50
2 )
2
≥ 105.70
Therefore, it can be said that sample size 106 would be taken into account.
Question 3
 For 95% confidence
Z value for 90% confidence = 1.96
Error E = 5%
Standard deviation S = 1 (assuming)
Now,
Sample ¿ ≥ ( ZS
E )2
n ≥ ( 1.96∗1
0.05 )
2
≥ 1536.64
Therefore, it can be said that sample size 1537 would be taken into account.
3

 For 98% confidence
Z value for 98% confidence = 2.33
Error E = 3%
Standard deviation S = 1 (assuming)
Now,
Sample ¿ ≥ ( ZS
E )
2
n ≥ ( 2.33∗1
0.03 )2
≥ 6032.1
Therefore, it can be said that sample size 6033 would be taken into account.
Question 4
α = .05
Hypotheses
H0 : μ1=μ2
4

H1 : μ1> μ2
Assuming – population have equal variance and normal distribution
The p value for one tail comes out to be 0.0003.
It is apparent that p value is lower than significance level (0.0003<0.05) and therefore, null
hypothesis would be rejected and alternative hypothesis would be accepted. Hence, the
population 1 has a higher mean as compared with the mean of population 2.
Question 5
(a) Hypotheses
H0 : μ1=μ2
H1 : μ1 ≠ μ2
α = .01
Assuming – population have equal variance and normal distribution
5

The p value for two tailed comes out to be 0.0001.
It is apparent that p value is lower than significance level (0.0001 <0.01) and therefore, null
hypothesis would be rejected and alternative hypothesis would be accepted. Hence, the claim
that average values for the population 1 and population 2 are different is correct.
Question 6
Hypotheses
μ ≥ 80
μ<80
Sample size = 37
6

Population standard deviation = 9.90 micrograms
It is apparent that population standard deviation is known and sample size is higher than 30 and
thus, according to the central limit theorem population distribution can be assumed to be normal.
Hence, z statistics would be taken into account.
Mean of given sample = 75.75
z= 75.75−80
( 9.9
√37 )
z=−2.607
The corresponding p value comes out to be 0.00456 with the help of excel function
NORMSDIST ( −2.607).
α = .05
It can be seen that p value is lower than level of significance and therefore, null hypothesis
would be rejected and alternative hypothesis would be accepted. Hence, the claim that “the city’s
average number of suspended particles is significantly lower than it was when the initial
measurements were taken” is correct.
Question 7
Hypotheses
μ ≥17.9
μ<17.9
Sample size = 32 weeks
Population standard deviation =2.25 orders per week
7

It is apparent that population standard deviation is known and sample size is higher than 30 and
thus, according to the central limit theorem population distribution can be assumed to be normal.
Hence, z statistics would be taken into account.
Sample mean = 15.5 orders per week
z= 15.5−17.9
( 2.25
√32 )
z=−6.0339
The corresponding p value comes out to be 0.00 with the help of excel function NORMSDIST¿).
α = .05
It can be seen that p value is lower than level of significance and therefore, null hypothesis
would be rejected and alternative hypothesis would be accepted. Hence, it can be concluded that
the average number of weekly orders decreased in the recession period. Further, when
significance level is taken as 0.10 then also, null hypothesis would be rejected because p value is
lower than level of significance.
Question 8
Hypotheses
H0:   1158
Ha:  > 1158
Level of significance = 5%
Sample size = 11
Sample mean = 1236.36
8

Standard deviation = 103.81
In this case, sample size is lower than 30 and also, the population standard deviation is unknown
and hence, t value would be taken into account in place of z value.
t= 1236.36−1158
( 103.81
√ 11 )
t=2.5036
Degree of freedom = n-1 = 11-1 =10
The corresponding p value for the inputs t = 2.5036 and dof =10 comes out to be 0.015627.
It can be seen that p value is lower than level of significance and therefore, null hypothesis
would be rejected and alternative hypothesis would be accepted.
Question 9
Hypotheses
μ ≤253
μ>253
Sample size = 21 families
Standard deviation =39.337
Sample mean = 280.476
It is apparent that population standard deviation is known and sample size is higher than 30 and
thus, according to the central limit theorem population distribution can be assumed to be normal.
Hence, z statistics would be taken into account.
9

z= 280.476−253
( 39.337
√ 21 )
z=3.2008
The corresponding p value comes out to be 0.999 with the help of excel function NORMSDIST¿
).
Level of significance = 5%
It can be seen that p value is higher than level of significance and therefore, null hypothesis
would not be rejected. Therefore, Sebastian believes the weekly bill for this type of family in
Glenn Osmond is less than or equal to the national average figure.
Level of significance = 10%
It can be seen that p value is higher than level of significance and therefore, null hypothesis
would not be rejected. Therefore, Sebastian believes the weekly bill for this type of family in
Glenn Osmond is less than or equal to the national average figure.
Question 10
Hypotheses
p ≥0.78
p<0.78
p0= 110
150 =0.733
The z value would be computed as shown below:
z= p− p0
√ p0 ( 1− p0 )
n
10

¿ 0.78−0.733
√ 0.733 ( 1−0.733 )
150
¿ 1.301
The p value comes out to be 0.9034.
Level of significance = 5%
It is apparent that p value is higher than significance level (0.9034 <0.05) and therefore, null
hypothesis would not be rejected. Hence, “there sufficient evidence to conclude that the
proportion is significantly less than the 78% claimed by the banking industry”
Question 11
Hypotheses
po=0.478
po ≠ 0.478
p= 163
378 =0.431
Sample size = 378
The z value would be computed as shown below:
z= p− p0
√ p0 ( 1− p0 )
n
¿ 0.431−0.478
√ 0.478 ( 1−0.478 )
378
¿−1.845
11

The p value comes out to be 0.0325.
Level of significance = 5%
It is apparent that p value is lower than significance level (0.0325 <0.05) and therefore, null
hypothesis would be rejected and alternative hypothesis would be accepted. Hence, it can be
concluded that the policy holder’s proportion is not equal to 47.8%.
Level of significance = 1%
It is apparent that p value is higher than significance level (0.0325 <0.01) and therefore, null
hypothesis would not be rejected. Hence, it can be concluded that the policy holder’s proportion
is equal to 47.8%. However, when the level of significance is 5%, then the null hypothesis
would not be rejected.
Question 12
Hypotheses
μ2008=μ2014
μ2008< μ2014
Two – sample z test would be used for comparing two means
Here, z= x1−x2
√ σ 1
2
n1
+ σ2
2
n2
Sample 1: 2008
n1=51trips
σ 1=$ 18.50
x1=$ 190 per day
12

Sample 2: 2014
n2 =47 trips
σ 2=$ 14.90
x2=$ 197 per day
Now,
z= 190−197
√ 18.502
51 + 14.902
47
z=−2.0701
The p value comes out to be -2.0701.
Level of significance = 0.01
It is apparent that p value is lower than significance level (-2.0701 <0.01) and therefore, null
hypothesis would be rejected and alternative hypothesis would be accepted. Hence, it can be
concluded per diem cost in 2008 is lower as compared with 2014.
13