P value in z statistic

2. Test Concerning Single Means
P- VALUE:
P- value is used in hypothesis testing to help you Accept or reject the
null hypothesis. The p value is the evidence against a null hypothesis.
The smaller the p-value, the stronger the evidence that you should reject
the null hypothesis.
Or
Supposing null hypothesis were true, What is the probability of getting a
value of 𝑥̅ this far from is called a P-Value.
MAKE DECISION BY USING P-VALUE IN Z - STATISTIC
IF P- Value ≤ α; Reject H0
IF P- Value > α; Fail to Reject H0 (Accept H0)
NOTE:
I would try to solve following examples by
1- Using classical approach
2- Using P-value approach


3. Example-1:
The yield of corn in kg/hectare is normally distributed with variance
. For a random sample of six plots the yields are 1430, 1190,
1280, 1270, 1310 and 1380 kg. Test the hypothesis that mean yield () is
greater than equal to 1350 kg at 5% level of significance.
Solution:
n=6, =120, =1310, =0.05, =1350
1 .Hypothesis
H0: ≥ 1350
H1: < 1350
2. Level of significance =0.05
3. Test of statistic
4. Crtical Region 𝑧α = 𝑧0.05 = -1.65
5. Computation 𝑍 =
1310−1350
120
√6
= - 0.82
6. Conclusion: Z = - 0.82
We should accept H0. The average life is significantly
greater than and equal to 1350 the hypothesized value.
144002
=
 X  



X
z
n


−
=
−1.65 –0– 

4. MAKE DECISION BY USING P-VALUE IN Z - STATISTIC
Example-1:
The yield of corn in kg/hectare is normally distributed with variance
. For a random sample of six plots the yields are 1430, 1190,
1280, 1270, 1310 and 1380 kg. Test the hypothesis that mean yield () is
greater than equal to 1350 kg at 5% level of significance.
Solution:
n=6, =120, =1310, =0.05, =1350
1 .Hypothesis
H0: ≥ 1350
H1: < 1350
2. Level of significance =0.05
3. Test of statistic
4. Computation P = P(𝑥̅ ˂ 1310)
P = 𝑃(𝑍˂
1310−1350
120
√6
)
P = P(Z˂ - 0.82)
P = 0.2061
5. P-Value P-Value = 0.2061
6. Conclusion: P- Value > α; Fail to Reject H0 (Accept H0)
We should accept H0. The average life is
significantly greater than and equal to 1350 the
hypothesized value.
144002
=
 X  



X
z
n


−
=

5. Example-2:
The average commission charged by full-service brokerage firms on a
sale of common stock is $144, and the standard deviation is $52.Joel
Freelander has taken a random sample of 121 trades by his clients and
determined that they paid an average commission of $151. At a 0.10
significance level,can Joel conclude that his clients’ commissions are
higher than the industry average?
Solution:
n=121, =52, =151, =0.10, =144
1.Hypothesis
H0: = 144
H1: > 144
2. Level of significance =0.10
3. Test of statistic
4. Crtical Region
5. Computation
Z = 1.48
6. Conclusion:
Reject H0.Their commission are significantly higher.
 X  



X
z
n


−
=
0.10 1.28Z Z
= =
151 144
52
121
Z =
−
− –0– 1.28

6. MAKE DECISION BY USING P-VALUE IN Z - STATISTIC
Example-2:
The average commission charged by full-service brokerage firms on a
sale of common stock is $144, and the standard deviation is $52.Joel
Freelander has taken a random sample of 121 trades by his clients and
determined that they paid an average commission of $151. At a 0.10
significance level,can Joel conclude that his clients’ commissions are
higher than the industry average?
Solution:
n=121, =52, =151, =0.10, =144
1.Hypothesis
H0: = 144
H1: > 144
2. Level of significance =0.10
3. Test of statistic
4.Computation P = P(𝑥̅ ˃ 151)
P = 𝑃(𝑍 ˃
151−144
52
√121
)
P = P(Z ˃ 1.48)
P = 1- P(Z ˂ 1.48)
P = 1- 0.9306
P = 0.0694
5. P-Value P-Value is 0.0694
6. Conclusion: P- Value ≤ α; Reject H0
Reject H0.Their commission are significantly
higher.
 X  



X
z
n


−
=

7. Example-3:
American Theaters knows that a certain hit movie ran an average of 84
days in each city, and the corresponding standard deviation was 10 days.
The manager of the southeastern district was interested in compairing
the movie’s popularity in his region with that in all of American’s other
theaters. He randomly chose 75 theaters in his region and found that they
ran the movie an average 81.5days. State appropriate hypotheses for
testing whether there was a significant difference in the length of the
picture’s run between theaters in the southeastern district and all
Americans’s other theaters at a 1 percent significance level.
Solution:
n=75, =10, =81.5, =0.01, =84
1. Hypothesis
H0: = 84
H1: ≠ 84
2. Level of significance =0.01
3. Test of statistic
4. Crtical Region
5. Computation = -2.17
6. Conclusion: The value of Zcal lies in the area of acceptance therefore
We accept H0.The length of run in the southeastern is
not significantly different from the length of run in
other regions.
 X  



Z=
X
n


−
0.01 0.005
2 2
2.58Z Z Z
= = = 
81.5 84
10
75
Z =
−
−2.58 –0– 2.58

8. MAKE DECISION BY USING P-VALUE IN Z - STATISTIC
Example-3:
American Theaters knows that a certain hit movie ran an average of 84 days in
each city, and the corresponding standard deviation was 10 days. The manager of
the southeastern district was interested in compairing the movie’s popularity in his
region with that in all of American’s other theaters. He randomly chose 75 theaters
in his region and found that they ran the movie an average 81.5days. State
appropriate hypotheses for testing whether there was a significant difference in the
length of the picture’s run between theaters in the southeastern district and all
Americans’s other theaters at a 1 percent significance level.
Solution:
n=75, =10, =81.5, =0.01, =84
1. Hypothesis
H0: = 84
H1: ≠ 84
2. Level of significance =0.01
3. Test of statistic
4. Computation P = 2P(𝑥̅ ˂ 81.5)
P = 2𝑃(𝑍 ˂
81.5−84
10
√75
)
P = 2P(Z ˂ - 2.17)
P = 2×(0.0150)
P = 0.03
5. P-Value P-Value is 0.03
6. Conclusion: P- Value > α; Fail to Reject H0 (Accept H0)
We accept H0.The length of run in the
southeastern is not significantly different from the
length of run in other regions.
 X  



X
z
n


−
=

9. Example-4:
In a Census the average wage of the factory workers in an industrial area
is found to be Rs.1240. If a random sample of 900 workers yields a
mean wage of Rs.1250 with a standard deviation of Rs.300, test the
significance of sample result. Use .
Solution:
n = 900, s =300, = 1250, =0.05, = 1240
1. Hypothesis
H0: = 1240
H1: ≠ 1240
2. Level of significance =0.05
3. Test of statistic 𝑍 =
𝑥̅− 𝜇
𝑠
√ 𝑛
4. Crtical Region 𝑍∝
2
=
𝑍0.05
2
=
𝑍0.025= ±1.96
5. Computation 𝑍 =
1250−1240
300
√900
= 1.00
6. Conclusion: The value of Zcal lies in the area of acceptance
therefore We accept H0.
05.0=
X  



−1.96 –0– 1.96

10. MAKE DECISION BY USING P-VALUE IN Z - STATISTIC
Example-4:
In a Census the average wage of the factory workers in an industrial area
is found to be Rs.1240. If a random sample of 900 workers yields a
mean wage of Rs.1250 with a standard deviation of Rs.300, test the
significance of sample result. Use .
Solution:
n = 900, s =300, = 1250, =0.05, = 1240
1. Hypothesis
H0: = 1240
H1: ≠ 1240
2. Level of significance =0.05
3. Test of statistic 𝑍 =
𝑥̅− 𝜇
𝑠
√ 𝑛
4. Computation P = 2P(𝑥̅ ˃ 1250)
P = 2𝑃(𝑍 >
1250−1240
300
√900
)
P = 2P(Z > 1.00)
P = 2{1- P(Z< 1.00)}
P = 2{1- 0.8413}
P = 2(0.1587)
P = 0.3174
5. P-Value P-Value is 0.3174
6. Conclusion: P- Value > α; Fail to Reject H0 (Accept H0)
05.0=
X  


