Answer:
Step-by-step explanation:
Given that:
The sample mean [tex]\overline x = 23[/tex]
The standard deviation [tex]\sigma[/tex] = 9
Population mean = 20
Null hypothesis:
[tex]H_o: \mu = 20[/tex]
Alternative hypothesis:
[tex]H_1 : \mu> 30[/tex]
(a)
When Sample size = 10
[tex]Test \ statistics=\dfrac{\overline x - \mu }{\dfrac{\sigma }{\sqrt{n}} }[/tex]
[tex]=\dfrac{23 -20}{\dfrac{9}{\sqrt{10}} }[/tex]
[tex]=\dfrac{3 \times \sqrt{10}}{9 }[/tex]
t = 1.0541
Degree of freedom df:
df = n -1
df = 10 - 1
df = 9
P(value) for t = 1.0541 at df = 9:
P(value) = P(Z > 1.0541)
P(value) = 1 - P(< 1.0541)
P(value) = 1 - 0.8403
P(value) = 0.1597
There is no enough evidence to infer at the 5% significance since p-value is greater than the level of significance.
(b) When sample size = 30
[tex]Test \ statistics=\dfrac{\overline x - \mu }{\dfrac{\sigma }{\sqrt{n}} }[/tex]
[tex]=\dfrac{23 -20}{\dfrac{9}{\sqrt{30}} }[/tex]
[tex]=\dfrac{3 \times \sqrt{30}}{9 }[/tex]
t = 1.8257
Degree of freedom df:
df = n -1
df = 30 - 1
df = 29
P(value) for t = 1.8257 at df = 29:
P(value) = P(Z > 0.9609)
P(value) = 1 - P(< 0.9609)
P(value) = 1 - 0.9609
P(value) = 0.0391
There is enough evidence to infer that the mean is greater than 20 at the 5% significance level as the p-value is less than the significance level.
(c) When sample size = 50
[tex]Test \ statistics=\dfrac{\overline x - \mu }{\dfrac{\sigma }{\sqrt{n}} }[/tex]
[tex]=\dfrac{23 -20}{\dfrac{9}{\sqrt{50}} }[/tex]
[tex]=\dfrac{3 \times \sqrt{50}}{9 }[/tex]
t = 2.3570
Degree of freedom df:
df = n -1
df = 50 - 1
df = 49
P(value) for t = 2.3570 at df = 49:
P(value) = P(Z > 0.9888)
P(value) = 1 - P(< 0.9888)
P(value) = 1 - 0.9888
P(value) = 0.0112
There is enough evidence to infer that the mean is greater than 20 at the 5% significance level as the p-value is less than the significance level.
