Answer:
Step-by-step explanation:
Given that :
population Mean = 15000
standard deviation= 1200
sample size n = 100
sample mean = 16000
The null and the alternative hypothesis can be computed as follows:
[tex]\mathtt{H_o : \mu = 15000 }\\ \\ \mathtt{H_1 : \mu > 15000}[/tex]
Using the standard normal z statistics
[tex]z = \dfrac{\overline X - \mu}{\dfrac{\sigma }{\sqrt{n}}}[/tex]
[tex]z = \dfrac{16000 -15000}{\dfrac{1200 }{\sqrt{100}}}[/tex]
[tex]z = \dfrac{1000}{\dfrac{1200 }{10}}[/tex]
[tex]z = \dfrac{1000\times 10}{1200}[/tex]
z = 8.333
degree of freedom = n - 1 = 100 - 1 = 99
level of significance ∝ = 0.05
P - value from the z score = 0.00003
Decision Rule: since the p value is lesser than the level of significance, we reject the null hypothesis
Conclusion: There is sufficient evidence that the Dean claim for his graduate students earn more than average salary of $15,000
Dean's Claim of Average Salary = 16000, ie greater than 15000 : is correct
Null Hypothesis [ H0 ] : Average Salary = 15000
Alternate Hypothesis [ H1 ] : Average Salary > 15000
Hypothesis is tested using t statistic.
t = ( x - u ) / ( s / √ n ) ; where -
x = sample mean , u = population mean , s = standard deviation, n = sample size
t = ( 16000 - 15000 ) / ( 1200 / √100 )
= 1000 / 120
t {Calculated} = 8.33,
Degrees of Freedom = n - 1 = 100 = 1 = 99
Tabulated t 0.05 (one tail) , at degrees of freedom 99 = 1.664
As Calculated t value 8.33 > Tabulated t value 1.664 , So we reject the Null Hypothesis in favour of Alternate Hypothesis.
So, conclusion : Average Salary > 15000
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