Answer:
Teachers in his school district have less than 5 yearſ of experience on average is true.
Step-by-step explanation:
Sample mean = [tex]\bar{x}=4[/tex]
Sample standard deviation = s = 2
We are supposed to to test Ha :u= 5 versus H : u < 5 using a sample of 25 teachers
[tex]H_a:\mu=5\\H_0:\mu < 5[/tex]
n = 25
Since n<30 and population standard deviation is unknown
So, we will use t test
Formula : [tex]t=\frac{x-\mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t=\frac{4-5}{\frac{2}{\sqrt{25}}}[/tex]
t=-2.5
[tex]\alpha = 0.05[/tex]
Degree of freedom = n-1 = 25-1 = 24
Refer the t table
[tex]t_{df,\frac{\alpha}{2}}=t_{24,\frac{0.05}{2}}=2.064[/tex]
t critical > t calculated
So, We are failed to reject null hypothesis
So, teachers in his school district have less than 5 yearſ of experience on average is true.
The statement is true that the teacher in Rory school have less than five years of experience on average.
According to the question,
Sample mean, [tex]\bar x[/tex] = 4
Standard deviation, s = 2
Sample number of teachers, n = 25
Test:
[tex]H_a[/tex] : μ = 5
[tex]H_0[/tex] : μ < 5
We know the formula,
→ t = [tex]\frac{x- \mu}{\frac{s}{\sqrt{n} } }[/tex]
By substituting the values,
= [tex]\frac{4- 5}{\frac{2}{\sqrt{25} } }[/tex]
= -2.5
Now, the degree of freedom be:
= n - 1
= 25 - 1
= 24
By using t-table,
[tex]t_{df,\frac{\alpha}{2} }[/tex] = [tex]t_{24,\frac{0.05}{2} }[/tex]
= 2.064
We can say that [tex]t_{critical}[/tex] > [tex]t_{calculated}[/tex]
Thus the response above is correct.
Find out more information about standard deviation here:
https://brainly.com/question/19243813
