Respuesta :
Answer:
|t| =| - 3.0303| = 3.03
Hence we rejected null hypothesis.
The claim that the population of student course evaluations has a mean is not equal to 3.75
Step-by-step explanation:
Step(i)
A data set includes data from student evaluations of courses.
Given n= 89
sample mean x⁻ = 3.58
Sample Standard deviation (S) = 0.53
Given data population of student course evaluations has a mean equal to 3.75.
μ = 3.75
Step(ii)
Null hypothesis:-H₀: the population of student course evaluations has a mean equal to 3.75
μ = 3.75
Alternate hypothesis :- H₁: μ ≠ 3.75
Level of significance :-∝=0.01
The test statistic
[tex]t = \frac{x^{-} -mean}{\frac{S}{\sqrt{n} } }[/tex]
Step(iii):
The degrees of freedom γ=n-1 = 89-1=88
tabulated value t = 2.3733
Calculated value
[tex]t = \frac{3.58-3.75}{\frac{0.53}{\sqrt{89} } }[/tex]
t = - 3.0303
|t| =| - 3.0303| = 3.03
The tabulated value of 't' for 88 degree of freedom at 1% of level of significance is 2.3733
since calculated value t=3.03 is greater than tabulated value t = 2.3733 of t
Hence we rejected null hypothesis.
the claim that the population of student course evaluations has a mean is not equal to 3.75
