Answer:
|t| = 2.235 > 2.093
we rejected null hypothesis.
It is not coming from population
Step-by-step explanation:
Step:-1
Let’s assume that, over the years, a paper and pencil test of anxiety yields a mean score of 35 for all incoming college freshmen.
μ = 35
small sample size n =20
mean of the sample x⁻ = 30
standard deviation of (S) = 10
Null hypothesis :- The difference between x⁻ and μ is not significant
Alternative hypothesis:- The difference between x⁻ and μ is significant
that is μ ≠ 35
Level of significance :-∝ =0.05
The test statistic:-
[tex]t = \frac{x^{-} - population mean}{\frac{S}{\sqrt{n} } }[/tex]
substitute all values in above equation
μ = 35 ,n =20
mean of the sample x⁻ = 30
standard deviation of (S) = 10
[tex]t = \frac{30 - 35}{\frac{10}{\sqrt{20} } }[/tex]
t= -2.235
|t| = 2.235
The degrees of freedom γ =n-1 =20-1 =19
By tabulated value of 't' for 19 degrees of freedom at 5% level of significance.= 2.093 for two tailed test.(see attached diagram below)
conclusion:-
|t| = 2.235 > 2.093
we rejected null hypothesis.
It is not coming from population
