Can someone help lol
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Answer:
|t| = 2.235 > 2.093
we rejected null hypothesis.
It is not coming from population
Step-by-step explanation:
<u>Step:-1</u>
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
<u>Null hypothesis</u> :- The difference between x⁻ and μ is not significant
<u>Alternative hypothesis</u>:- The difference between x⁻ and μ is significant
that is μ ≠ 35
<u>Level of significance </u>:-∝ =0.05
<u>The test statistic:- </u>
<u></u><u></u>
<u>substitute all values in above equation</u>
μ = 35 ,n =20
mean of the sample x⁻ = 30
standard deviation of (S) = 10
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)
<u>conclusion</u>:-
|t| = 2.235 > 2.093
we rejected null hypothesis.
It is not coming from population
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