Question

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. We wish to determine whether the scores of a random sample of 20 new freshmen, with a mean of 30 and a standard deviation of 10, can be viewed as coming from this population. Test at the .05 level of significance.

Answer 1

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:-

$t = \frac{x^{-} - population mean}{\frac{S}{\sqrt{n} } }$

substitute all values in above equation

μ = 35 ,n =20

mean of the sample   x⁻ = 30

standard deviation of  (S) = 10

$t = \frac{30 - 35}{\frac{10}{\sqrt{20} } }$

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