Question

Calculate the test statistic and p-value.

Short answer

The value of test statistics is $5.421$and the p-value is $0$

Step-by-step solution

Given information

Given following data

Simplify 

Finding standard error

$SE=\sqrt{\frac{{s_1}^2}{n1}}+\frac{{s_2}^2}{n_2}=\sqrt{4.26853}=2.06604$

Test statistics is $t=\frac{\mu \omega -\mu \eta \omega }{SE}=\frac{45.3-34.1}{2.06604}=5.421$

let us determine the number of degrees of freedom using the formula from 10.1:

$df=\frac{S{E}^4}{{\displaystyle \frac{{S_1}^4+{s_2}^4}{{n_1}^2(n_1-1)+n{\displaystyle \frac{2}{2}}(n2-1)}}}=\frac{2.{06604}^4}{0.12249}=148.749$

using the above formula we get p-value is$0$

Graphical representation 

Student's t distribution

$X~t_{\left(df\right)}$

$\mu =E\left(X\right)=0$$\alpha =SD\left(X\right)=1.007$${\alpha }^2=Var\left(X\right)=1.014$