We are asked to determine the test statistic for two populations. To do that we will use the following formula:
[tex]z=\frac{\bar{x_2}-\bar{x_1}}{\sqrt[]{\frac{SD^2_2}{n_2}+\frac{SD^2_1}{n_1}}}[/tex]
Where:
[tex]\begin{gathered} \bar{x_1},\bar{x_2}=\text{ population means} \\ SD_1,SD_2=\text{ standard deviations} \\ n_1,n_2=\text{ population sizes} \end{gathered}[/tex]
Substituting the values we get:
[tex]z=\frac{83.3_{}-75.4}{\sqrt[]{\frac{(17.8)^2_{}}{19}+\frac{(9.7)^2_{}}{12}}}[/tex]
Solving the operations we get:
[tex]z=1.596[/tex]
Therefore, the test statistic is 1.596.
To determine the P-value we will determine the probability that the test statistic is less than the value we determined. This is:
[tex]p-\text{value}=P(z<1.596)[/tex]
The value of the probability we find it in the z-table using the value z = 1.596, we get:
[tex]p-\text{value}=0.9441[/tex]
Therefore, the p-value is 0.9441.
