Given data:
Data are normally distibuted. Unknown standard deviation but variances are assumed to be equal.
FIrst Population Data
Sample size (n₁) = 12
Sample mean (x₁) = 75.4
Sample SD (s₁) = 9.7
Second Population Data
Sample size (n₂) = 19
Sample Mean (x₂) = 83.3
Sample SD (s₂) = 17.8
Find: test statistic value and p-value
Solution:
Since both population have small sample size (less than 30) and the population standard deviation is unknown as well, the appropriate test statistic that must be use is the t-test. Here is the formula:
[tex]t=\frac{\bar{x}_1-\bar{x}_2}{\sqrt[]{\frac{(n_1-1)(s^2_1)+(n_2-1)(s^2_2)}{n_1+n_2-2}(\frac{1}{n_1}+\frac{1}{n_2})}};df=n_1+n_2-2[/tex]
Let's plug those given data that we have above to the formula.
[tex]t=\frac{75.4-83.3}{\sqrt[]{\frac{(12-1)(9.7^2)+(19-1)(17.8)^2}{12+19-1}(\frac{1}{12}+\frac{1}{19})}}[/tex][tex]undefined[/tex]
