Given data:
Normally distributed and unknown population SD, variances are equal (assumed)
First Population
Sample Size = 12
Sample Mean = 75.4
Sample SD = 9.7
Second Population
Sample Size = 19
Sample Mean = 83.3
Sample SD = 17.8
Find: test statistic and p-value
Solution:
Since the sample size is small for both population and its standard deviation is unknown, we can use t-test. The formula for this is:
[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})}};withdf=n_1+n_2-2[/tex]
Substitute the given data that we have above to the formula.
[tex]\begin{gathered} t=\frac{75.4-83.3}{\sqrt[]{\frac{(12-1)(9.7)^2+(19-1)(17.8)^2}{12+19-2}(\frac{1}{12}+\frac{1}{19})}} \\ t=\frac{-7.9}{\sqrt[]{\frac{(11)(94.09)+(18)(316.84)}{29}(\frac{31}{228})}} \\ t=\frac{-7.9}{\sqrt[]{\frac{1034.99+5703.12}{29}(\frac{31}{228})}} \\ t=\frac{-7.9}{\sqrt[]{\frac{6738.11}{29}(\frac{31}{228})}} \\ t=\frac{-7.9}{5.6206} \\ t\approx-1.406 \end{gathered}[/tex]
Hence, the test statistic is approximately -1.406.
The p-value for this test statistic is equal to 0.0852.
