Answer:
t = 3.016
P-value = 0.0061
Step-by-step explanation:
Let us assume that the population variance is unequal
Null hypothesis :
[tex]H_0 :[/tex] μ[tex]_1[/tex] = μ[tex]_2[/tex]
[tex]H_a :[/tex] μ[tex]_1[/tex] ≠ μ[tex]_2[/tex]
The alternative hypothesis :
Here we have
[tex]n_1=14 \\n_2 = 12\\x_1 = 23.1\\s^2_1 = 5.3 \\x_2 = 20.6\\s^2_2 = 3.7[/tex]
Since it is not given that variances are equal, degrees of freedom of the test is given as
[tex]df=\frac{(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2} )^2}{\frac{(\frac{s^2_1}{n_1})^2}{n_1-1}+ \frac{(\frac{s^2_2}{n_2})^2}{n_2-1}} =23[/tex]
[tex]t=\frac{x_1-x_2}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}} = 3.016[/tex]
The P-value of the test is = 0.0061
Since the p-value is less than 0.05, we will reject the null hypothesis