Respuesta :
Answer:
There is enough evidence to claim that population mean 1 is less than population 2 at the 0.05 significance level.
Step-by-step explanation:
We are given the following in the question:
Group 1:
[tex]\mu_1 = 63.3\\\sigma_1 = 3.7\\n_1 = 11[/tex]
Group 2:
[tex]\mu_2 = 70.2\\\sigma_2 = 6.6\\n_2 = 13[/tex]
Alpha, α = 0.05
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu_1 \geq \mu_2\\H_A: \mu_1 < \mu_2[/tex]
Since, the population variances are equal and that the two populations are normally distributed, we use t-test(pooled test) for difference of two means.
Formula:
Pooled standard deviation
[tex]s_p = \sqrt{\displaystyle\frac{(n_1-1)\sigma_1^2 + (n_2-1)\sigma_2^2 }{n_1 + n_2 - 2}}[/tex]
[tex]t_{stat} = \displaystyle\frac{\mu_1-\mu_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}[/tex]
[tex]\text{Degree of freedom} = n_1 + n_2 - 2[/tex]
Putting all the values we get:
[tex]s_p = \sqrt{\displaystyle\frac{(11-1)(3.7)^2 + (13-1)(6.6)^2 }{11 + 13 - 2}} = \sqrt{29.9827} = 5.48[/tex]
[tex]t_{stat} = \displaystyle\frac{63.3-70.2}{5.48\sqrt{\frac{1}{11}+\frac{1}{13}}} = -3.073[/tex]
[tex]\text{Degree of freedom} = 11 + 13 - 2 = 22[/tex]
Now,
[tex]t_{critical} \text{ at 0.05 level of significance, 22 degree of freedom } = -1.717[/tex]
Since,
[tex]t_{stat} < t_{critical}[/tex]
We fail to accept the null hypothesis and reject the null hypothesis. We accept the alternate hypothesis.
We conclude that the mean score for Group 1 is significantly lower than the mean score for Group 2.
There is enough evidence to claim that population mean 1 is less than population 2 at the 0.05 significance level.
