Question

Two researchers conducted a study in which two groups of students were asked to answer 42 trivia questions from a board game. The students in group 1 were asked to spend 5 minutes thinking about what it would mean to be a​ professor, while the students in group 2 were asked to think about soccer hooligans. These pretest thoughts are a form of priming. The 200 students in group 1 had a mean score of 21.9 with a standard deviation of 3.4​, while the 200 students in group 2 had a mean score of 19.8 with a standard deviation of 3.5.
(a) Determine the 95% confidence interval for the difference in scores μ1 - μ2.

Answer 1

Answer:

$(21.9-19.8) -1.966\sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 1.422$

$(21.9-19.8) -1.966 \sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 2.778$

And we are 95% confident that the true difference means are between $1.422 \leq \mu_1 -\mu_2 \leq 2.778$

Step-by-step explanation:

We know the following info:

$\bar X_1 = 21.9$ sample mean for group 1

$\bar X_2 = 19.8$ sample mean for group 2

$s_1 = 3.4$ sample standard deviation for group 1

$s_2 = 3.5$ sample standard deviation for group 2

$n_1 = 200$ sample size group 1

$n_2 = 200$ sample size group 2

We want to find a confidence interval for the difference of means and the correct formula to do this is:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

Now we just need to find the critical value. The confidence level is 0.95 then the significance is $1-0.95 =0.05$ and $\alpha/2 =0.025$. The degrees of freedom are given by:

$df= n_1 +n_2 -2= 200+200-2= 398$

The critical value for this case would be :$t_{\alpha/2}=1.966$  

And replacing into the confidence interval formula we got:

$(21.9-19.8) -1.966\sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 1.422$

$(21.9-19.8) -1.966 \sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 2.778$

And we are 95% confident that the true difference means are between $1.422 \leq \mu_1 -\mu_2 \leq 2.778$