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

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

`(21.9-19.8) -1.966 \sqrt{(3.4^2)/(200) +(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`

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{(s^2_1)/(n_1) +(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{(3.4^2)/(200) +(3.5^2)/(200)}= 1.422`

`(21.9-19.8) -1.966 \sqrt{(3.4^2)/(200) +(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`