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 24.7 with a standard deviation of 4, while the 200 students in group 2 had a mean score of 16.1 with a standard deviation of 3. complete parts (a) and (b) below
(a) determine the 90% confidence interval for the difference in scores, \\( mu _ { 1 } - mu _ { 2 } \\). interpret the interval
the lower bound is
the upper bound is
(round to three decimal places as needed.)

Explanation:

Step1: Calculate the difference in means

The difference in means $\bar{x}_1-\bar{x}_2 = 24.7 - 16.1=8.6$

Step2: Calculate the standard error

The formula for the standard error (SE) of the difference between two means for independent samples is $SE=\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}$.
Here, $s_1 = 4$, $n_1=200$, $s_2 = 3$, $n_2 = 200$.
$SE=\sqrt{\frac{4^{2}}{200}+\frac{3^{2}}{200}}=\sqrt{\frac{16 + 9}{200}}=\sqrt{\frac{25}{200}}=\sqrt{0.125}\approx0.354$

Step3: Find the critical value

For a 90% confidence interval, the significance level $\alpha=1 - 0.90=0.10$, and $\alpha/2=0.05$.
Using the standard normal distribution (since $n_1\geq30$ and $n_2\geq30$), the critical value $z_{\alpha/2}=z_{0.05}\approx1.645$

Step4: Calculate the margin of error

The margin of error (ME) is $ME = z_{\alpha/2}\times SE$
$ME=1.645\times0.354\approx0.582$

Step5: Calculate the lower and upper bounds

The lower bound is $(\bar{x}_1-\bar{x}_2)-ME$
$8.6- 0.582=8.018$
The upper bound is $(\bar{x}_1-\bar{x}_2)+ME$
$8.6 + 0.582=9.182$

Answer:

The lower bound is $8.018$
The upper bound is $9.182$