Question

Two researchers conducted a Muty 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 atents in group 2 were asked to think about soccer hooligans. These protest thoughts are a form of priming. The 200 students in group 1 had a mean score of 232 with a standard deviation of 3.6,while the 200 students in group 2 had a mean score of 15.4 with a standard deviation of 35. Complete parts (a) and (b) below (*) Determine the confidence interval for the difference in scores, 19. Interpret the interval The lower bound is The upper bound is (Round to three decimal places as needed.) Interpret the interval Choose the correct answer below OA. There is a 95% probability that the difference of the means is in the interval O B. The researchers are 95% confident that the difference of the means is in the Worval O C. The researchers are 95% confident that the difference between randomly selected individuals will be in the interval OD. There is a 95% probability that the difference between randomly selected individuals will be in the interval (b) What does this say about priming? OA Since the 95% confidence interval does not contain zero, the results suggest that priming does have an effect on scores O B. Since the 95% confidence interval contains zero, the results suggest that priming does have an effect on scores OC. Since the 95% confidence interval does not contain zero, the results suggest that priming does not have an effect on scores OD. Since the 9% confidence interval contains zero, the results suggest that priming does not have an effect on scores.

Answer 1

Given
X̅1= 23.2           ....... Sample 1 Mean
n1 = 200           ....... Sample 1 Size
σ1 = 3.6             ....... Population/sample 1 Standard Deviation
X̅2 = 15.4           ....... Sample 2 Mean
n2 = 200           ....... Sample 2 Size
σ2 = 3.5             ....... Population/sample 2 Standard Deviation

For 95% Confidence interval

α = 0.05,      α/2 = 0.025
From z tables of Excel function NORM.S.INV(α/2) we find the z value
z = NORM.S.INV(0.025) = 1.96
We take the positive value of z

Confidence interval is given by
$\newline \bar{x_{1}}-\bar{x_{2}}\pm z\sqrt{\frac{(s_{1})^{2}}{n_{1}}+\frac{(s_{2})^{2}}{n_{2}}} \newline 23.2 - 15.4\pm1.96\times\sqrt{\frac{(3.6)^{2}}{200}+\frac{(3.5)^{2}}{200}}$
= (7.104, 8.496)

Answer :

The lower bound is $\mathbf{7.104}$
The upper bound is
8.496

(Note to student : Since there can be an error due to rounding up, if the above answer does not work, try the following answer).

The lower bound is $\mathbf{7.102}$
The upper bound is $\mathbf{8.498}$

Interpretation of the interval

Answer : (Option B)

• B.   The researchers are 95% confident that the difference of the means is in the interval.

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(b)

About priming

Answer : (Option A)

• A.   Since the 95% confidence interval does not contain zero, the results suggest that priming does have an effect on scores.

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