Two researchers conducted a study in which two groups of students were asked to answer 42 trivia questions from a board game. The students i

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 22.7 with a standard deviation of 4.5, while the 200 students in group 2 had a mean score of 19.7 with a standard deviation of 4.3. Complete parts (a) and (b) below.(a) Determine the 95% confidence interval for the difference in scores, μ1 – μ2. (Round to three decimal places as needed.)Interpret the interval. Choose the correct answer below.A. The researchers are 95% confident that the difference of the means is in the interval.B. There is a 95% probability that the difference of the means is in the interval.C. The researchers are 95% confident that the difference between randomly selected individuals will be in the interval.D. There is a 95% probability that the difference between randomly selected individuals will be in the interval.(b) What does this say about priming?A. Since the 95% confidence interval contains zero, the results suggest that priming does have an effect on scores.B. Since the 95% confidence interval contains zero, the results suggest that priming does not have an effect on scores.C. Since the 95% confidence interval does not contain zero, the results suggest that priming does have an effect on scores.D. Since the 95% confidence interval does not contain zero, the results suggest that priming does not have an effect on scores.

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Alexandra 3 days 2021-10-11T16:48:42+00:00 1 Answer 0

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    2021-10-11T16:50:17+00:00

    Answer:

    (1) Correct option is B.

    (2) Correct option is C.

    Step-by-step explanation:

    The information provided is:

    n_{1}=200,\ \bar x_{1}=22.7,\ s_{1} = 4.5\\n_{2}=200,\ \bar x_{2}=19.7,\ s_{2} = 4.3

    The (1 – α)% confidence interval for the difference between two mean is:

    CI=\bar x_{1}-\bar x_{2}\pm t_{\alpha/2, (n_{1}+n_[2}-2}\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}} }

    The critical value of t is:

    \alpha /2=0.05/2=0.025

    degrees of freedom =n_{1}+n_{2}-2=200+200-2=398

    t_{\alpha/2, n_{1}+n_{2}-2}=t_{0.025, 398}=1.96

    Compute the 95% confidence interval for the difference between two mean as follows:

    CI=22.7-19.7\pm 1.96\sqrt{\frac{4.5^{2}}{200}+\frac{4.3^{2}}{200} }\\=3\pm0.8624\\=(2.1376, 3.8624)\\\approx(2.14, 3.86)

    Thus, the 95% confidence interval, (2.14, 3.86) implies that the true mean difference value is contained in this interval with probability 0.95.

    Correct option is B.

    The null value of the difference between means is 0.

    As the value 0 is not in the interval this implies that there is a difference between the two means, concluding that priming does have an effect on scores.

    Correct option is C.

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