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11 August, 05:18

An experiment is conducted with 15 seniors who are taking Spanish at Oak View High School. A randomly selected group of eight students is first tested with a written test and a day later with an oral exam. To avoid order effects, the other seven students are tested in reverse order. The instructor is interested in the difference in grades between the two testing methods. The table below contains the descriptive statistics for the grades of the two tests.

n Mean Std. Deviation

Talk 15 1.523 1.530

Write 15 4.166 2.047

Difference (Talk - Write) 15 - 2.643 2.182

Calculate a 98% confidence interval for the difference in mean grades between the two testing methods.

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  1. 11 August, 05:21
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    Step-by-step explanation:

    The formula for determining the confidence interval for the difference of two population means is expressed as

    Confidence interval = (x1 - x2) ± z√ (s²/n1 + s2²/n2)

    Where

    x1 = sample mean grade for written test

    x2 = sample mean grade for oral test

    s1 = sample standard deviation for written

    s2 = sample standard deviation for oral test

    n1 = number of students for written test

    n1 = number of students for oral test

    For a 98% confidence interval, we would determine the z score from the t distribution table because the number of samples are small

    Degree of freedom =

    (n1 - 1) + (n2 - 1) = (15 - 1) + (15 - 1) = 28

    z = 2.467

    x1 - x2 = 1.523 - 4.166 = - 2.643

    Margin of error = z√ (s1²/n1 + s2²/n2) = 2.467√ (1.530²/15 + 2.047²/15) = 2.467√0.43540726667

    = 1.036

    The 98% confidence interval is - 2.643 ± 1.036
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