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.

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

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632

i think there was a typo. you wrote pat brings 3 quarts of ice-cream to the party. each serving of ice-cream is 1 6 quart. how many total servings of ice-cream does pat bring to the party? i think the bold is supposed to be 1/6. so 1 quart of ice cream = 6 servings. 6 * 3 = 18.