A teaching assistant collected data from students in one of her classes to investigate whether study time per week (average number of hours) differed between students in the class who planned to go to graduate school and those who did not. Complete parts (a) through (C). Data Full data set Graduate school: 13, 7, 15, 10, 5, 5, 2, 3, 12, 16, 15, 39, 8 14, 10, 17, 3, 27, 15, 5, 5 No graduate school: 6, 8, 14, 6, 5, 13, 10, 10, 13, 5 Find a 95% confidence interval comparing the population means. Interpret. Find a 95% confidence interval comparing the population means. The 95% confidence interval for (mu _1 - mu _2) is (1.6, 7.0) (Round to the nearest tenth as needed.)

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 of group 1

x2 = sample mean of group 2

s1 = sample standard deviation for data 1

s2 = sample standard deviation for data 2

For data 1(graduate)

number of samples, n1 = 21

Mean, x1 = (13 + 7 + 15 + 10 + 5 + 5 + 2 + 3 + 12 + 16 + 15 + 39 + 8 + 14 + 10 + 17 + 3 + 27 + 15 + 5 + 5)/21 = 11.7

Standard deviation = √(summation(x - mean)/n

Summation(x - mean) = (13 - 11.7)^2 + (7 - 11.7)^2 + (15 - 11.7)^2 + (10 - 11.7)^2 + (5 - 11.7)^2 + (5 - 11.7)^2 + (2 - 11.7)^2 + (3 - 11.7)^2 + (12 - 11.7)^2 + (16 - 11.7)^2 + (15 - 11.7)^2 + (39 - 11.7)^2 + (8 - 11.7)^2 + (14 - 11.7)^2 + (10 - 11.7)^2+ (17 - 11.7)^2 + (3 - 11.7)^2 + (27 - 11.7)^2 + (15 - 11.7)^2 + (5 - 11.7)^2 + (5 - 11.7)^2 = 1533.25

s = √1533.25/21 = 8.54

For data 2,

n2 = 10

Mean, x2 = (6 + 8 + 14 + 6 + 5 + 13 + 10 + 10 + 13 + 5)/10 = 9

Summation(x - mean) = (6 - 9)^2 + (8 - 9)^2 + (14 - 9)^2 + (6 - 9)^2 + (5 - 9)^2 + (13 - 9)^2 + (10 - 9)^2 + (10 - 9)^2 + (13 - 9)^2 + (5- 9)^2 = 110

s2 = √110/10 = 3.32

For a 95% 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) = (21 - 1) + (10 - 1) = 29

z = 2.045

x1 - x2 = 11.7 - 9 = 2.7

√(s1²/n1 + s2²/n2) = √(8.54²/21 + 3.32²/10) = √(3.473 + 1.10224)

= 2.14

Margin of error = 2.045 × 2.14 = 4.38

The upper boundary for the confidence interval is

2.7 + 4.38 = 7.08

The lower boundary for the confidence interval is

2.7 - 4.38 = - 1.68

We are confident that the difference in population means between students in the class who planned to go to graduate school and those who did not is between - 1.68 and 7.08

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