Question

A certain test preparation course is designed to help students improve their scores on the GRE exam. A mock exam is given at the beginning and end of the course to determine the effectiveness of the course. The following measurements are the net change in 5 students' scores on the exam after completing the course: 6,14,12,23,0 Using these data, construct a 95% confidence interval for the average net change in a student's score after completing the course. Assume the population is approximately normal. Step 2 of 4 : Calculate the sample standard deviation for the given sample data. Round your answer to one decimal place.

Answer 1

Answer:

(0.25 ; 21.75)

8.7 (1 decimal place)

Step-by-step explanation:

Net change in scores : X = 6,14,12,23,0

Sample mean, xbar = (6 + 14 + 12 + 23 + 0) /5 = 55 /5 = 11

Sample standard deviation, s = 8.66 ( from calculator)

Sample standard deviation = 8.7( 1 decimal place)

Sample size, n = 5

The 95% confidence interval; we use t, because n is small

Tcritical at 95%, df = 4 - 1 = 3 ; Tcritical = 2.776

Xbar ± standard Error

Standard Error = Tcritical * s/√n

Standard Error = 2.776 * 8.66/√5

Standard Error = 10.751086 = 10.75

Lower boundary = (11 - 10.75) = 0.25

Upper boundary = (11 + 10.75) = 21.75

(0.25 ; 21.75)

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