Question

A certain test preparation course is designed to help students improve their scores on the GMAT 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 55 students' scores on the exam after completing the course:

14,29,24,10,17

Using these data, construct a 90%confidence interval for the average net change in a student's score after completing the course. Assume the population is approximately normal.

1.Calculate the sample mean for the given sample data. Round your answer to one decimal place.

2. Calculate the sample standard deviation for the given sample data. Round your answer to one decimal place.

3.Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

4.Construct the 90% confidence interval. Round your answer to one decimal place.

Answer 1

Solution:

Given: The following measurements are the net change in 5 students' scores on the examination after completing the course:

14,29,24,10,17

We have to construct a 90%confidence interval for the average net change in a student's score after completing the course.

Part 1) Calculate the sample mean for the given sample data

T= n

x

14

29

24

10

17

> = 94

Thus

T= n

$\bar{x} = \frac{94}{5}$

T = 18.8

Part 2) Calculate the sample standard deviation for the given sample data

8 = 14 12 – (Στ)2 η η - 1

x	x^2
14	196
29	841
24	576
10	100
17	289
> = 94	Σ2 = 2002

Thus

8 = 14 12 – (Στ)2 η η - 1

S = 2002 - (94)2/5 5-1 V

2002 – 1767.2 S =

234.8 S = V 4

s= V58.7

2 2 = $

Part 3.Find the critical value that should be used in constructing the confidence interval.

df = n - 1 = 5 - 1 = 4

Two tail area = 1 -c = 1 - 0.90 = 0.10

t Table cum. probl ( 0.50 1.00 0.250 0.50 .20 0.40 0.15 0.30 0.10 0.20 0.05 0.10 two-tails 0.000 0.000 0.000 0000 0.000 1.000 0.816 0.765 0.741 0.727 1.376 1.061 0.978 0.941 0.920 1.963 1.386 1.250 1190 1.156 3.078 1.886 1.638 1522 1.476 6.314 2.92 2.359 2.132 2.015

t_c = 2.132

Part 4) Construct the 90% confidence interval.

(1 – E <μ< + Ε)

where

E = te x s/n

E = 2.132 x 7.7/15

E = 2.132 x 7.7/2.23607

E = 7,3

Thus

(1 – E <μ< + Ε)

(18.8 – 7.3 <μ< 18.8 +7.3)

(11.5 <μ< 26.1)

Thus a 90%confidence interval for the average net change in a student's score after completing the course is:

(11.5 <μ< 26.1)