Question

The grade point averages (GPA) of 18 randomly selected college students are used to estimate the mean GPA of the college students. The GPAs from the sample are as follows:

2.3       3.3       2.6       1.8 0.2 3.1       4.0       0.7 3.1

2.3       2.0       3.1       3.4       1.3       2.6       2.6 3.7 2.2

1. (5 points) Is it justified to use of the standard normal distribution to construct the confidence interval? Explain.
2. (10 points) If the population is assumed to be normally distributed, construct a 98% confidence interval for the population mean GPA. Show your work in detail and use 3 decimal digits.

Answer 1

Solution:

x	x2
2.3	5.29
3.3	10.89
2.6	6.76
1.8	3.24
0.2	0.04
3.1	9.61
4	16
0.7	0.49
3.1	9.61
2.3	5.29
2	4
3.1	9.61
3.4	11.56
1.3	1.69
2.6	6.76
2.6	6.76
3.7	13.69
2.2	4.84
∑x=44.3	∑x2=126.13

Mean ˉx=∑xn

=2.3+3.3+2.6+1.8+0.2+3.1+4+0.7+3.1+2.3+2+3.1+3.4+1.3+2.6+2.6+3.7+2.2/18

=44.3/18

=2.4611

Sample Standard deviation S=√∑x2-(∑x)2nn-1

=√126.13-(44.3)218/17

=√126.13-109.0272/17

=√17.1028/17

=√1.006

=1.003

Degrees of freedom = df = n - 1 = 18 - 1 = 17

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

t /2,df = t0.01,17 =2.567

Margin of error = E = t/2,df * (s /n)

= 2.567 * (1.003 / 18)

= 0.607

Margin of error = 0.607

The 98% confidence interval estimate of the population mean is,

- E < < + E

2.461 - 0.607 < < 2.461 + 0.607

1.854 < < 3.068

(1.854, 3.068 )