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 0.9 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

a- Would it be justified if the standard normal distribution is used to construct the confidence interval? Explain.

b-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 ? Explain

Answer 1

Solution:

x	x2
2.3	5.29
3.3	10.89
2.6	6.76
1.8	3.24
0.9	0.81
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=45	∑x2=126.9

Mean ˉx=∑xn

=2.3+3.3+2.6+1.8+0.9+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

=45/18

=2.5

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

=√126.9-(45)218/17

=√126.9-112.5/17

=√14.4/17

=√0.8471

=0.9204

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 =2.797

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

= 2.797 * (0.92 / 18)

= 0.457

Margin of error = 0.457

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

- E < < + E

2.5 - 0.457 < < 2.5+ 0.457

2.042 < < 2.957

(2.042, 2.957)