Question

3. The Graduate Record Examination (GRE) is a test required for admission to many U.S. graduate schools. Students’ scores on the quantitative portion of the GRE follow a normal. (Source: http://www.ets.org/.) Suppose a random sample of 10 students took the test. 152, 126, 146, 149, 152, 164, 139, 134, 145, 136 A normal probability plot verifies that the data are normally distributed.

a) Determine the sample standard deviation.

b) Construct a 95% confidence interval for the population standard deviation. Interpret interval.

Answer 1

a)

S.No		X	(X-x̄)	(X-x̄)2
1		152	7.700	59.29000
2		126	-18.300	334.89000
3		146	1.700	2.89000
4		149	4.700	22.09000
5		152	7.700	59.29000
6		164	19.700	388.09000
7		139	-5.300	28.09000
8		134	-10.300	106.09000
9		145	0.700	0.49000
10		136	-8.300	68.89000
Σx		1443	Σ(X-x̄)2=	1070.1000
x̄=Σx/n		144.30	s2=Σ(x-x̄)2/(n-1)=	118.90000
			s=√s2    =	10.9041

from above:

  sample standard deviation =10.9041

b)

	here n =	10
	s2=	118.900
Critical value of chi square distribution for n-1=9 df and 95 % CI
Lower critical value χ2L=	2.700
Upper critical valueχ2U=	19.023

for Confidence interval of standard deviation:
Lower bound =√((n-1)s2/χ2U)=sqrt((10-1)*118.9/19.023)=	7.500
Upper bound =√((n-1)s2/χ2L)==sqrt((10-1)*118.9/2.700)=	19.908

from above 95% confidence interval for population standard deviation =(7.5<σ<19.908)