Question

Consider the heights (inches) for a simple random sample of ten (10) supermodels
listed below:
70, 71, 69.25, 68.5, 69, 70, 71, 70, 70, 69.5

There is a claim that supermodels have heights that have much less variation than
the heights of women in the general population.
(a) Use a Hypothesis test with a significance level of 0.01 to verify the claim that
supermodels have heights with a standard deviation that is less than 2.6 inches.

(b) Use the Confidence Interval Method with a significance level of 0.01 to verify the
claim that supermodels have heights with a standard deviation that is less than 2.6
inches.

PLEASE HELP WITHOUT EXCEL OR ANY OTHER STATS PACKAGE

Answer 1

We want to test that supermodels have heights with a standard deviation that is less than 2.6 inches.

σ = 2.6 therefore σ2 = (2.6)^2 = 6.76

We want to test that supermodels have heights with a Variance that is less than 2.6 inches.

The following null and alternative hypotheses need to be tested:

Ho: σ2 =6.76

Ha: σ2 < 6.76

A)

Observation:	x	x^2
1	70	4900
2	71	5041
3	69.25	4795.5625
4	68.5	4692.25
5	69	4761
6	70	4900
7	71	5041
8	70	4900
9	70	4900
10	69.5	4830.25
Sum =	698.25	48761.06

Rejection Region

Based on the information provided, the significance level is α=0.01,df = n-1 = 10-1 = 9

the critical value is χ2df = χ2 0.01,9 = 2.088 from χ2 table

and then the rejection region for this left-tailed test is R={χ2:χ2<2.088}.

The Chi-Squared statistic is computed as follows:

Decision about the null hypothesis

Since it is observed that χ2 = 0.852 < χ2df = 2.088  it is then concluded that the null hypothesis is rejected.

Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population variance σ2 is less than 6.76, at the 0.01 significance level.

Therefore we may conclude that supermodels have heights with a standard deviation that is less than 2.6 inches.

B)

The 99 % confidence interval for standard deviation

sample variance is s^2 = 0.6396s2=0.6396.

The critical values for α=0.01 and df=9 degrees of freedom are Lower & upper critical values

χ2L ​=χ21−α/2,n−1​ = χ20.995,9​ = 1.7349

χ2L ​=χ2α/2​​​​​​​,n−1​ = χ2​​​0.005,9​ = 23.5894.

The 99 % confidence interval for variance is

Now that we have the limits for the confidence interval, the limits for the 99% confidence interval for the population standard deviation are obtained by simply taking the squared root of the limits of the confidence interval for the variance, so then:

CI(Standard Deviation) =

  

The confidence interval for standard deviation for supermodel height does not contain null hypothesis value ( 2.6) so we reject Ho at1% l.o.s.

Therefore we may conclude that supermodels have heights with a standard deviation that is less than 2.6 inches.