Question

A transect is an archaeological study area that is 1/5 mile wide and 1 mile long. A site in a transect is the location of a significant archaeological find. Let x represent the number of sites per transect. In a section of Chaco Canyon, a large number of transects showed that x has a population variance σ2 = 42.3. In a different section of Chaco Canyon, a random sample of 25 transects gave a sample variance s2 = 51.3 for the number of sites per transect. Use a 5% level of significance to test the claim that the variance in the new section is greater than 42.3. Find a 95% confidence interval for the population variance.

Answer 1

Answer: $31.28\leq\sigma^2\leq99.29$

Step-by-step explanation:

Confidence interval formula for population variance :-

$\dfrac{(n-1)s^2}{\chi^2_{\alpha/2}}\leq \sigma^2\leq\dfrac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

Given : $s^2 = 51.3$ and $n= 25$

Significance level : $\alpha=0.05$

Using chi-square distribution, the Critical values are:

$\chi^2_{n-1, \alpha/2}=\chi^2_{24,0.025}=39.36$

$\chi^2_{n-1, 1-\alpha/2}=\chi^2_{24,0.025}=12.40$

Then, the confidence interval for population variance is given by :-

$\dfrac{(24)( 51.3)}{39.36}\leq \sigma^2\leq\dfrac{(24)( 51.3)}{12.40}\\\\\approx31.28\leq\sigma^2\leq99.29$

Hence, the 95% confidence interval for the population variance :

$31.28\leq\sigma^2\leq99.29$