Question

the confidence interval for the population variance is

Answer 1

(730.72, 4569.53) is the confidence interval for the population variance σ² given that c = 0.98, s = 39 and n = 15. This can be obtained by using formula for confidence interval and using statistical table for values.

Find the confidence interval for population variance:

The formula for finding the confidence interval for population variance is:

$\frac{(n-1)s^{2} }{\chi^{2} _{\frac{\alpha }{2} } } < \sigma^{2} < \frac{(n-1)s^{2} }{\chi^{2} _{1-\frac{\alpha }{2} } }$

where n is the size of the sample, (n - 1) is the degrees of freedom, s is the standard deviation, α is the significance level.

Here it is given that,

• confidence level = 1 - α = 0.98

• significance level = α = 0.02

• standard deviation = 39

• sample size n = 15

• degrees of freedom (n - 1) = 15 - 1 = 14

α/2 = 0.02/2 = 0.01

1 - α/2 = 1 - 0.01 = 0.99

For (n - 1) degrees of freedom,

$\chi^{2}_{\frac{\alpha}{2} }$ = 29.141 , and  $\chi^{2}_{1-\frac{\alpha}{2} }$ = 4.660 (Using statistical table)

Using the formula we get,

$\frac{(n-1)s^{2} }{\chi^{2} _{\frac{\alpha }{2} } } < \sigma^{2} < \frac{(n-1)s^{2} }{\chi^{2} _{1-\frac{\alpha }{2} } }$

$\frac{(14)39^{2} }{29.141} < \sigma^{2} < \frac{(14)39^{2} }{4.660}$

$730.72 < \sigma^{2} < 4569.53$

The confidence interval for the population variance is (730.72, 4569.53)

Hence (730.72, 4569.53) is the confidence interval for the population variance σ² given that c = 0.98, s = 39 and n = 15.

Learn more about confidence interval here:

brainly.com/question/13807706

#SPJ1