Question

The number of hours of reserve capacity of 10 1 .75 1.88 randomly selected automotive batteries is shown to the 409 42 right. 1.51 1.68 1.77 1432 Assume the sample is taken from a normally distributed population Construct 90% confidence intervals for (a) the population variance of and (b) the population standard deviation o. (a) The confidence interval for the population variance is (Round to three decimal places as needed.) Interpret the results. Select the correct choice below and fill in the answer box(es) to complete your choice. (Round to three decimal places as needed.) O A. With 10% confidence, it can be said that the population variance is between and O C. With 90% confidence, it can be said that the population variance is less than B. With 90% confidence, it can be said that the population variance is between and OD. With 10% confidence, it can be said that the population variance is greater than Click to select your answer(s).

Answer 1

Now, we need to square all the sample values as shown in the table below:

Observation:	X	X^2
1	1.75	3.0625
2	1.88	3.5344
3	1.51	2.2801
4	1.68	2.8224
5	1.77	3.1329
6	1.99	3.9601
7	1.36	1.8496
8	1.56	2.4336
9	1.43	2.0449
10	2.06	4.2436
Sum =	16.99	29.3641

Therefore, the sample variance is computed as shown below:

$s^2 = \displaystyle \frac{1}{n-1} \left( \sum_{i=1}^n X_i^2 - \frac{1}{n}\left(\sum_{i=1}^n X_i\right)^2 \right)$

$s^2 =\displaystyle \frac{1}{10 - 1} \left( 29.3641 - \frac{16.99^2}{10} \right)$

$s^2$ =  0.0553

Therefore, based on the data provided, the sample variance is $s^2$= 0.0553.

The critical values for $\alpha$= 0.1 and df = 9 degrees of freedom are $\chi_L^2 = \chi^2_{1-\alpha/2,n-1}$ = 3.3251 and$\chi_U^2 = \chi^2_{\alpha/2,n-1}$ = 16.919. The corresponding confidence interval is computed as shown below:

$CI(\text{Variance}) =\displaystyle \left( \frac{(n-1) s^2}{\chi^2_{\alpha/2,n-1}}, \frac{(n-1) s^2}{\chi^2_{1-\alpha/2,n-1} } \right)$

$CI(\text{Variance}) = \displaystyle \left( \frac{(10-1) \times 0.0553}{16.919}, \frac{(10-1) \times 0.0553}{3.3251} \right)$

$CI(\text{Variance})$= (0.0294, 0.1498)

B) With 90% confidence, it can be said that the population variance is between 0.0294 and 0.1498.

Now that we have the limits for the confidence interval, the limits for the 90% 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:

$CI(\text{Standard Deviation}) = \displaystyle \left( \sqrt{0.0294} , \sqrt{0.1498} \right)$

$CI(\text{Standard Deviation}) = \left( 0.1716 , 0.387\right)$

B) With 90% confidence, you can say that the population standard deviation is between 0.1716 and 0.387 hours of reserve capacity.