Question

magazine includes a report on the energy costs per year for 32-inch liquid arystal display (LCD) televisions. The article states that 14 randomly selected 32 inch LCD televisions have a sample standa intervals for (a) the population variance and (b) the population standard deviation o. deair d St S Amume te sa ple l' taken fenn a normally dishbded population. Construct 90% oorndence Interpret the resuts Round to two decimal places as needed Interpret the resuts Select the comect choice below and fill in the answer box(es) to complete your choice Round to two decimal places as needed ) oc. Wie, 90% oorddene,you can saythat the population vananoe is less than e, D. Wit, 90% confidence, you can say that the population variance between 40 and 4.60 b) The conidence interval for the population standard deviation is (Round to two decimal places as needed ) Interpret the results Select the corect choioe below and fill in the answer box(es) to complete your Round to two decimal places as needed ) choice. OA. WO, 10% oordderoe, you can say Pat te population standard devason is loss that , wit, 10% oorfderce, you can say that the population starda ddeviati ris betwoo, and dollars per year dolars per year. with eos ortorm you can say that the ppotaton standard devation is greater than □ dolars per year OD, with 90% contderee, you can say that he populato standard do anon is beseen land c. dolianm per year

Answer 1

a) n = 14

DF = 14 - 1 = 13

With 13 degrees of freedom and 90% confidence interval the critical values are $\chi^{2}$_{$\alpha/2$, n - 1} = $\chi^{2}$_{0.05, 13} = 22.36

                                                                                                                               $\chi^{2}$_{1 - $\alpha/2$, n - 1} = $\chi^{2}$_{0.95, 13} = 5.89

The confidence interval for variance is

(n - 1)s²/ $\chi^{2}$$\alpha/2$, n - 1 < $\sigma^{2}$ < (n - 1)s²/ $\chi^{2}$_{1 - $\alpha/2$, n - 1}

or, 13 * (3.15)^2/22.36 < $\sigma^{2}$ < 13 * (3.15)^2/5.89

or, 5.8 < $\sigma^{2}$ < 21.9

Option - D is correct.

b) The confidence interval for standard deviation is

2.4 < $\sigma$ < 4.7

Option - D is correct.

c) n = 18

df = 18 - 1 = 17

With 17 degrees of freedom and 90% confidenced interval the critical value is t_{0.05, 17} = 1.740

The confidence interval is

$\bar x$ +/- t_{0.05, 17} * s/sqrt(n)

= 0.0353 +/- 1.74 * 0.0031/sqrt(18)

= 0.0353 +/- 0.0013

= 0.034, 0.0366

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