Question

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Multi part question set

If you can’t finish all of it, that’s ok!

Would really appreciate any and all help as soon as possible

Thank you so much in advance!

you!! Commuting - 1, MULTIPLE CHOICE Thank There are 5 questions in this set. As you proceed through these questions, use necessary information in the previous questions. A transportation economist wants to compare the mean number of hours spent commuting in the population of commuters in 2018 with the mean number of hours spent commuting in the population of commuters ten years earlier, in 2008. In a random sample of 350 commuters from 2018, the mean number of hours spent commuting was 225. and the standard deviation was 12 hours. In an independent random sample of 250 commuters from 2008, the mean number of hours spent commuting was 208, and the standard deviation was 11 hours We begin by calculating the 95% confidence interval for the difference between the population meanin 2018 (the "Xpopulation) and the population mean in 2008 (the "Y" population). Which of the following is the correct confidence interval? *-*-*-*8+usy ONE - 1-0 TWO X-Y IS Vm my THREE CX-Y- 1 ng mg Vnx y FOUR cx-% -2, <HxWx<X-T+ = 1-6 TWO O ONE - THREE O FOUR Give the table value (value of, or to that you will use. Four decimals. 2 Calculate the lower (smaller endpoint) of the 95% confidence interval. Two decimals 3 Calculate the upper (larger endpoint) of the 95% confidence interval. Two decimals J The economist reads an article saying that the mean number of hours spent commuting increased by 19 hours from 2008 to 2018. Use the confidence interval you estimated above to test the null hypothesis that the mean number of hours spent commuting increased by 19 hours from 2008 to 2018, against the alternative hypothesis that it did not at a 5% significance level. What is your conclusion? ACCEPT Ho (fail to reject Hal REJECT Ho in favor of HA 5 ONCCEPT O REJECT

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Answer 1

Using f test for equality of two variances, we see Homogeneity assumption is valid.

Then we can use pooled standard deviation s in the standard error term.

Hence, the confidence interval correct form is

ONE.

For 95% confidence interval, $\alpha$ =0.05

The value of t used is t_{(0.05/2),(350+250-2)}

=t_0.025,598=1.9639(using calculator)

s = |(3498 129+(249 X 112) 598 = 11.5941 The 95% confidence interval! (225-208) tt x 11.59x350 01025;598 f 250 =14351.9639x 1.59x/350 sro -/15.1152,1818848) The loveer (smaller end endpoint) L- 15.1 (3 AM) The upper (larger endpoint) :- 18.88 (4 Aus) The confidence interval is (15.12, 18+88) value 19 does not lie in this seeject Since interval, wee Ho. 5) Amy Reject

If there is any understanding problem regarding the sum, please feel free to ask any doubt in comment box. THANKS.