Question

15 In this problem, assume that the distribution of differences is approximately normal. Note: For degrees of freedom dr. not in the Student's t table, use the closest d.. that is smaller. In some situations, this choice of d.f. may increase the P-value by a small amount and therefore produce a slightly more conservative" answer. A survey of houses and traditional hogans was made in a number of different regions of the modern Navajo Indian Reservation. The following table is the result of a random sample of eight regions on the Navajo Reservation. Area on Navajo Reservation Number of Tnhabited Houses Number of inhabited Hogans Bitter Springs 16 Rainbow Lodge 18 12 Kayenta 68 Red Mesa 9 Black Mesa 11 Canyon de Chelly 28 Cedar Point Burnt Water Does this information indicate that the population mean number of inhabited houses is greater than that of hogans on the Navajo Reservation? Use a 5 level of significance. Let d = houses - hogans.) (a) What is the level of significance? 46 32 15 47 17 50 50 State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? Mol H= 0; H> 0; right-tailed Hot Hd > 0; H M = 0; right-tailed Ho! = 0; HTH i two-tailed HO HO HEMO:left-talled (6) What sampling distribution will you use? What assumptions are you making? The standard normal. We assume that has an approximately normal distribution The student'st Wessume that has an approximately normal distribution The standard normal. We assume that has an approximately uniform distribution

Answer 1

Formulas: 1) right-tailed test is a test in which we test the null hypothesis Ho: Il = Mo against the alternative hypothesis H : > Mo. Given that difference (houses-hogans) in data 1,6,22,-23,-4,-19,33,32. Find the Sample mean(). By using excel formula =AVERAGE(1,6,22,-23,-4,-19,33,32) We got Sample mean() = 6. Find the Sample standard deviation(s). By using excel formula ESTDEV.S(1,6,22,-23,-4,-19,33,32) We got Sample standard deviation(s) = 21.5539.

We have Number of random samples = n = 8 Sample mean = ž= 6 Sample standard deviation=s= 21.5539 Significance level = a = 5% = 0.05 Claim: population mean is less than 0. a) :: Significance level = 0.05 Claim: population mean is less than 0. H:Hd = 0; H:Md > 0; right – tailed :: Ho: Mld = 0; H:Hd > 0; right – tailed

b) :: The student's t, we assume that d has an approximately normal distribution. The value of the test statistic t X-MO Vn 6-0 21.5539 V& = 0.787 : test statistic = t = 0.787.

c) Now we have to find the p-value. Degrees of freedom df = n-1 = 8-1 = 7. Using excel formula: =T.DIST.RT(0.787,7), we got p-value = 0.1084. :: 0.125 <p-value < 0.250

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