Question

A group of researchers on Sable Island are wondering whether the grey seals pups on the west side of the Island grow at different rates than those on the east side of the island. To test this they measured the weight of 21 one week old grey seal pups on the west side of the island and 28 one week old grey pups on the east side of the island. They found that the average weight of one week old pups on the west of the island is 22.571 and the average weight of one week old pups on the eas side of the island is 32.493. They found that the sample standard deviations were 17.166 and 14.697 for the west and east sides respectively. The researchers assume that the population variance of weight for one week old pups is the same on bot the east and west sides of the island. Which of the following is the correct null and alternative hypotheses? Let , be the average weight of one week old grey seal pups on the west side of the island and 2 to be the average weight of one week old grey seal pups on the cast side of th island. HONI = 2 and Ma: H1 CH2 Ho! Mia and Hi-V2 He: P1 - pard 1 >> HO: Hi = 42 and H: HI # 2 [1 pts)] Satriar Tries 0/1 What is the correct distribution for the test statistic? Tails distribution x2 distribution t distribution F distribution Strul Are Tries 0/1 What are the degrees of freedom of the test statistic [1 pt() SAT Tries 0/3 Calculate the standard error of the test statistic [1 pt(s)] Banca Tries 0/3 Calculate the test statistic 1 ptsil SatulAra Tries 0/3 Do we reject the null hypothesis at significance level a = 0.17 Yes, we reject the null hypothesis No, we fail to reject the null hypothesis [1 pt(s)] Gatwira Tries 0/1 This discussion is closed. Send Feede

Answer 1

Solution The null and alternatore hypothers are Hill, = ", option is correct. Hail, Ella t costribution because population variance are unberoun. Ostion ces correct. Degree of freedom df = n, th -2 = 2 +88-2 = 07. standard Erot; S* 1,-21 n=28. + 1 *, -21671 S = 170166 Y2 = 322193 Sa Cr, -1) 5,²+2 12 2 ² 32 =14697 S- 20617,166)*+ [ 28-1) (14.697) * 017 S. 15.7949 standard Error . sht 5 15.7949x 1 of = 4,5596 28 tal 22.571-321493 si fata -9.922 4.5596 4.5596 - 2.17607 -2. (161

117 significance level 2+011 pralue! - t = 2.1876 Pralue = 0.034616, pratice (0.03U6 COID), reject to of tertical value Cof-u7, « L.) et central = 1.67774 It caule tentical value then we reput nuuhypother option an corret Yes we repent the null hypothes,