DEVEMEM_0321500458_SE.qxd x e Module 6 (page 12 of 20) X + .ca/mod/quiz/attempt.php?attempt= 10364974&cmid=5824486&page=11 Beartracks sign on @ eClass Portal ) Sessions List - MS... Lobby | Top Hat sos SOS review . Beyond Labz Portal Chem 261 V ass HELP EMAIL Question 12 It is believed that animals living in northern regions have shorter appendages than animals living in southern regions. Random samples of birds were selected from each group and wing lengths for birds (in millimeters) were measured. Tries remaining 2 The following table gives summary statistics. Marked out of 3.00 Region n Average Standard deviation Flag Northern (N) 33 115.1 1.1 question Southern (S) 33 116.9 2.4 Difference (d = N - S) 33 -1.8 1.0 (a) Find a 98% confidence interval for the mean difference of wing lengths of birds between the northern regions and the southern regions. (Round your answers to 4 decimal places, if needed.) Fin (-2.2766 -1.3234 Your last answer was interpreted as follows: 2.2766 Your last answer was interpreted as follows: -1.3234 (b) Based on this confidence interval, is there enough evidence to support the belief? Yes, since the interval contains 0. No, since the interval is completely below 0. No, since the interval contains 0. Yes, since the interval is completely below 0 Check Last saved at 18:29:23 Save Progress Previous page Next page IN W X 18.C PrtSon Home End F3 F4 F5 6 F8 F9 $10 % &quiz/attempt.php?attempt= 10364974&cmid=5824486&page=12 artracks sign on @ eClass Portal ) Sessions List - MS... Lobby | Top Hat sos SOS review 8 Beyond Labz Portal Chem 261 W15 Pre. HELP EMAIL oard / My courses / STAT 151 (LEC P1 Q1 R1 $1 T1 12 U1 V1 Winter 2022) / Online Homework Assignment / Module 6 Quiz nav on 13 A clinic compared healthy dogs it owned with healthy pets brought to clinic to be neutered. The summary statistics for blood cholesterol levels (milligrams deciliter of blood) appear in the following table: 1 2 maining: v V d out of Types n Average Standard deviation Pets (P) 61 198 64.2 Clinic (C) 61 177 43.4 13 Difference (d = P - C) 61 21.0 51.3 19 20 (a) Find a 90% confidence interval for the difference in mean blood cholesterol levels dogs brought to the clinic (P) and v V dogs owned by the clinic (C). Round your answers to 4 decimal places, if needed.) Finish atte (b) Based on this confidence interval, can we conclude there is some differences in mean blood cholesterol levels between the two groups? No, since the interval contains 0. Yes, since the interval is completely above 0. O No, since the interval is completely above 0. O Yes, since the interval contains 0. Check Last saved at 18:29:28 ave Progress Previous page Next page O M IN w X 18.C O X PrtScn Home End F3 F4 F5 F6 F7 F8 F10 A & 5 O 8 9 0d/quiz/attempt.php?attempt= 10364974&cmid=5824486&page=14 Beartracks sign on e eClass Portal > Sessions List - MS... Lobby | Top Hat sos SOS review . Beyond Labz Portal Chem 261 W15 Pre HELP EMAIL tion 15 Quiz na A medical researcher wanted to compare two different methods (A and B) for measuring cardiac output (L/min). Method A is the "gold standard" but is invasive (involves inserting a catheter into the main heart artery). Method B is 1 remaining: less accurate but is non-invasive. To control for differences between persons, both methods are used on each person V ed out of participating in the study. This allows measurements to be directly compared for each method, per individual. Cardiac output for 10 subjects was measured by both methods and some summary statistics of the data are summarized in the table below. Is there enough evidence to conclude that the cardiac output for method B is less than the cardiac output 13 ag tion for method A on average at a significance level of 0.005. 19 20 Cardiac Method n Average Standard deviation v A 10 5.81 1.84 10 5.16 2.08 Finish att Difference (d = A - B) 10 0.65 0.72 Perform a statistical test. (a) What are the null and alternative hypotheses? Ho: P 0 HA: P 4 (b) What is the test statistic? (Round your answer to 3 decimal places, if needed.) (c) Using the statistical table, the p-value is 0.01