Question

Question 3. (L.0.1.1, L.0.1.2, L.0.1.3, L.O.2.1, L.0.2.3). Total: 25 points A study examined the influence of the drug succinylcholine on the circulation levels of androgens in the blood. Blood samples were taken from wild, free-ranging deer immediately after they had received an intramuscular injection of succinylcholine administered using darts and a capture gun. A second blood sample was obtained from each deer 30 minutes after the first sample, after which the deer was released. The levels of androgens at time of capture and 30 minutes later, measured in ng/mL, for several deers are given below. Deer 1 2 3 4. 5 At Time of Injection 2.76 5.182.683.05 4.10 30 Minutes after Injection 7.02 3.10 5.44 3.99 5.21 Assuming that the populations of androgen levels at time of injection and 30 minutes later are normally distributed, (a) (10 points) Determine a 95% confidence interval for the mean level of androgens at time of capture (b) (15 points) Test at the 1% level of significance whether the androgen concentrations are altere- after 30 minutes.

Answer 1

Solution: 2 3 J 3.05 3.10 5 Deer At time of Injection 2.76 5.18 2.68 4:10 30 Minutes after Injection 7.02 5:44 3.99 5621 Determine a 95% confidence interval for the mean level of androgens at time of captuve. Ans: confidence level = 959%. = 0.95 = 1 -0.95 a = 0.05 significance level 95% confidence Interval for mean level of androgens at time of capture is given by: 0 + toros s un

where 2 χ= Σχ n [{(x-769² (n-0) and s= رس ملح x={x D 2.76 + 5:18 + 2-68 +3.05 + 4.10 5 3.554 17.77 5 20 = 3.554 =(x-3)? = (2.76–3,554) +(5.18-3,554) + (2.68- 3.554)?+(3.05–3.554)+7(4.10– 3.554)? Elx-)? - 0.6304 + 2.6439 +0.7639+0.2540 +0.2981 E(2-)= 4.5903 Se 4.5903 4.5903 5-1 -F-147575 4 = 1.07125

Critical value of t at 0.05 level of stgnificance for (5-1)= 4 degree of freedom toros (4) 2.776 I from 2 tailed tables] we get same value if we divide 0.05 = 0.025 to.25(4)= 2.776 [ from 1 tai led table] 95% confidence Interval is : 3.554 + 2.776 x 1.07125 5 3. 554 I 2.776 X 0:4791 3.554 + 1:32 (3.554 - 1.33, 3.554 + 1.33) (1 = (2.224, 4.884)

Dale (b) Test at 1% level of significance whether the androgen concentrations are altered after 30 minutes Deerl At time of Injection 30 minutes after (X) Injection (Y) d=x-Y d? 2.76 7.02 -4.26 18.1476 2 5.18 3.10 2.08 4.3264 3 2.68 5.44 -2.76 716176 y 3.05 3.99 -0.94 5 4.10 -1011 1.2321 Ed=-6.99 22232.20 73 We get Ed= -6:99 and Ed2 320 2073 0.8836 5.21 Ed b -6.99 5 -1.398 T = -1.398 Ho: Ma=0 Hi : Md ²0 Androgen concentrations are altered after 30 minutes This is two tailed test. Under Ho, the test Statistic ta d slun with degree of freedom = n-1 where. Sa I Fed2 (5d27 2 n-1 n

S- 1 5-1 32.2073 - (-6099227 5. 5 [32-2073 – 48.86017 Vu [32-2013 – 9:27 202] 1 x 22. 43528 5. 60 882 = 2.3683 S= 2.3683 Now t= te d -10398 -1.396 slun 2.3683/5 1.0591 t = -1.32 Significance level = 0.01 degree of freedom = 5-1 = 4 Now, critical value of t at oool level of Significance for 4 degree of freedom for two tailed test tovar (4) = 4,604 [ from 2 tailed tables Our test is two tailed, so critical value is -4.604 ana +4.604

Date Test Criteria : Reject Ho if calculated t calculated t < -4.604 Reject Ho if calculated t >+4.604 In Our Resulte : Calulated t > -4.604 -1.32 > -4.604 -4.604 So, we fail to reject Ho at 1% level of significance and conclude that Androgen concentrations are not altered after 30 minutes.