Question

Lab 3: Homework Assignment (30 pts.) Using the data from Table 2 below, calculate the statistics and fill in the cells of Worksheet 1 below. Use these data to make a table in MS Excel that lists the population name (Lake) and the sample statistics for each population including mean ducklings fledged, median, DF, SS, s2, SD, SE, and tos(SE) for each of the populations. (10 pts) Use Table 3 below as a template to construct your table in MS Word (follow the Homework Guidelines and the format we used in Lab 1). Also in MS Excel or SigmaPlot, make a bar graph that illustrates the mean with the 95% CI represented as "error bars. (10 pts) Table 2 (Sample data for homework). The number of ducklings that fledged (survived) from randomly located nests at five prairie pothole lakes in southern Minnesota during the summer of 2017 Hemlock Loon Saranac Eagle Middle Lake Lake Lake Lake Lake 8 6 12 7 16 9 5 8 7 18 19 9 Upper and lower quartile (25% and Table 3. Use this sheet to set up your table for this homework75% assignment in Excel. (Note: format your table following the 12 3 5 16 6 15 6 4 13 3 3 11 10 15 4 6 9 5 11 3 10 3 8 format provided in the Homework Guidelines) SD SE (SE x Median DF SS Lake Hemlock Loon Saranac Eagle Middle

Answer 1

Solutions for the above questions are given below.

At first we have to compute table 3... Lake T Hemlock 13. 5 Median DF SS s? SD SE tgs (SE} 7 3.16 2.33 NE 1:32 Loon Saranae 7.78 509 2016 1.98 Eagle 1 6 Middle 6 6 . We have computed all the necessary values to solve the given problems and 1. The mean of Hemlock lake is = 13.5 The 95%. Condas Confidence Interval is given by = ( x - I tu sa an + h on this) =(135 = 98 47; 0.025 ,13.5+ VS718 17:0-028) Bo -442-5-0-620X-2-345 35+062747-365) =(135='1-12*2*365;13+5+ 1*12x225From Biometrie n . . 1 L Table t7;025 22-365 +(10-85, 16:15.) Herencz 8 S2 Sample SD Tri 3:16

E 2. To check, which lakes had a differsonel in wеали пло. Дел оf du c ): 1 әр. have to calculate Con 95% confidence Intervals for all the cases, :) Hemlock Lake : (10 185, 16 15 ) E Calculeled in part 1 11) Loon Lake: 52.233 die $ 1.32 951. CI ( * - fra 20:0-025, + im Vez 002) = (2-33-193 X2:306,2-33, + 132 x 2-36) ? (1•315, 3'345) ili ) Saranae Lake! 27.72 *** 205 227698 nag 854 CI ! (7-es ta;0-025, *tag #9;0023) +(7:73 - 1.99 *2*306, 7-78+1.98 x 2300) 3.-> (6:26, 9.30) ; iv) Eagle Lake : ña 11 s 25.09 uz 7

85% CI: (8 a 1850-025, T + 6:0-025) 2(11-6-5 x 2 447,11 + 504 x24+7) 3(629, 15.70 . . . 8) Meddle Lace X26 52 2.16 1 z 7 954. CI! (T6 - t6; 0.025, 7 + en tar 3 0 -025) 3 (6* *** * 2:447,67 2 4+7) z (4,71997) For each of the cases 95%. Confidence Interval hos has been calculated. If population means die in between the confidence intervals then there is no significant difference in between the means. 3. We assume, . for Saranae a Lake, X ; ld N(.M.,0,?] izl(on, and for Middle Lake , Yi id N (M2.022) Les og i 21(1) ne

Our Testing Problem, Ho: Mo = M2 Vs Hi: M,&Hz Test Statistie : 4. X- . tnitla-2 Rejection Criteria We reject the the hull hypothesis Ho against t at 15% level of significance if and only it on 1 Observed ?] > tothz-2; to ķ otherwise we accept it. Calculation s de se nà - Debre Here n. 29 . N₂ 27 X 27.78 726lands 8²2 Sample Variance for Saranae Lave - 3.92 S? - Sample Variance for Middle Lake 24.66

$? Combined Variouice 2. (1,-1).s.+ Me 1 ) 2 Notu 2 8 X 3.02 + 6X466 27-2 ? 4.237 522.058 = 727178-6 2.058 XI 1'78 1037 Now, titi 0.025 2 2.145 2 1.716 =Observed T151-716 Here 22 0·05 | L Here, 10bserved al < thithe -2; & = We accept the null hypothesis ie there is no significant difference in means fose Saranae and Middle Lakes