Safari File Edit View History Bookmarks Window Help C 0 4) 53%D Sat 9:42 PM Alli
ID: 3314153 • Letter: S
Question
Safari File Edit View History Bookmarks Window Help C 0 4) 53%D Sat 9:42 PM Allis Adimey a E webassign.net 0 User Mail -alli s.ucf.edu Quiz 09-Modulc 09 Lactation promotes a temporary loss of bone m Damage to grapes from bird predation is a serious problem for grape growers. An article reported on an experiment involving a bird-feeder table, time-lapse video, and artificial foods. Information was collected for two different bird species at both the experimental location and at a natural vineyard setting. Consider the following data on time (sec) spent on a single visit to the location. (Note: Assume the populations are normally distributed.) Species Blackbirds Blackbirds Silvereyes Silvereyes Location Exptl Natural Exptl Natural 13.8 9.5 49.3 38.3 SE mean 2.05 1.79 4.76 5.08 50 25 29 (a) Calculate an upper confidence bound for the true average time that blackbirds spend on a single visit at the experimental location. (Use = 0.05. Round your answer to two decimal places.) sec (b) Does it appear that true average time spent by blackbirds at the experimental location exceeds the true average time birds of this type spend at the natural location? Carry out a test of appropriate hypotheses. (Use = 0.05.) State the relevant hypotheses. (Use for blackbirds at the experimental location and 2 for blackbirds at the natural location.) Ho: 1-2 = 0 Compute the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.)Explanation / Answer
95% confidence interval for the true average time that blackbirds spend on a single visit at the experimental location : x ± t1-,df*standard error
Where = 0.05 and df= n-1 = 49
t0.95,49=1.67655 (t critical value for the right tail because only the upper limit is calculated)
The upper confidence bound: 13.8 +1.67655 *2.05 = 17.2369 ~ 17.24
H0 : µ1 - µ2 = 0
H1: µ1 - µ2 < 0
t = [ (x1 - x2) ] / SE
SE = sqrt[ (s12/n1) + (s22/n2) ] = sqrt(2.052 + 1.792) = 2.721507
t = ( 13.8-9.5)/2.7121507 = 1.585458
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
= (2.052+1.792)2 / (2.054/49)+(1.794/24) ~ 69
p-value is calculated used R software:
> pt(-1.585458,df=69)
[1] 0.05871752
Since p-value(0.0587) > 0.05, we fail to reject H0. The data suggests the true average time spend by blackbirds at the experimental location does not exceed that of blackbirds at the natural location.
95% Confidence interval for the difference between the true average time blackbirds spend at the natural location and true average time that silvereyes spend at the natural location:
µblackbord - µsilvereyes ± t0.975,df * standard error
DF = (1.792+5.082)2 / (1.794/24)+(5.084/28) ~ 34
t0.975,df =1.69092 ( t critical value for two tail )
(9.5-38.3) ± 1.69092*sqrt(1.792+5.082) = (-33.2333, -24.3667)
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.