Question

In: Statistics and Probability

Groups of dolphins were observed off the coast of Iceland near Keflavik in 1998. The data...

Groups of dolphins were observed off the coast of Iceland near Keflavik in 1998. The data in the file dolphin_dat on the course website give the time of the day and the main activity of the group, whether travelling quickly, feeding, or socializing. The dolphin groups varied in size. Usually feeding or socializing groups were larger than travelling groups.

Source of data: Marianne Rasmussen, Department of Biology, University of Southern Denmark, Odense, Denmark.

Activity        Period  Groups
Travel  Morning  6
Feed    Morning 28
Social  Morning 38
Travel  Noon     6
Feed    Noon     4
Social  Noon     5
Travel  Afternoon       14
Feed    Afternoon        0
Social  Afternoon        9
Travel  Evening 13
Feed    Evening 56
Social  Evening 10

A) Find a 90% confidence interval for the difference between morning and evening for the proportion of dolphins feeding assuming that the data is a result of two simple random samples and that the samples for both time periods are independent of each other.

B) Does there appear to be a significant difference in the proportion of dolphins engaged in feeding between morning and evening? Conduct the appropriate test of significance and discuss your results.

Solutions

Expert Solution

A.
TRADITIONAL METHOD
given that,
sample one, x1 =6, n1 =28, p1= x1/n1=0.214
sample two, x2 =6, n2 =38, p2= x2/n2=0.158
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.214*0.786/28) +(0.158 * 0.842/38))
=0.098
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.64
margin of error = 1.64 * 0.098
=0.16
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.214-0.158) ±0.16]
= [ -0.104 , 0.216]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample one, x1 =6, n1 =28, p1= x1/n1=0.214
sample two, x2 =6, n2 =38, p2= x2/n2=0.158
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.214-0.158) ± 1.64 * 0.098]
= [ -0.104 , 0.216 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ -0.104 , 0.216] contains the difference between
true population proportion P1-P2
2) if a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the difference between
true population mean P1-P2

B.
Given that,
sample one, x1 =6, n1 =28, p1= x1/n1=0.214
sample two, x2 =6, n2 =38, p2= x2/n2=0.158
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
reject Ho, if zo < -1.645 OR if zo > 1.645
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.214-0.158)/sqrt((0.182*0.818(1/28+1/38))
zo =0.587
| zo | =0.587
critical value
the value of |z α| at los 0.1% is 1.645
we got |zo| =0.587 & | z α | =1.645
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.587 ) = 0.5572
hence value of p0.1 < 0.5572,here we do not reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 0.587
critical value: -1.645 , 1.645
decision: do not reject Ho
p-value: 0.5572
we do not have enough evidence to support the claim that significant difference in the proportion of dolphins engaged in feeding between morning and evening


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