3) The \"jumping seal\" lives in Coconut Island. Because this is an endangered s

Question

3) The "jumping seal" lives in Coconut Island. Because this is an endangered species its population has been studied intensively. Scientists have determined that the body size of adult jumping seals has a parametric mean of 1.83 m (measured from the tip of the snout to the tip of the tail) and a parametric standard deviation of 0.21 m. Recently, a new population of jumping seals was discovered in a small island south of Coconut Island. A research team sampled 9 individuals from the new population and calculated a sample average of 2.11 m and a standard deviation of 0.15 m. a. Conduct a hypothesis test to examine if the small island population has a different population mean for body size compared to the population in Coconut Island b. Conduct a hypothesis test to examine if the two population have different variances.

Explanation / Answer

a.

Given that,
mean(x)=1.83
standard deviation , s.d1=0.21
number(n1)=9
y(mean)=2.11
standard deviation, s.d2 =0.15
number(n2)=9
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.306
since our test is two-tailed
reject Ho, if to < -2.306 OR if to > 2.306
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =1.83-2.11/sqrt((0.0441/9)+(0.0225/9))
to =-3.255
| to | =3.255
critical value
the value of |t | with min (n1-1, n2-1) i.e 8 d.f is 2.306
we got |to| = 3.25493 & | t | = 2.306
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -3.2549 ) = 0.012
hence value of p0.05 > 0.012,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -3.255
critical value: -2.306 , 2.306
decision: reject Ho
p-value: 0.012

we have enough evidence to support a small island has different population mean for body size compared to the population in coconut island

b.

Given that,
mean(x)=1.83
standard deviation , s.d1=0.21
number(n1)=9
y(mean)=2.11
standard deviation, s.d2 =0.15
number(n2)=9
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.12
since our test is two-tailed
reject Ho, if to < -2.12 OR if to > 2.12
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (8*0.0441 + 8*0.0225) / (18- 2 )
s^2 = 0.0333
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=1.83-2.11/sqrt((0.0333( 1 /9+ 1/9 ))
to=-0.28/0.086023
to=-3.254934
| to | =3.254934
critical value
the value of |t | with (n1+n2-2) i.e 16 d.f is 2.12
we got |to| = 3.254934 & | t | = 2.12
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != -3.2549 ) = 0.0047
hence value of p0.05 > 0.0047,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -3.254934
critical value: -2.12 , 2.12
decision: reject Ho
p-value: 0.0047

we have enough evidence to support different population variances

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