To demonstrate flavor aversion learning (that is, learning to dislike a flavor t

Question

To demonstrate flavor aversion learning (that is, learning to dislike a flavor that is associated with becoming sick), researchers gave one group of laboratory rats an injection of lithium chloride immediately following consumption of saccharin-flavored water. Lithium chloride makes rats feel sick. A second control group was not made sick after drinking the flavored water. The next day, both groups were allowed to drink saccharin-flavored water. The amounts consumed (in milliliters) for both groups during this test are given below. Amount Consumed by Rats That Were Made Sick (n = 4) Amount Consumed by Control Rats (n = 4) 5 8 1 7 3 12 5 11

(a) Test whether or not consumption of saccharin-flavored water differed between groups using a 0.05 level of significance. State the value of the test statistic. (Round your answer to three decimal places.)

State the decision to retain or reject the null hypothesis. Retain the null hypothesis. Reject the null hypothesis.

(b) Compute effect size using eta-squared (2). (Round your answer to two decimal places.) 2 =

Explanation / Answer

PART A.

Given that,
mean(x)=3.5
standard deviation , s.d1=1.9149
number(n1)=4
y(mean)=9.5
standard deviation, s.d2 =2.3805
number(n2)=4
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.447
since our test is two-tailed
reject Ho, if to < -2.447 OR if to > 2.447
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (3*3.6668 + 3*5.6668) / (8- 2 )
s^2 = 4.6668
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=3.5-9.5/sqrt((4.6668( 1 /4+ 1/4 ))
to=-6/1.5275
to=-3.9279
| to | =3.9279
critical value
the value of |t | with (n1+n2-2) i.e 6 d.f is 2.447
we got |to| = 3.9279 & | t | = 2.447
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != -3.9279 ) = 0.0057
hence value of p0.05 > 0.0057,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -3.9279
critical value: -2.447 , 2.447
decision: reject Ho
p-value: 0.0057

PART B.

Grand Mean = G / N = 3.5+9.5 / 2 = 6.5

SST = ( Xi - GrandMean)^2 = (5-6.5)^2 + (1-6.5)^2 + (3-6.5)^2 + ……..& so on = 100

SS Within = (Xi - Mean of Xi ) ^2 =,(5-3.5)^2 + (1-3.5)^2 + (3-3.5)^2 + ……..& so on = 28

SS Between = SST - SS Within = 100 - 28 = 72

eta-sqaured= SSbetweenSStotal = 72/100 = 0.72

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