Question

In: Statistics and Probability

To demonstrate flavor aversion learning (that is, learning to dislike a flavor that is associated with...

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)
1 12
5 8
5 7
3 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 (η2). (Round your answer to two decimal places.)
η2 = ___________

Solutions

Expert Solution

For Treatment: = 3.5, s1 = 1.9149, n1 = 4

For Control: = 9.5, s2 = 2.3805, n2 = 4

Since s1/s2 = 1.9149 / 2.3805 = 0.8 (it lies between 0.5 and 2) we used the pooled variance.

The degrees of freedom used is n1 + n2 - 2 = 4 + 4 -2 = 6 (since pooled variance is used)

_____________________________________

The Hypothesis:

This is a Two tailed test.

______________________________________

The Test Statistic:We use the students t test as population standard deviations are unknown.

The p Value:    The p value (2 Tail) for t = -4.567, df = 6, is; p value = 0.0014

The Decision Rule: If tobservedis >tcriticalor If tobserved is < -tcritical, Then Reject H0.

Also If the P value is < (0.05), Then Reject H0

___________________________________________________

The Decision: Since P value (0.0014) is < (0.05), We Retain the null hypothesis.

_________________________________________________

The Effect Size = SS Between / SS total

Grand Mean = (3.5 + 9.5) / 2 = 6.5

SS between = SUM[n * (Mean - Grand Mean)2] = 4 * (3.5 - 6.5)2 + 4 * (9.5 - 6.5)2 = 72

SS error = SUM [(n - 1) * Variance] = 3 * 1.91492 + 3 * 2.38052 = 28

SS Total = 72 + 28 = 100

Therefore = 72 / 100 = 0.72

___________________________________________________

_________________________________________________


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