Question

A team of psychologists hypothesize that feeling sick is associated with liking a flavor. They design a study in which rats are randomly split into two group. In one group the rats are given an injection of lithium chloride immediately following consumption of saccharin-flavored water. Lithium chloride makes rats feel sick. The rats in the other group were not given Lithium chloride after drinking the flavored water. The next day, all rats could drink saccharin-flavored water. The amount of saccharin-flavored water consumed (in milliliters) for all rat for the next day are given below. What can the psychologists conclude with α= 0.01?

lithium	no lithium
4 9 6 7 9 7 5 6 7	4 4 1 3 2 6 8 6 5

a) What is the appropriate test statistic?
---Select--- na z-test One-Sample t-test Independent-Samples t-test Related-Samples t-test

b)
Condition 1:
---Select--- disliking a flavor no lithium chloride saccharin-flavored water lithium chloride the rats
Condition 2:
---Select--- disliking a flavor no lithium chloride saccharin-flavored water lithium chloride the rats

c) Compute the appropriate test statistic(s) to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
p-value =  ; Decision:  ---Select--- Reject H0 Fail to reject H0

d) Using the SPSS results, compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and/or select "na" below.
d =  ;   ---Select--- na trivial effect small effect medium effect large effect
r² =  ;   ---Select--- na trivial effect small effect medium effect large effect

e) Make an interpretation based on the results.

Rats that were given lithium chloride drank significantly more saccharin-flavored water than those that were not given lithium chloride.Rats that were given lithium chloride drank significantly less saccharin-flavored water than those that were not given lithium chloride.    There is no significant difference in saccharin-flavored water drank between rats that were give

Answer 1

A.

Independent-Samples t-test

B.

I could not understand the options as they are not properly formatted. I think you can interpret them by looking at the whole solution I am providing below and If you can't, comment the options in the proper format I will help you.

C.

n₁ = 9

₁ = 6.667

s₁ = 1.658

n₂ = 9

₂ = 4.333

s₂ = 2.179

Let's conduct a hypothesis for equality of variances of the two population:

H₀: ₁² = ₂²

H₁: ₁²    ₂²

Formula Used:

F = 0.579

p-value for F_8,8= 0.77 > 0.01 i.e H₀ can't be rejected and hence we can say that variances of the two populations' are equal.

Now,

Let's test the hypothesis that the mean saccharin-flavoured water consumed in one group is different from the other group or not.

H₀: ₁ =  ₂

H₁: _1  2

Formula Used:

Assumptions: Populations are normally distributed and the populations' variances are equal (proved above)

t = 2.557

df = 9 + 9 - 2

= 16

p-value = 0.0211 > 0.01 i.e. We fail to reject H₀.

D.

d = (₂ - ₁) ⁄ SD_pooled

SD_pooled = √((SD₁² + SD₂²) ⁄ 2) = 1.936105

d = 1.2055

Large effect

r² = d²/d²+4

= 0.2665

Large

E.

There is no significant difference in saccharin-flavoured water drank between rats that were given lithium chloride and those who were not.

Please upvote if you have liked my answers, would be of great help. Thank you.