Question

QUESTION 16. A researcher wants to test if temperature has an effect on eating behavior. Under room temperature conditions, rats eat an average of μ = 10 grams of food each day. The researcher selects a random sample of rats and places them in a controlled habitat with the temperature maintained at 90°F (much hotter than room temperature). The daily food consumption for each of the rats in the sample is as follows:

Food Consumption (in grams)                 

9.1 8.3 7.6 10.2 8.4      

6.9 9.3 10.7 11.2 9.8      

8.5 12.1 8.4 7.7 9.2

A.State the hypotheses (i) formally with symbols and (ii) in words explain what the hypotheses predict in terms of the independent and dependent variables.

B.Sketch the distribution and locate the critical region.

C.Compute the test statistic.

D.Make a decision THREE WAYS: (i) Make a decision regarding H0 (ii) State the results formally, using symbols and statistical jargon (iii) State the results in jargon-free, regular English -- what would you tell your grandmother?

E.Calculate the effect size (Cohen's d)

F.Construct a 95% Confidence Interval

Answer 1

A) NULL HYPOTHESIS H0:

ALTERANTIVE HYPOTHESIS Ha:

NULL HYPOTHESIS H0: Rats eat an average of 10 grams of food each day.

ALTERNATIVE HYPOTHESIS Ha: Rats do not eat an average of 10 grams of food each day.

INDEPENDENT VARIABLE : Room temperature

DEPENDENT VARIABLE: EATING HABBITS OF RABITS.

C Test statistic:

D: Based on the information provided, the significance level is α=0.05, and the critical value for a two-tailed test is tc​=2.145.

The rejection region for this two-tailed test is R={t:∣t∣>2.145}.

Since it is observed that ∣t∣=2.275>tc​=2.145, it is then concluded that the null hypothesis is rejected.

ii) Using the P-value approach: The p-value is p=0.0392, and since p=0.0392<0.05, it is concluded that the null hypothesis is rejected.

iii) We can conclude that there is an effect of temperature on eating habbit of Rats

E) Effect size: =

F) sample Mean = 9.16
t critical = 2.14
sM = √(1.432/15) = 0.37
μ = M ± t(sM)
μ = 9.16 ± 2.14*0.37
μ = 9.16 ± 0.7918

95% CI [8.3682, 9.9518].

You can be 95% confident that the population mean (μ) falls between 8.3682 and 9.9518.