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) 1 7 5 12 4 11 3 9 (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.) Incorrect: Your answer is incorrect. State the decision to retain or reject the null hypothesis. Retain the null hypothesis. Reject the null hypothesis. Correct: Your answer is correct. (b) Compute effect size using eta-squared (η2). (Round your answer to two decimal places.) η2 = Incorrect: Your answer is incorrect.

Answer 1

	Amount Consumed by Rats That Were Made Sick (n = 4)	Amount Consumed by Control Rats (n = 4)
	1	4
	7	11
	5	3
	12	9
Mean	6.25	6.75
SD	4.573474245	3.862210075

Testing for Equality of Variances AF-test is used to test for the equality of variances. The following F-ratio is obtained: 4.572 F 1.402 3.862 The critical values are F 0.065 and Fu 15.439, and since F 1.402, then the null hypothesis of equal variances is not rejected

The provided sample means are shown below: X1 6.25 X2 =6.75 Also, the provided sample standard deviations are 814.57 82 = 3.86 4 4 and n2 and the sample sizes are n1 = (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: 2 Ha: 2 This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used

(2) Rejection Region Based on the information provided, the significance level is a 0.05, and the degrees of freedom are df 6. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal: 6 Hence, it is found that the critical value for this two-tailed test is te = 2.447, for a = 0.05 and df The rejection region for this two-tailed test is R {t: \t> 2.447} (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: X- X2 t = (21-1)s+(n2-1) 1 1) nn2-2 n2 6.25 6.75 0.167 (4-1)4.572+(4-1)3.86 4+4-2 () (4) Decision about the null hypothesis Since it is observed that t= 0.167 < te=2.447, it is then concluded that the null hypothesis is not rejected Using the P-value approach: The p-value is p 0.8727, and since p = 0.8727 > 0.05, it is concluded that the null hypothesis is not rejected

(5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean 1 is different than H2, at the 0.05 significance level

Effect size = Mean diff / Sp

s²_p = (SS₁ + SS₂) / (df₁ + df₁)

s²_p = ((df₁)(s²₁) + (df₂)(s²₂)) / (df₁ + df₂) = 35.78 / 6 = 17.892

Sp =  4.2299

Effect size = (6.75-6.25)/ 4.2299 = 0.1182