Conduct an independent samples t-test (a.k.a., between-subjects design) comparing two independent groups on any variable of your choice.Each sample should have at least n=10 individuals (thus, 20 participants in the study total, at minimum)What to turn in:State your research question (e.g., Is there a difference in coffee consumption between males and females?).State the null and alternative hypotheses, and then conduct the hypothesis test using an alpha level of .05.Calculate and interpret a measure of effect size (estimated Cohen’s d or r2).Calculate a 95% confidence interval.Write a conclusion statement, as would appear in a published research report.Critical analysis: Write a short paragraph (2-3 sentences) explaining why your research question is important and where your findings could be applied. This is very open ended. For example, if one were studying coffee consumption differences between males and females, they could discuss the importance of this information for Starbucks marketing purposes.Additional notes:Please submit well-organized and clearly indicated values (e.g., SS, M, etc.).

Null and alternate hypotheses: The investigator is specifically interested to test whether the brain chemical serotonin is related to aggression. For this the investigator has selected 2 independent samples of rats. one sample of rats is given placebo and the other sample of rats is given a drug (Serotonin). The number of aggression responses are recorded for each sample to test whether the drug have a significant effect on aggression or not. Denote µ1 as the population mean number of aggression responses for the sample of rats that received Serotonin drug and µ2 as the population mean number of aggression responses for the sample of rats that received Placebo (Control). Null hypothesis: H0 : μ1 = μ2 That is, there is no significant difference between the mean number of aggression responses of treatment group and control group. Alternative hypothesis: H1 : μ1 ≠ μ2 That is, there is a significant difference between the mean number of aggression responses of treatment group and control group. In order to test the hypothesis regarding the significant difference between the means of two independent samples, when the population standard deviations are unknown an independent sample t-test is appropriate.Degrees of freedom: The size of the treatment group is n1 = 15. The size of the control group is n2 = 10. Assume that the population variances are equal. The degrees of freedom is obtained as 23 from the calculations given below:Critical value: The level of significance is α = 0.05. The critical value is obtained as ±2.0687 from the calculation given below: Since, the t-distribution is symmetric, the two critical values are –t(α/2) = -2.0687 and +t(α/2) = +2.0687.Pooled sample variance: The sum of squares for treatment group is SS1 = 130, The sum of squares for control group is SS2 = 180.5. The pooled sample variance is obtained as 13.5 from the calculation given below:Standard error: The standard error of the difference of means is obtained as 1.5 from the calculation given below:Test statistic: The test statistic is obtained as 3.333 from the calculations given below: Decision rule: Denote t as test statistic value and t(α/2) as the critical value. Decision rule based on critical approach: If t ≤ –t(α/2) (or) t ≥ t(α/2), then reject the null hypothesis H0. If –t(α/2) < t < t(α/2), then fail to reject the null hypothesis H0. Conclusion: Conclusion based on critical value approach: The test statistic value is 3.333 and critical value is ±2.0687. Here, t ≥ t(α/2). That is, 3.333 (=t) > 2.0687 (= t(α/2)). By the rejection rule, reject the null hypothesis H0. Therefore, there is a significant difference between the mean number of aggression responses of treatment group and control group. Hence, at the 0.05 level of significance, there is enough evidence to conclude that the drug have a significant effect on aggression.Compute the Cohen’s d to measure the size of treatment effect: Effect size can be calculated using Cohen’s d as given below: Therefore, the cohen’s d is 1.3608.