How do I solve to answer this question? I don't understand how to perform a t-test.

tion to achieve dental efficacy. The authors score, as early as week 2 of treatment. This propose a strategy of adding creatine mono- differential improvement favoring crea- hydrate (creatine) to a selective serotonin tine was maintained at weeks 4 and 8. reuptake inhibitor (SSRI) in the treatment of There were no differences between treat- patients with major depressive disorder. Such ment groups in the proportion of patients augmentation may lead to a more rapid who discontinued treatment prematurely onset of antidepressant effects and a greater (creatine: N=8, 32.0%. placebo: N=5, treatment response, potentially by restoring 18.5%%) or in the overall frequency of all brain bioenergetics at the cellular level. reported adverse events (creatine: 36 Method: Filly-two women with major events; placebo; 45 events). depressive disorder were enrolled in an Conclusions: The current study suggests B-week double-blind placebo-controlled that creatine augmentation of SSRI treat- clinical trial and randomly assigned to ment may be a promising therapeutic receive escitalopram in addition to either approach that exhibits more rapid and creatine (5 g/day. N=25) or placebo (N=27). efficacious responses in women with ma- Efficacy was primarily assessed by changes jor depressive disorder. in the Hamilton Depression Rating Scale (HAM-D) score. (Am ] Psychiatry 2012; 169:937-945) In this study, the outcome of interest was the change in women's responses on the Hamilton Depression Rating Scale bet the treatment (antidepressant + oral creatinine supplementation) and control (antidepressant + oral placebo supplement Because of randomization, the Hamilton Depression Rating Scales (HSRD) were nearly identical on average between the t and control groups. As such, the Hamilton scores at the end of the eight-week follow-up period can be compared to asses differences, if any, in the average Hamilton Score change. In Homework #3, you reported sample mean difference in Hamilton Score changes between the Treated and Control grou -4.4, with a 95% CI of (-6.5 to -2.3). 1 = 0.05 The group specific means and sd are as follows: mean differ Group mean standard deviation Treatment 17 5.4 3.0 Control 22 9.8 3.5 Suppose you plan to perform a two-sample t-test for this study. What is the null hypothesis for this test? a. The mean HDRS scores are different in the treatment and control populations option and b. The mean difference in HDRS scores between the treatment and control populations is 1. t = c. The mean HDRS scores are equal in the treatment and control populations. d. The mean difference in HDRS scores between the treatment and control populations is not 0. a 15. standa