Question

Most vertebrates have testosterone, and have behaviors that are mediated by this hormone. Testosterone can be helpful to animals, by enhancing (e.g.) territory acquisition, or harmful, by (e.g.) causing physiological stress. When male blackbirds are exposed to other male blackbirds, their testosterone levels change. In order to understand the impacts of testosterone on male blackbirds, researchers followed a few individual males, to monitor testosterone changes after encountering another male.   A pre-exposure measurement was made (in nanograms/deciliter) and a post-exposure measurement was taken, data below. Researchers will test the hypothesis that pre-exposure testosterone levels are the same as post-testosterone levels.

Before exposure	After exposure
105	85
50	74
136	145
90	86
122	148
132	148
131	150
119	142
145	151
130	113
116	118
119	99
138	150

(2 points) 1. Does this hypothesis test depend on a t distribution, a Z distribution, or a χ² distribution?

(2 points) 2. What are the df for the test you will do?

(2 points) 3a. Is this a one-tailed, or a two-tailed test?

(2 points) 3b. How would you rephrase the hypothesis test to make it the other way (for instance, if you chose a one-tailed test in 3a, how would you re-phrase my original question to make it a two-tailed test?)

(2 points) 4. What is the hypothesized difference?

(2 points) 5. Calculate your test statistic here:

(2 points) 6. Do you reject or fail to reject the null hypothesis?

(2 points) 7. How would you articulate your conclusion to my grandmother, who would not like to hear about rejecting (or failing to reject) null hypotheses, but would be interested to know about blackbirds?

!!!!!!ANSWER ALL!!!!

Answer 1

Using R software,

> x=c(105,50,136,90,122,132,131,119,145,130,116,119,138)
> y=c(85,74,145,86,148,148,150,142,151,113,118,99,150)
> n=13
> d=x-y
> dbar=mean(d);dbar1=rep(dbar,n)
> ss=sum((d-dbar1)^2)
> s=sqrt(ss/(n-1))
> t=dbar*sqrt(n)/s
> t
[1] -1.266549

> t.test(x,y,paired=T)

Paired t-test

data: x and y
t = -1.2665, df = 12, p-value = 0.2293
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-15.903144 4.210836
sample estimates:
mean of the differences
-5.846154
Conclusion:

here p-value= 0.2293 > 0.05

hence we may accept null hypothesis.