Question

Mercury pollution is a serious ecological problem. It typically becomes dangerous once it falls into large bodies of water. At this point microorganisms change it into methylmercury (CH₃²⁰³). The fish consume these microorganisms which makes them contaminated and hence anyone eating those fish are at risk.

Because of this, investigators are interested in research around mercury poisoning. In particular they want to investigate the methlymercury metabolism and whether it proceeds at a different rate for women than for men. The table below captures the half-life (in days) of an oral administration of protein-bound methlymercury among six females and nine males. Round all the numbers to 2 decimal places.

1. (2 points) Create and interpret a 95% confidence interval for the average methylmercury half-life in males fish. Note: Do the calculations one time using R and the other time by hand.
2. (2 points) Create and interpret a 95% confidence interval for the average methylmercury half-life in females. Note: Do the calculations one time using R and the other time by hand.
3. (2 points) In comparing the confidence intervals created in (a) and (b) does there appear to be a difference in the average methylmercury half-life between males and females? Explain.

MUST SHOW HOW TO DO IT IN R AND BY HAND

	Methlymercury half-lives (in days)
	Females	Males
	52	72
	69	88
	73	87
	88	74
	87	78
	56	70
		78
		93
		74
Mean	70.83	79.33
Standard Deviation	15.09	8.08

Answer 1

using R

female <- c(52,69,73,88,87,56)
male <- c(72,88,87,74,78,70,78,93,74)

#a
t.test(male)
#b
t.test(female)

a)

95% confidence interval for male is (73.1242,85.5424)

b)

95% confidence interval for female is (54.9953,86.6135)

c)

since confidence interval overlap with each other, there does not appear to be a difference in the average methylmercury half-life between males and females

by hand

> mean(male)
[1] 79.33333
> sd(male)
[1] 8.077747
> mean(female)
[1] 70.83333
> sd(female)
[1] 15.09194

(Xbar +- t * sd/sqrt(n))
for female

n = 6

alpha = 0.05

df =n-1 = 5

t= 2.570582   {using table or in R qt(0.975,5) }

Xbar = 79.33

sd = 8.08  

hence
(70.8333 - 2.570582 * 15.09/sqrt(6))

= ( 54.9973 , 86.6693)

for male

n = 9

df =n-1 = 8

t =qt(0.975,8) = 2.306

(79.33 +- 2.306 * 8.08/sqrt(9))

= ( 73.119 , 85.5408 )