Question

Rosene (1950) studied how quickly hairs on radish roots absorbed water when they were immersed.
For each of eleven radishes, she measured the rate of influx of water for a young root hair and an
old root hair on that radish. The data is given below.
Radish Old Young
A ---------0.89------ 2.13
B -------- 0.49-------1.16
C----------0.91------ 2.60
D----------0.80 ------1.58
E-------- 0.56------ 1.53
F--------- 0.79 -----1.70
G --------0.47------- 2.67
H ---------0.50 ------2.64
I ----------1.08 -------2.19
J---------- 1.65 -------2.54
K ---------1.94 -------4.46
Table 1: Radish root hair absorption data. Rates are in cubic microns per square micron per
minute.
For each pair, the \Young" number is bigger than the \Old" number, so even without a test
it's clear that young roots take in water more quickly. But how much more quickly?
1. Explain what test we should use and why: one sample test, paired samples test or
two independent samples test .
2. (a) Use a normal probability (qqnorm) plot of the di erences (old minus young)
and the sample size to explain why: We should be hesitant to do a t-test on these
di erences (old minus young);
(b) Explain why: We should not take the logs of these di erences (old minus
young.)
3. Instead of using the differences, we can look at the ratio: old divided by young.
This ratio looks to come from a much closer to normal distribution. Write R code to find a
90% con dence interval for the average value of this ratio.

Answer 1

1. We should use : paired samples test because a paired t-test is used to compare two population means. where one sample observation paired to other sample observation.

2(a): Using R:

Use a normal probability (qqnorm) plot of the difference (old minus young) and the sample size.

We can not use t.test because the difference (old minus young) does not follow normality.

2(b): We should not take the logs of these difference (old minus young.)

3: After take division of Old by Young. our data follow normality then we can use one sample t test for 90% Confidence Interval.

90% CI = (0.331, 0.480)