Question

T is a pesticide banned in the United States for its danger to humans and animals. In an experiment on the impact of DDT, six rats were exposed to DDT poisoning and six rats were not. For each rat in the experiment, a measurement of nerve sensitivity was recorded. The researchers suspected that the mean nerve sensitivity for rats exposed to DDT is greater than that for rats not poisoned. The data follow:Poisoned rats: 12.207 16.869 25.050 22.429 8.456 20.589Unpoisoned rats: 11.074 9.686 12.064 9.351 8.182 6.642Let μ1 be the mean nerve sensitivity for rats poisoned with DDT. Let μ2 be the mean nerve sensitivity for rats not poisoned with DDT.The numerical value of the standard error of the difference in sample means is?

Answer 1

Answer:

$SE =\sqrt{\frac{s^2_{P}}{n_{P}}+\frac{s^2_{UP}}{n_{UP}}}$

And if we replace we got:

$SE =\sqrt{\frac{(6.34)^2}{6}+\frac{(1.95)^2}{6}}}=2.708$

Step-by-step explanation:

1) Data given and notation

P:[12.207 ,16.869, 25.050, 22.429, 8.456, 20.589]

UP:[11.074, 9.686 ,12.064, 9.351, 8.182, 6.642]

$\bar X_{P}=17.6$ represent the mean for the sample poisoned

$\bar X_{UP}=9.50$ represent the mean for the sample unpoisoned

$s_{P}=6.34$ represent the sample standard deviation for the sample poisoned

$s_{UP}=1.95$ represent the sample standard deviation for the sample unpoisoned

$n_{P}=6$ sample size for the group poisoned

$n_{UP}=6$ sample size for the group unpoisoned

t would represent the statistic (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for rats exposed to DDT is greater than that for rats not poisoned , the system of hypothesis would be:

Null hypothesis:$\mu_{P} \leq \mu_{UP}$

Alternative hypothesis:$\mu_{P} > \mu_{UP}$

If we analyze the size for the samples both are less than 30 and the population deviations are not given, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{P}-\bar X_{UP}}{\sqrt{\frac{s^2_{P}}{n_{P}}+\frac{s^2_{UP}}{n_{UP}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

The standard error is given by:

$SE =\sqrt{\frac{s^2_{P}}{n_{P}}+\frac{s^2_{UP}}{n_{UP}}}$

And if we replace we got:

$SE =\sqrt{\frac{(6.34)^2}{6}+\frac{(1.95)^2}{6}}}=2.708$

3) Calculate the statistic

We can replace in formula (1) the results obtained like this:

$t=\frac{17.6-9.5}{\sqrt{\frac{(6.34)^2}{6}+\frac{(1.95)^2}{6}}}}=2.99$  

4) Statistical decision

For this case we don't have a significance level provided $\alpha$, but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:

$df=n_{P}+n_{UP}-2=6+6-2=10$

Since is a unilateral test the p value would be:

$p_v =P(t_{(10)}>2.99)=0.0067$

So the p value is a very low value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to don't reject the claim that the group with DDT have a mean greater than the group for rats not poisoned.