Respuesta :
Answer:
[tex]SE =\sqrt{\frac{s^2_{P}}{n_{P}}+\frac{s^2_{UP}}{n_{UP}}}[/tex]
And if we replace we got:
[tex]SE =\sqrt{\frac{(6.34)^2}{6}+\frac{(1.95)^2}{6}}}=2.708[/tex]
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]
[tex]\bar X_{P}=17.6[/tex] represent the mean for the sample poisoned
[tex]\bar X_{UP}=9.50[/tex] represent the mean for the sample unpoisoned
[tex]s_{P}=6.34[/tex] represent the sample standard deviation for the sample poisoned
[tex]s_{UP}=1.95[/tex] represent the sample standard deviation for the sample unpoisoned
[tex]n_{P}=6[/tex] sample size for the group poisoned
[tex]n_{UP}=6[/tex] 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:[tex]\mu_{P} \leq \mu_{UP}[/tex]
Alternative hypothesis:[tex]\mu_{P} > \mu_{UP}[/tex]
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:
[tex]t=\frac{\bar X_{P}-\bar X_{UP}}{\sqrt{\frac{s^2_{P}}{n_{P}}+\frac{s^2_{UP}}{n_{UP}}}}[/tex] (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:
[tex]SE =\sqrt{\frac{s^2_{P}}{n_{P}}+\frac{s^2_{UP}}{n_{UP}}}[/tex]
And if we replace we got:
[tex]SE =\sqrt{\frac{(6.34)^2}{6}+\frac{(1.95)^2}{6}}}=2.708[/tex]
3) Calculate the statistic
We can replace in formula (1) the results obtained like this:
[tex]t=\frac{17.6-9.5}{\sqrt{\frac{(6.34)^2}{6}+\frac{(1.95)^2}{6}}}}=2.99[/tex]
4) Statistical decision
For this case we don't have a significance level provided [tex]\alpha[/tex], but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:
[tex]df=n_{P}+n_{UP}-2=6+6-2=10[/tex]
Since is a unilateral test the p value would be:
[tex]p_v =P(t_{(10)}>2.99)=0.0067[/tex]
So the p value is a very low value and using any significance level for example [tex]\alpha=0.05, 0,1,0.15[/tex] always [tex]p_v<\alpha[/tex] 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.
