Respuesta :
Answer:
a) [tex]t=\frac{1.51-0.87}{\sqrt{\frac{0.25^2}{6}+\frac{0.31^2}{6}}}=3.936[/tex]
"=T.INV(1-0.025,10)", and we got [tex]t_{critical}=\pm 2.28 [/tex]
Statistical decision
Since our calculated value is higher than our critical value,[tex]z_{calc}=3.936>2.28=t_{critical}[/tex], we have enough evidence to reject the null hypothesis at 5% of significance.
b) [tex] (\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}[/tex]
The degrees of freedom are given:
[tex] df = n_1 + n_2 -2 = 6+6-2 = 10[/tex]
[tex] (1.51 -0.87) - 2.28\sqrt{\frac{0.25^2}{6}+\frac{0.31^2}{6}}= 0.269[/tex]
[tex] (1.51 -0.87) + 2.28\sqrt{\frac{0.25^2}{6}+\frac{0.31^2}{6}}= 1.010[/tex]
Step-by-step explanation:
Part a
Data given and notation
[tex]\bar X_{1}=1.51[/tex] represent the mean for scent of pre ovulatory
[tex]\bar X_{2}=0.87[/tex] represent the mean for post ovolatory
[tex]s_{1}=0.25[/tex] represent the sample standard deviation for preovulatory
[tex]s_{2}=0.31[/tex] represent the sample standard deviation for postovulatory
[tex]n_{1}=6[/tex] sample size for the group preovulatory
[tex]n_{2}=6[/tex] sample size for the group postovulatory
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value
Concepts and formulas to use
We need to conduct a hypothesis in order to check if the mean's are different, the system of hypothesis would be:
H0:[tex]\mu_{1} = \mu_{2}[/tex]
H1:[tex]\mu_{1} \neq \mu_{2}[/tex]
If we analyze the size for the samples both are lower than 30, so for this case is better apply a t test to compare means, and the statistic is given by:
[tex]t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}[/tex] (1)
z-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.
Calculate the statistic
We have all in order to replace in formula (1) like this:
[tex]t=\frac{1.51-0.87}{\sqrt{\frac{0.25^2}{6}+\frac{0.31^2}{6}}}=3.936[/tex]
Find the critical value
We find the degrees of freedom:
[tex] df = n_1 + n_2 -2 = 6+6-2 = 10[/tex]
In order to find the critical value we need to take in count that we are conducting a two tailed test, so we are looking for thwo values on the t distribution with df =10 that accumulates 0.025 of the area on each tail. We can us excel or a table to find it, for example the code in Excel is:
"=T.INV(1-0.025,10)", and we got [tex]t_{critical}=\pm 2.28 [/tex]
Statistical decision
Since our calculated value is higher than our critical value,[tex]z_{calc}=3.936>2.28=t_{critical}[/tex], we have enough evidence to reject the null hypothesis at 5% of significance.
Part b
For this case the confidence interval is given by:
[tex] (\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}[/tex]
The degrees of freedom are given:
[tex] df = n_1 + n_2 -2 = 6+6-2 = 10[/tex]
[tex] (1.51 -0.87) - 2.28\sqrt{\frac{0.25^2}{6}+\frac{0.31^2}{6}}= 0.269[/tex]
[tex] (1.51 -0.87) + 2.28\sqrt{\frac{0.25^2}{6}+\frac{0.31^2}{6}}= 1.010[/tex]