in tilapia, an important freshwater food fish from Africa, the males actively court females. They have more incentive to court a female who has already laid all of her eggs, but can they tell the difference? an experiment was done to measure the male tilapia's response to the smell of female fish. Water containing feces from females that were either pre-ovulatory (they still had eggs) or post-ovulatory (they had already laid their eggs) was washed over the gills of males hooked up to an electro-olfactogram machine which measured when the senses of the males were excited. The amplitude of the electro-olfactogram was used as a measure of the excitability of the males in the two different circumstances. Six males were exposed to the scent of pre-ovulatory females; their readings average 1.51 with a standard deviation of .25. Six different males were exposed to post-ovulatory females; their average readings of 0.87 with standard deviation is .31. Assume that the electro-olfactogram readings were approximately normally distributed within the groups.

(A) test for a difference in the excitability of the males with exposure to these two types of females
(B) what is the estimated average difference in electro-olfactogram readings between the two groups? What is the 95% confidnece limit for the difference between population means?

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]