Question

A researcher is interested in seeing if negative political ads against an opponent (group one) are more effective than positive ads for the original candidate (group two). If the mean for group one is 7.00, the mean for group two is 10.00, the n for group one is 20, the n for group two is 20, the variance for group one is 2.50, and the variance for group two is 4.5.
What is the correct write up for this study in a results section?A. t(40) = 5.07, p < .01B. t(38) = .592, p > .05C. t(19) = 5.07, p < .01D. t(38) = 2.59, p > .05E. None of the above (it should be t(38) = 5.07, p < .01)

Answer 1

Answer:

$t=\frac{10-7}{\sqrt{\frac{1.581^2}{20}+\frac{2.121^2}{20}}}}=5.07$  

The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=20+20-2=38$

Since is a one side test the p value would be:

$p_v =P(t_{(38)}>5.07)=5.33x10^{-6}$

And the best option for this case would be:

None of the above (it should be t(38) = 5.07, p < .01)

Step-by-step explanation:

Data given and notation

$\bar X_{1}=7$ represent the mean for the sample 1

$\bar X_{2}=10$ represent the mean for the sample 2

$s_{1}=\sqrt{2.5}= 1.581$ represent the sample standard deviation for the sample 1

$s_{2}=\sqrt{4.5}= 2.121$ represent the sample standard deviation for the sample 2

$n_{1}=20$ sample size selected for 1

$n_{2}=20$ sample size selected for 2

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the group 1 is higher than the mean for group 2, the system of hypothesis would be:

Null hypothesis:$\mu_{2} \leq \mu_{1}$

Alternative hypothesis:$\mu_{2} > \mu_{1}$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{2}-\bar X_{1}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (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".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{10-7}{\sqrt{\frac{1.581^2}{20}+\frac{2.121^2}{20}}}}=5.07$  

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=20+20-2=38$

Since is a one side test the p value would be:

$p_v =P(t_{(38)}>5.07)=5.33x10^{-6}$

And the best option for this case would be:

None of the above (it should be t(38) = 5.07, p < .01)