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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.

Question #1:

What is the correct write up for this study in a results section?

A. t(40) = 5.07, p < .01

B. t(38) = .592, p > .05

C. t(19) = 5.07, p < .01

D. t(38) = 2.59, p > .05

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

Question #2:

What is your degrees of freedom for this data set?

A. 20

B.19

C.40

D.38

E.None of the above

Respuesta :

Answer:

Question #1:

What is the correct write up for this study in a results section?

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

Question #2:

D.38

Step-by-step explanation:

Data given and notation  

[tex]\bar X_{1}=7.00[/tex] represent the mean for 1  

[tex]\bar X_{2}=10.00[/tex] represent the mean for 2  

[tex]s_{1}=\sqrt{2.5}=1.58[/tex] represent the sample standard deviation for 1  

[tex]s_{2}=\sqrt{4.5}=2.12[/tex] represent the sample standard deviation for 2

[tex]n_{1}20[/tex] sample size for the group 1  

[tex]n_{2}=20[/tex] sample size for the group 2  

[tex]\alpha[/tex] Significance level provided  

t would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the population mean for a group is higher than the other group:  

Null hypothesis:[tex]\mu_{2}-\mu_{1}\leq 0[/tex]  

Alternative hypothesis:[tex]\mu_{2} - \mu_{1}> 0[/tex]  

We don't have the population standard deviation's, we can apply a t test to compare means, and the statistic is given by:  

[tex]t=\frac{(\bar X_{2}-\bar X_{1})-\Delta}{\sqrt{\frac{s^2_{2}}{n_{2}}+\frac{s^2_{1}}{n_{1}}}}[/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.  

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

[tex]t=\frac{(10-7.0)-0}{\sqrt{\frac{4.5}{20}+\frac{2.5}{20}}}}=5.07[/tex]  

P value  

We need to find first the degrees of freedom given by:  

[tex]df=n_1 +n_{2}-2=20+20-2=38[/tex]  

Since is a one right tailed test the p value would be:  

[tex]p_v =P(t_{38}>5.07)=5.33x10^{-6}[/tex]  

Comparing the p value with a significance level for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the population mean for group 2 is significantly higher than the population mean for group 2.

Question #1:

What is the correct write up for this study in a results section?

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

Question #2:

D.38

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