Mathematics, 06.12.2019 23:31 thawkins79

11112
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

Answers: 2

Show answers

Answers

Answer from: christylivingsowzxa2

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

$\bar X_{1}=7.00$ represent the mean for 1

$\bar X_{2}=10.00$ represent the mean for 2

$s_{1}=\sqrt{2.5}=1.58$ represent the sample standard deviation for 1

$s_{2}=\sqrt{4.5}=2.12$ represent the sample standard deviation for 2

$n_{1}20$ sample size for the group 1

$n_{2}=20$ sample size for the group 2

$\alpha$ 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: $\mu_{2}-\mu_{1}\leq 0$

Alternative hypothesis: $\mu_{2} - \mu_{1} 0$

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

$t=\frac{(\bar X_{2}-\bar X_{1})-\Delta}{\sqrt{\frac{s^2_{2}}{n_{2}}+\frac{s^2_{1}}{n_{1}}}}$ (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:

$t=\frac{(10-7.0)-0}{\sqrt{\frac{4.5}{20}+\frac{2.5}{20}}}}=5.07$

P value

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

$df=n_1 +n_{2}-2=20+20-2=38$

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

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

Comparing the p value with a significance level for example $\alpha=0.05$ we see that p_v 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

Answer from: Quest

the law of definite proportions states samples of a compound will always contain the same proportion of elements by mass. in other words, a given compound will always contain the same elements in the same proportions by mass.the law of multiple proportions states when two elements combine with each other to form more than one compound, then the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers.

Answer from: Quest

the answer is y = 11.54

hope this : )