Mathematics, 20.04.2020 19:55 jayjay5246

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)

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Answer from: crystalhoff6176

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

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)

Answer from: Quest

when functions are first introduced, you will probably have some simplistic "functions" and relations to deal with, usually being just sets of points. these won't be terribly useful or interesting functions and relations, but your text wants you to get the idea of what the domain and range of a function are. small sets of points are generally the simplest sorts of relations, so your book starts with those.

step-by-step explanation:

state the domain and range of the following relation. is the relation a function?

{(2, –3), (4, 6), (3, –1), (6, 6), (2, 3)}

the above list of points, being a relationship between certain x's and certain y's, is a relation. the domain is all the x-values, and the range is all the y-values. to give the domain and the range, i just list the values without duplication:

domain: {2, 3, 4, 6}

range: {–3, –1, 3, 6}

(it is customary to list these values in numerical order, but it is not required. sets are called "unordered lists", so you can list the numbers in any order you feel like. just don't duplicate: technically, repetitions are okay in sets, but most instructors would count off for this.)

while the given set does indeed represent a relation (because x's and y's are being related to each other), the set they gave me contains two points with the same x-value: (2, –3) and (2, 3). since x = 2 gives me two possible destinations (that is, two possible y-values), then this relation is not a function.

note that all i had to do to check whether the relation was a function was to look for duplicate x-values. if you find any duplicate x-values, then the different y-values mean that you do not have a function. remember: for a relation to be a function, each x-value has to go to one, and only one, y-value.

state the domain and range of the following relation. is the relation a function?

{(–3, 5), (–2, 5), (–1, 5), (0, 5), (1, 5), (2, 5)}

i'll just list the x-values for the domain and the y-values for the range:

Answer from: Quest

your answer is c

step-by-step explanation: