2. I stand outside the student union and conduct a poll ofpassing students about their favorite fast food place. There are 6response options = McDonald's, Wendy's, Burger King, Taco Bell,Popeye's, and Arby's. I also collect the student's year in schoolfor demographic purposes - 5 levels - First Year, Sophomore,Junior, Senior, and Grad Student. I build a two-way table of thisdata to prepare to conduct a Chi-square analysis. How many degreesof freedom would my analysis have?
Group of answer choices:
6
35
20
5
1
3. The expected values for a Chi-Square Test of Independencecome from:
Group of answer choices:
the population values
the marginals
a chi-square table
4. I conduct the Chi-Square test of independence for my FastFood poll and obtain an observed Chi-square value of 22.55. Thechisq-test() in R also reports a p-value of 0.3114. How do Iinterpret this result if my alpha is 0.05?
Group of answer choices
I fail to reject the null hypothesis and thereforedetermine that there is a significant association between fast foodpreference and class year.
I fail to reject the null hypothesis and thereforeconclude that there is no association between class year andfavorite fast food restaurant.
I reject the null hypothesis and conclude that there isa significant association between fast food preference and classyear.
I reject the null hypothesis and conclude that there isno association between the variables fast food preference and classyear.
5. My student union poll included another question regarding thepreference for different dog breeds. I find a significantassociation between preferred dog breeds and gender of thestudents. I calculate a Cramer's V test and get a result of 0.05.What conclusion would I make about this result?
Group of answer choices:
The Cramer's V score disproves our statisticalsignificant finding.
The Cramer's V value further proves that the result issignificant.
The result was statistically significant, but notsubstantively significant.
6. I decide to conduct another poll outside the student union,and I want to ensure that my poll will have a low probability ofType II error and will be able to detect a difference with a largeeffect size. I run the following code:
pwr.chisq.test(w = 0.3, N=NULL, df = 20, sig.level = 0.05, power= 0.8)
I get the following output in R:
Chi squared power calculation
w = 0.3
N = 232.8977
df = 20
sig.level = 0.05
power = 0.8
What does this output tell me about how I need to design my nextpoll.
Group of answer choices
My new poll needs a power of 0.8 to have an effect sizeof 0.3.
A sample size of 230 should be sufficient for mypoll.
Since I set my sample size at 233 I will achieve a powerof 0.8.
I need a sample size of 233 students to obtain a resultwith the power I desire to have in my analysis.
7. The area under the curve of a normal distribution is equalto:
Group of answer choices
the mean of the distribution
a probability of 1.0
the standard deviation of the distribution.
the z-score
8. In my student union poll I asked students what they scored onthe SAT. I know that the mean score of the UMD population is 1340with a standard deviation of 222. My friend wants to know how herscore of 1280 stacks up to the distribution of all scores atUMD.
What is her z-score?
Group of answer choices
-0.53
-1
0.27
-0.27
9. Another friend asked me to calculate his z-score so he couldsee how he compared to the distribution of SAT scores among UMDstudents. I found that his z-score was 0.33. What is theinterpretation of his z-score?
Group of answer choices
He scored 3 SDs higher than the mean.
He scored better than 33% of students atUMD.
His SAT score shows he was 1/3 of an SD above the meanscore.
He did worse than 33% of students at UMD.
10. We have more fitness test data from Vitor (who is male) andManuela (who is female), who are applying to a military academy.Vitor did 50 push-ups in a minute, while Manuela only did 45.
We know that among previous applicants to the academy, thedistribution of number of sit-ups is as follows:
Males have a mean of 60 and a standard deviation of 6.5.
Females have a mean of 40 and a standard deviation of 4.3.
What is the z-score for Manuela's result on the test?
Group of answer choices
1.16
1.69
-0.92
3.49
11. Vitor did 50 push-ups in a minute, while Manuela only did45.
We know that among previous applicants to the academy, thedistribution of number of sit-ups is as follows:
Males have a mean of 60 and a standard deviation of 6.5.
Females have a mean of 40 and a standard deviation of 4.3.
What is the z-score for Vitor's result on the test?
Group of answer choices:
0
2.33
-1.16
-1.54
12. Vitor did 50 push-ups in a minute, while Manuela only did45.
We know that among previous applicants to the academy, thedistribution of number of sit-ups is as follows:
Males have a mean of 60 and a standard deviation of 6.5.
Females have a mean of 40 and a standard deviation of 4.3.
Relative to their gender, who did better on the push-up test,Vitor or Manuela?
Group of answer choices:
Manuela
Vitor
