Question

2. I stand outside the student union and conduct a poll of passing students about their favorite fast food place. There are 6 response options = McDonald's, Wendy's, Burger King, Taco Bell, Popeye's, and Arby's. I also collect the student's year in school for demographic purposes - 5 levels - First Year, Sophomore, Junior, Senior, and Grad Student. I build a two-way table of this data to prepare to conduct a Chi-square analysis. How many degrees of freedom would my analysis have?

Group of answer choices:

6

35

20

5

1

3. The expected values for a Chi-Square Test of Independence come 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 Fast Food poll and obtain an observed Chi-square value of 22.55. The chisq-test() in R also reports a p-value of 0.3114. How do I interpret this result if my alpha is 0.05?

Group of answer choices

I fail to reject the null hypothesis and therefore determine that there is a significant association between fast food preference and class year.

I fail to reject the null hypothesis and therefore conclude that there is no association between class year and favorite fast food restaurant.

I reject the null hypothesis and conclude that there is a significant association between fast food preference and class year.

I reject the null hypothesis and conclude that there is no association between the variables fast food preference and class year.

5. My student union poll included another question regarding the preference for different dog breeds. I find a significant association between preferred dog breeds and gender of the students. 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 statistical significant finding.

The Cramer's V value further proves that the result is significant.

The result was statistically significant, but not substantively 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 of Type II error and will be able to detect a difference with a large effect 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 next poll.

Group of answer choices

My new poll needs a power of 0.8 to have an effect size of 0.3.

A sample size of 230 should be sufficient for my poll.

Since I set my sample size at 233 I will achieve a power of 0.8.

I need a sample size of 233 students to obtain a result with the power I desire to have in my analysis.

7. The area under the curve of a normal distribution is equal to:

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 on the SAT. I know that the mean score of the UMD population is 1340 with a standard deviation of 222. My friend wants to know how her score of 1280 stacks up to the distribution of all scores at UMD.

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 could see how he compared to the distribution of SAT scores among UMD students. I found that his z-score was 0.33. What is the interpretation of his z-score?

Group of answer choices

He scored 3 SDs higher than the mean.

He scored better than 33% of students at UMD.

His SAT score shows he was 1/3 of an SD above the mean score.

He did worse than 33% of students at UMD.

10. We have more fitness test data from Vitor (who is male) and Manuela (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, the distribution 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 did 45.

We know that among previous applicants to the academy, the distribution 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 did 45.

We know that among previous applicants to the academy, the distribution 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

Answer 1

2. Degree of freedom = (6-1)*(5-1) = 20

Answer: 20

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3. The expected values for a Chi-Square Test of Independence come from: the marginals

Answer : the marginals

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4. p-value = 0.3114 > 0.05, fail to reject the null hypothesis

I fail to reject the null hypothesis and therefore conclude that there is no association between class year and favorite fast food restaurant.

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5. Answer :

The result was statistically significant, but not substantively significant.

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7. Answer:

a probability of 1.0

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8. z-score = (1280 - 1340)/222 = -0.27

Answer : -0.27

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9. Answer :

His SAT score shows he was 1/3 of an SD above the mean score.

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10. z-score for Manuela's result on the test = (45 - 40)/4.3 = 1.16

Answer : 1.16

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11. z-score for Vitor's result on the test = (50 - 60)/6.5 = -1.54

Answer : -1.54

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12. z score for Vitor = (50 - 60)/6.5 = -1.54

z score for Manuela = (45 - 40)/4.3 = 1.16

The z score for Manuela is closer to the mean.

Answer : Manuela