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Conducting a One-Sample T-Test and Z-Test for Analyzing Experimental Data

Objectives

1.You will conduct a one-sample t-test for Writing for the null hypothesis in the population H0: μ = 50, with α = .05, two tailed. Answer the following questions (report to two decimal places)

a.Obtain the t-statistic, by hand computation.

b.Obtain the critical value from the t table.

c. Obtain the 95% confidence interval around , by hand computation.

d. Interpret the confidence interval, according to its definition. (You are not asked to make a decision on H0 here yet.)

e. Obtain the p-value from SPSS output.

f.Interpret the p-value, according to its definition. (You are not asked to make a decision on H0 here yet.)

g.Based on each of the three evidences: t-statistic, CI, and p-value, would you reject H0? Why?

h.Based on each of the three evidences: t-statistic, CI, and p-value , what can you conclude about mean Writing scores for the population represented by this sample?

2.Mathew believes that Power Physics will increase the players’ creativity, but doesn’t think that there is any mechanism by which it could decrease a players creativity. What are the null and alternative hypothesis for the one-sample z-test in symbols and in words?

The null alternative hypothesis is that the mean posttest score of Creativity is greater than 7 in the population. The alternative hypothesis is that the mean posttest score of Creativity is equal to 7 in the population.

The null hypothesis is that the mean posttest score of Creativity is equal to 7 in the population. The alternative hypothesis is that the mean posttest score of Creativity is greater than 7 in the population.

The null alternative hypothesis is that the mean posttest score of Creativity is greater than 7 in the population. The alternative hypothesis is that the mean posttest score of Creativity is equal to 7 in the population.

The null hypothesis is that the mean posttest score of Creativity is equal to 7 in the population. The alternative hypothesis is that the mean posttest score of Creativity is greater than 7 in the population.

3.What is the standard error for the mean posttest creativity score of the pilot sample?

.11

.17

.50

2.3

4.Suppose the mean of the pilot sample follows a normal distribution with a mean of 7 (from the null hypothesis). What is the probability of observing a mean posttest score as big or bigger than the one that Mathew saw in the pilot test (i.e., what is the p-value)?

.10

.12

.38

.40

5.Mathew is pleased to see a positive difference in the pilot sample, but Russell is cautious about committing a Type I error.  What is the Type I error, and what is its probability of occurrence?

The Type I error is the probability of failing to reject a false hypothesis. Its probability of occurrence is equal to the p-value.

The Type I error is the probability of failing to reject a false hypothesis. Its probability of occurrence is equal to the alpha level.

The Type I error is the probability of rejecting a correct null hypothesis. Its probability of occurrence is equal to the p-value.

The Type I error is the probability of rejecting a correct null hypothesis. Its probability of occurrence is equal to the alpha level.

6..Based on results from 4 and 5, at the alpha level of .05, what should Mathew conclude about the mean posttest score based on the pilot data?

We fail to reject the null hypothesis and conclude that the mean posttest score for the population represented by the pilot data is significantly different from 7.

We fail to reject the null hypothesis and conclude that the mean posttest score for the population represented by the pilot data is not significantly different from 7.

We reject the null hypothesis and conclude that the mean posttest score for the population represented by the pilot data is significantly different from 7.

We reject the null hypothesis and conclude that the mean posttest score for the population represented by the pilot data is not significantly different from 7.

7.Mathew is planning to run the full-scale study with 350 students. Assume that the full-scale sample follows a normal distribution with population mean 7 and standard deviation 2.3. Mathew expects that the mean posttest score on the full sample will be about 8.5. What is the z-score corresponding to the full sample mean?

.65

3.30

12.20

228.26

8..Russell says he is willing to accept a 5% chance of making a Type I error in this study.  What is the z critical value for the population mean for rejecting the null hypothesis (Caution: review your hypothesis: is the test one- or two-tailed)?

1.64

1.73

1.96

2.09

9.What should Mathew conclude about the mean posttest score based on the full sample data? Is the conclusion based on the full sample data different from the conclusion based on the pilot data?

Mathew should conclude that the mean posttest score for the population represented by the full sample data is not significantly different from 7. The conclusions based on the two samples differ.

Mathew should conclude that the mean posttest score for the population represented by the full sample data is not significantly different from 7. The conclusions based on the two samples are the same.

Mathew should conclude that the mean posttest score for the population represented by the full sample data is significantly different from 7. The conclusions based on the two samples differ.

Mathew should conclude that the mean posttest score for the population represented by the full sample data is significantly different from 7. The conclusions based on the two samples are the same.

10.What is the effect size of posttest mean on the full sample?

65

3.30

12.20

228.26

11.Mathew determines that the “power” for detecting a posttest score of the size Mathew expects with his experiment is .9999.  Explain what the power means in words.

9999 is the probability that Mathew concludes the mean posttest score is greater than 7 when the actual mean gain score is greater than 7.

9999 is the probability that Mathew fails to conclude the mean posttest score is greater than 7 when the actual mean gain score is greater than 7.

9999 is the probability that Mathew concludes the mean posttest score is greater than 7 when the actual mean gain score is not greater than 7.

9999 is the probability that Mathew fails to conclude the mean posttest score is greater than 7 when the actual mean gain score is not greater than 7