Question

For my PSY230 course, I started a new practice of offering extra credit for making an appointment with the Statistics and Methods Lab. The goal is to help students feel more positive toward statistics through one-on-one assistance and therefore feel more comfortable in the course.

In order to see if this new element was effective, I compared results from a statistics attitude survey given to my current class to the existing pool of past statistics attitude surveys from all my previous sections of PSY230 prior to implementing SAM Lab extra credit. In other words, the new class with the SAM Lab extra credit is my sample, which will be compared to the population of all of my past students.

According to my records, the population of all past statistics attitude surveys have a mean (μ) survey score of 71 points and standard deviation (σ) of 6 points. The new class of 64 students had a mean (M) survey score of 73 points. I conducted a hypothesis test to see if the new class would have significantly higher attitude scores from the population of past students. My review of the previous literature showed interventions such as this to have a positive effect, and so I was interested in whether there is an improvement in attitude score. The significance level for my Z test was set at α= .05.

1. Identify the dependent variable for this study (1 points)

2. State your null hypothesis and alternative hypothesis using both words and statistical notation (4 points: 1 for each notation, 1 for each written hypotheses)

Hint: The hypotheses should be directional

3. Calculate standard error (SE, which is the standard deviation of the sampling distribution) (2 points total: 1 for process/work, 1 for result)

4. Calculate the Z statistic (which indicates where our sample mean is located on the sampling distribution) (2 points total: 1 for process/work, 1 for result)

5. Specify whether the hypothesis test should be a two-tailed or a one-tailed test, and explain the rationale for the choice. (2 points total: 1 for chosen test, 1 for rationale)

6. Determine the critical value for Z (1 point)

7. Comparison of obtained Z and critical Z and then make a decision about the result of the hypothesis test: Explicitly states “reject” or “fail to reject” the null hypothesis (2 points: 1 for comparing the correct numbers and saying which one is more “extreme”; 1 for making the decision about null hypothesis.)

8. Write a 1-2 sentences to conclude the results (you can simply restate the accepted hypothesis or explain it in another way) (1 point)

1. Calculate the raw effect size and the standardized effect size for this test. (4 points total: 2 points for each measure: 1 for process/work, 1 for result)

10. In order to perform a more rigorous test, I could also set up a non-directional research hypothesis. What would the written hypotheses and the notations be?  (4 points total: 1 for each notation, 1 for each written hypotheses)

11. Using the alpha level α= .05, determine the critical Z value for the non-directional hypothesis. (1 point)

50. In this statistical test, how high does the mean statistics attitudinal survey score from the new class have to be, at least, to be considered “significantly” higher than the pool of past statistics attitudinal survey scores?

Hint: In other words, when does the calculated Z equal the critical Z? What needs to be the sample mean for that to happen?(2 points total: 1 for answer, 1 for formula/work)

Answer 1

a.

Dependent variable: Attitude Points(scores) obtained by the students​​​​​​.

b.

Let given population mean =$\mu0$​​​​​​. So, $\mu0$ =71

Let the population mean for new class =$\mu1$

Null Hypothesis(H0):

The new population mean score is not significantly higher than that of old one. μ1 $\leq$$\mu0$ (that is, $\mu1 \leq 71$)

Alternative Hypothesis(H1):

The new population mean score is significantly higher than that of old one. μ1 > $\mu0$ (that is, ​$\mu1 >71$​​​​​​) (right tailed test).

c.

Given: sample size, n =64 and std.deviation, $\sigma$ =6

Standard error, SE =$\sigma/\sqrt{n}$ =$6/\sqrt{64}$ =6/8 =0.75

(assumed that both populations have equal variances).

d.

Given: Sample mean, $\bar{X}$ =73

The test statistic, Z =$(\bar{X} - \mu0)/SE=(73-71)/0.75$ =2.67

e.

The hypothesis test should be a one-tailed test. It's because you are testing if the population mean score of new class is greater than the population mean score of old class but not testing if they are equal where the test is two-tailed.

f.

Given: Significance level, $\alpha =$0.05 and the test is one-tailed (right tailed).

At 0.05 significance level, for a right-tailed test, the critical value of Z is Z_crit =1.645

g.

The obtained or calculated Z-score (test statistic) of 2.67 is greater than the critical Z-score of 1.645 (Z>Z_crit). So, reject the null hypothesis at 0.05 significance level.

h.

Conclusion: We have a sufficient statistical evidence to claim that the population mean score of new class is greater than the population mean score of old class being 71.

i.

Raw Effect size =Mean 1 - Mean 2 =73 - 71 =2

Standardised Effect size, d =$Z/\sqrt{n} =2.67/\sqrt{64} =$0.33

(OR) d =(mean 1 - mean 2)/Std.deviation =2/6 =0.33

j.

Non-directional (two-tailed) test hypotheses:

Null Hypothesis(H0):

The new population mean score is not significantly different from that of old one. $\mu1 = \mu0\ (that is, \mu1 =71)$

Alternative Hypothesis(H1):

The new population mean score is significantly different from that of old one. $\mu1 \neq \mu0\ (that is, \mu1 \neq71)$

k.

At 0.05 significance level, for non-directional hypothesis (two-tailed test), the critical value of Z is Z_crit =$\pm$1.96

ax.

Z = Z_crit

$\implies$$(\bar{X} - \mu0)/(\sigma/\sqrt{n})$ =1.96

$\implies$$(\bar{X} - 71)/(6/\sqrt{64})$=1.96

$\implies$$\bar{X}$ =72.47

Thus, in this two-tailed statistical test, the mean statistics attitudinal survey score from the new class have to be , at least, 72.47 to be considered “significantly” higher than the pool of past statistics attitudinal survey scores.