Writink Services ## Hypothesis, Effect Size, and Power

#### Problem Set 3.1: Sampling Distribution of the Mean Exercise

Criterion: Interpret population mean and variance.

Suppose a researcher wants to learn more about the mean attention span of individuals in some hypothetical population. The researcher cites that the attention span (the time in minutes attending to some task) in this population is normally distributed with the following characteristics: 20 36. Based on the parameters given in this example, answer the following questions:

2. #### Sketch the distribution of this population. Make sure you draw the shape of the distribution and label the mean plus and minus three standard deviations.

Standard deviation of population = σ = 36 = 6

### PSYC FPX4700 Assessment 3 Hypothesis, Effect Size, Power, and t Tests

μ-3σ = 20 – (3×6) = 2

μ-2σ = 20 – (2×6) = 8

μ-σ = 20 – (6) = 14

μ+σ = 20 + (6) = 26

μ+2σ = 20 – (2×6) = 32

μ+3σ = 20 – (3×6) = 38

#### Problem Set 3.2: Effect Size and Power

Criterion: Explain effect size and power.

Instructions: Read each of the following three scenarios and answer the questions.

Two researchers make a test concerning the effectiveness of a drug use treatment. Researcher A determines that the effect size in the population of males is d = 0.36; Researcher B determines that the effect size in the population of females is d = 0.20. All other things being equal, which researcher has more power to detect an effect? Explain.

Researcher A is more likely to have higher power to detect the effect because of the larger discrepancy between the null and alternative means. This is attributable to the large effect size, which results in a more powerful test.

Two researchers make a test concerning the levels of marital satisfaction among military families. Researcher A collects a sample of 22 married couples (n = 22); Researcher B collects a sample of 40 married couples (n = 40). All other things being equal, which researcher has more power to detect an effect? Explain.

Researcher B’s ability to detect an effect is higher because they have a larger sample size, and power is positively associated with sample size.

Two researchers make a test concerning standardized exam performance among senior high school students in one of two local communities. Researcher A tests performance from the population in the northern community, where the standard deviation of test scores is 110 (); Researcher B tests the performance from the population in the southern community, where the standard deviation of test scores is 60 (). All other things being equal, which researcher has more power to detect an effect? Explain. _

Researcher B’s ability to detect an effect is higher because they have a larger sample size, and power is positively associated with sample size.

__

#### Problem Set 3.3: Hypothesis, Direction, and Population Mean

Criterion: Explain the relationship between hypothesis, tests, and population mean.

Directional versus nondirectional hypothesis testing. Cho and Abe (2013) provided a commentary on the appropriate use of one-tailed and two-tailed tests in behavioral research. In their discussion, they outlined the following hypothetical null and alternative hypotheses to test a research hypothesis that males self-disclose more than females:

• H0: µmales − µfemales ≤ 0
• H1: µmales − µfemales > 0

What type of test is set up with these hypotheses, a directional test or a nondirectional test? ___

The hypothesis is formulated as a directional test because the alternative hypothesis (H1) is expressed in a manner that permits only one direction of effect, which is that males engage in greater self-disclosure than females.

Do these hypotheses encompass all possibilities for the population mean? Explain. Absolutely, the offered hypotheses cover all potential outcomes for the population mean, as male and female means can be equal, male means can be lower than female means, and male means can be higher than female means.

Problem Set 3.4: Hypothesis, Direction, and Population Mean

Criterion: Explain decisions for p values.

Instructions: Read the following and respond to the prompt.

The value of a p-value. In a critical commentary on the use of significance testing, Lambdin (2012) explained, “If a p < .05 result is ‘significant,’ then a p = .067 result is not ‘marginally significant’” (p. 76).

Explain what the author is referring to in terms of the two decisions that a researcher can make. ___________

We will reject the null hypothesis if the p-value is less than 0.05, and we will fail to reject the null hypothesis and reject the alternative hypothesis if the p-value is greater than 0.05.

## t-Tests

#### Problem Set 3.5: One-Sample t test in JASP

Criterion: Calculate a one-sample t-test in JASP.

Data: Use the dataset minutes reading. jasp. The dataset minutes reading.jasp is a sample of the reading times of Riverbend City online news readers (in minutes).  Riverbend City online news advertises that it is read longer than the national news. The mean for national news is 8 minutes per week.

Instructions: Complete the steps below.

2. In the Toolbar, click T-tests. In the menu that appears, under Classical, select One-sample t-test.
3. Select Time and then click Arrow to send it over to the Variables box.
4. Make sure the box is checked for Students. In the box labeled Test value, enter 8. Hit enter.
5. Copy and paste the output into the Word document.

## One Sample T-Test

One Sample T-Test

t

df

p

Time

-0.493

14

0.629

Note.  For the Student t-test, the alternative hypothesis specifies that the mean is different from 8.

Note.  Student’s t-test.

### PSYC FPX4700 Assessment 3 Hypothesis, Effect Size, Power, and t Tests

1. State the non-directional hypothesis.

The mean of the Riverbend City online news viewing length differs from the national news mean of 8 minutes per week in the population.

1. State the critical t for a = .05 (two tails).

Critical t for a = .05 (two tails) with df= 14 is ±2.145

1. Answer the following: Is the length of viewing for Riverbend City online news significantly different than the population mean? Explain

Based on the t-value of -0.493 and the p-value of 0.629, we fail to reject the null hypothesis that the population mean of the length of viewing for Riverbend City online news is equivalent to the national news mean of 8 minutes per week. The p-value is greater than the predetermined level of significance (α = 0.05). Therefore, we cannot claim that the length of viewing for Riverbend City online news is significantly distinct from the population mean.

Note: You will continue to use this dataset for the next problem

#### Problem Set 3.6: Confidence Intervals

Criterion: Calculate confidence intervals using JASP.

Data: Continue to use the dataset minutes reading. jasp.

Instructions: Based on the output from Problem Set 6.2, including a test value (population mean) of 8, calculate the 95% confidence interval by following the steps below.

1. Check the box Location Estimate.
2. Check the box Confidence interval. Fill in the box with 95.0%.
3. Copy and paste the output into the Word document.

One Sample T-Test

95% CI for Mean Difference

t

df

p

Mean Difference

Lower

Upper

Time

-0.493

14

0.629

-0.667

-3.564

2.231

Note.  For the Student t-test, location difference estimate is given by the sample mean difference d

Note.  For the Student t-test, the alternative hypothesis specifies that the mean is different from 8.

Note.  Student’s t-test.

#### Problem Set 3.7: Independent Samples t Test

Criterion: Calculate an independent samples t-test in JASP.

Data: Use the dataset scores. jasp. Dr. Z is interested in discovering if there is a difference in depression scores between those who do not watch or read the news and those who continue with therapy as normal. She divides her clients with depression into 2 groups. She asks Group 1 not to watch or read any news for two weeks while in therapy and asks Group 2 to continue with therapy as normal. The dataset scores.jasp is a record of the results of the measure, administered after 2 weeks.

Instructions: Complete the steps below.

1. Download scores. jasp. Double-click the icon to open the dataset in JASP.
2. In the Toolbar, click T-tests. In the menu that appears, under Classical, select Independent-samples T-test.
3. Select Score and then click the top Arrow to send it over to the Dependent Variables box.
4. Select Group and then click the bottom Arrow to send it over to the Grouping Variable box.
5. Make sure the Student box is selected. Also, select Descriptives and deselect any other boxes.
6. Copy and paste the output into the Word document.

Independent Samples T-Test

t

df

p

Score

-2.580

12

0.024

Note.  Student’s t-test.

#### Problem Set 3.8: Independent t Test in JASP

Criterion: Identify IV, DV, and hypotheses and evaluate the null hypothesis for an independent samples t test.

Data: Use the information from Problem Set 3.7.

Instructions: Complete the following:

1. Identify the IV and DV in the study. __

IV: Group (news abstention vs. therapy as normal); _ DV: Depression scores__

1. State the null hypothesis and the directional (one-tailed) alternative hypothesis. ___

Null hypothesis: There is no significant difference in depression scores between individuals who refrain from watching news and those who continue with therapy as usual.

Alternative hypothesis: The alternative hypothesis is that there is a significant difference in depression scores between individuals who abstain from watching the news and those who continue with therapy as usual.

1. Can you reject the null hypothesis at α = .05? Explain why or why not. _

Yes, you can reject the null hypothesis at α = .05. The null hypothesis is typically that there is no significant difference between the means of the two independent groups being compared. In this case, the t-value is negative (-2.580), which suggests that the

mean of group 1 is less than the mean of group 2. The p-value is 0.024, which is less than the alpha level of .05.

When the p-value is less than the alpha level, it means that the probability of observing such a significant difference between the groups by chance alone is less than 5%. In other words, there is strong evidence to suggest that the difference between the two groups is not due to chance alone.

Therefore, based on these results, we can reject the null hypothesis and conclude that there is a significant difference between the means of the two independent groups being compared.

#### Problem Set 3.9: Independent t Test using Excel

Criterion: Calculate an independent samples t test in Excel.

Data: Use this data:

Depression Scores:

Group 1: 34, 25, 4, 64, 14, 49, 54

Group 2: 24, 78, 59, 68, 84, 79, 57

### PSYC FPX4700 Assessment 3 Hypothesis, Effect Size, Power, and t Tests

Instructions: Complete the following steps:

1. Open Excel.
2. On an empty tab, enter the data from above. Use column A for group 1 and column B for Group 2. In Cell A1, enter 1. In cell B1, enter 2.
3. Enter the data for each group below the label.
4. Click Data Analysis, and select t-Test: Two-Sample Assuming Equal Variances. Click OK.
5. In Variable 1 Range enter \$A\$2:\$A\$8. (Or, click the graph icon at the right of the box and highlight your data for Group 1. Then, click the graph icon.)
6. In Variable 2 Range enter \$B\$2:\$B\$8.
7. Then click OK. Your results will appear on a new tab to the left.
8. Return to your data. Click Data Analysis, and select t-Test: Two-Sample Assuming Unequal Variances. Then click OK.
9. In Variable 1 Range enter \$A\$2:\$A\$8. (Or, click the graph icon at the right of the box and highlight your data for Group 1. Then, click the graph icon.)
10. In Variable 2 Range enter \$B\$2:\$B\$8.
11. Then click OK. Your results will appear on a new tab to the left.
12. Copy the results from both t-tests below.

 T-Test: Two-Sample Assuming Equal Variances Variable 1 Variable 2 Mean 34.85714 64.14286 Variance 483.4762 418.4762 Observations 7 7 Pooled Variance 450.9762 Hypothesized Mean Difference 0 df 12 t Stat -2.57996 P(T<=t) one-tail 0.01205 t Critical one-tail 1.782288 P(T<=t) two-tail 0.0241 t Critical two-tail 2.178813

 T-Test: Two-Sample Assuming Unequal Variances Variable 1 Variable 2 Mean 34.85714 64.14286 Variance 483.4762 418.4762 Observations 7 7 Hypothesized Mean Difference 0 df 12 t Stat -2.57996 P(T<=t) one-tail 0.01205 t Critical one-tail 1.782288 P(T<=t) two-tail 0.0241 t Critical two-tail 2.178813 ### WritinkServices.com 