Question 1

What is the average value expected for the independent-measures t statistic if the null hypothesis is true?

Accepted Answer

If the null hypothesis is true, then the means of the two independent groups are equal. Therefore, the difference between the means (M1 - M2) would be zero, resulting in an average value of zero for the independent-measures t statistic.

Question 2

Which of the following confidence intervals also indicates a significant difference between treatments with α = .05?

Accepted Answer

A 95% confidence interval that does not include 0 indicates a significant difference at α = .05. Option C's interval (2 to 10) does not include 0, suggesting a significant difference.

Question 3

Two separate samples are being used to estimate the population mean difference between two treatment conditions.Which of the following would produce the widest confidence interval?

Accepted Answer

The width of a confidence interval for the difference between two means is influenced by both the sample sizes and the variances of the samples. A larger variance (pooled variance in this case) leads to a wider confidence interval. Among the options, the same sample sizes with a higher variance (option B: n1 = n2 = 10 with a pooled variance of 100) would produce the widest confidence interval.

Question 4

For the independent-measures t statistic,what is the effect of increasing the sample variances?

Accepted Answer

Increasing the sample variances increases the denominator of the t statistic formula, leading to a smaller t value. This decreases the likelihood of rejecting the null hypothesis (H0) and also decreases measures of effect size, as larger variances imply less reliable differences between group means.

Question 5

One sample has M = 18 and a second sample has M = 14.If the pooled variance for the two samples is 16,what is the value of Cohen's d?

Accepted Answer

Cohen's d = (M1 - M2) / sqrt((s1^2 + s2^2) / 2) = (18 - 14) / sqrt((16 + 16) / 2) = 4 / 4 = 1.00

Question 6

An independent-measures research study uses two samples,each with n = 15 participants.If the data produce a t statistic of t = 2.760,which of the following is the correct decision for a two-tailed hypothesis test?

Accepted Answer

With a t statistic of 2.760 and 28 degrees of freedom (df = n1 + n2 - 2 = 15 + 15 - 2 = 28), using a t-table or calculator, the p-value is less than .05 but greater than .01. Therefore, we would reject the null hypothesis with α = .05 but fail to reject it with α = .01. Option A is the only choice that reflects this decision.

Question 7

A researcher reports t(22)= 5.30,p < .01 for an independent-measures experiment.How many individuals participated in the entire experiment?

Accepted Answer

The "t(22)" indicates the degrees of freedom for the t-test, which is equal to the total sample size minus 2. Therefore, degrees of freedom (df) = 22.

Since the p-value is less than .01, we know that the results are statistically significant at the .01 level. This means that the probability of obtaining these results by chance is less than 1%, or p < .01.

Based on the information provided, we can use a t-table or statistical software to determine the critical t-value for df = 22 and p < .01. The critical t-value is approximately 2.823.

The reported t-value of 5.30 is much larger than the critical t-value of 2.823, which indicates a strong effect size. We can conclude that the independent variable had a significant impact on the dependent variable.

To calculate the total sample size, we need to know the number of participants in each group. Since this information is not provided in the question, we cannot determine the exact sample size. However, we can make a reasonable estimation based on the available information.

Assuming that the experiment had two groups of equal size, we can calculate the sample size as follows:

t = (M1 - M2) / (s / sqrt(n))
n = [(s / (M1 - M2)) * t]^2

Let's assume that the standard deviation (s) is equal to 1 (this is just an example for illustration purposes). Then, we can re-arrange the formula to solve for n:

n = (t * s / (M1 - M2))^2

Using the reported t-value of 5.30 and assuming a hypothetical difference between the means (M1 - M2) of 1, we get:

n = (5.30 * 1 / 1)^2 = 28.09

Since we assumed that the two groups were equal in size, the total sample size would be twice as large, or approximately:

sample size = 2 * n = 2 * 28.09 = 56.18

This is not an integer, so we cannot directly use it as the answer. However, we can round it up or down to the nearest integer based on our assumptions and interpretation of the results.

If we assume that the group sizes were integers and equal to each other, we can try different combinations of group size and see if they give us a total sample size of approximately 56.18. For example:

- If each group had 24 participants, the total sample size would be 48, which is too small.
- If each group had 25 participants, the total sample size would be 50, which is still too small.
- If each group had 26 participants, the total sample size would be 52, which is close but still too small.
- If each group had 27 participants, the total sample size would be 54, which is very close.
- If each group had 28 participants, the total sample size would be 56, which is just right.

Therefore, the best answer is D) 24.

Question 8

The results of an independent-measures research study are reported as "t(22)= 2.12,p < .05,two tails." For this study,what t values formed the boundaries for the critical region?

Accepted Answer

The answer of The results of an independent-measures research study...

Question 9

A researcher predicts that scores in treatment A will be higher than scores in treatment B.If the mean for the 10 participants in treatment A is 4 points higher than the mean for the 10 participants in treatment B and the data produce t = 2.095,what decision should be made?

Accepted Answer

The researcher predicts a directional difference between treatment A and treatment B (i.e. treatment A will have higher scores). Therefore, a one-tailed test is appropriate. The t-value of 2.095 indicates a significant difference between the two groups. With α = .05, the critical t-value for a one-tailed test with 18 degrees of freedom is 1.734. Since the obtained t-value is greater than the critical t-value, we reject the null hypothesis for a one-tailed test. However, for a two-tailed test, the critical t-value is 2.101 (with α = .05 and 18 degrees of freedom). The obtained t-value of 2.095 is not greater than the critical t-value for a two-tailed test, so we fail to reject the null hypothesis for a two-tailed test. Therefore, the correct answer is C: With α = .05, reject H0 for a one-tailed test but fail to reject for a two-tailed test.

Question 10

An independent-measures research study with n = 5 participants in each treatment produces sample variances of 8 and 10,and a 2-point difference between the two treatment means.What is the value of Cohen's d?

Accepted Answer

Cohen's d is calculated as the difference between two means divided by the standard deviation for the data. In this case, the pooled standard deviation (sqrt[(8+10)/2]) is 3, and the difference between means is 2. Thus, Cohen's d = 2/3.

Question 11

An independent-measures study produces sample means of M₁ = 20 and M₂ = 17.If both samples have n = 18 scores and Cohen's d = 0.50,what is the value for the pooled variance?

Accepted Answer

The answer of An independent-measures study produces sample means of...

Question 12

Two samples,each with n = 8,produce an independent-measures t statistic of t = -2.15.Which of the following decisions is justified for a two-tailed test?

Accepted Answer

Since the t-statistic is negative, we know that the sample mean difference is negative. For a two-tailed test with n = 8 and alpha = .05, the critical t-value is approximately ±2.306. Since our calculated t-value of -2.15 falls within the critical range of -2.306 to +2.306, we fail to reject the null hypothesis with alpha = .05. However, for alpha = .01, the critical t-value is approximately ±3.355, which is outside our calculated t-value range, so we reject the null hypothesis with alpha = .01. Therefore, we can justify rejecting the null hypothesis with alpha = .01 but failing to reject it with alpha = .05. Choice A is the only option that reflects this conclusion.

Question 13

An independent-measures study produces sample means of M₁ = 20 and M₂ = 17,and a pooled variance of 9.For this study,what is the value of Cohen's d?

Accepted Answer

Cohen's d is calculated as (M1 - M2) / sqrt(pooled variance). In this case, (20 - 17) / sqrt(9) = 3/3.

Question 14

An independent-measures study comparing two treatment conditions with n = 6 in each sample produces t = 4.00.What is the value of r² for this study?

Accepted Answer

To find r², we need to use the formula: t² / (t² + df), where df = n1 + n2 - 2. 
Substituting the given values, we get: 
r² = (4.00)² / [(4.00)² + (6+6-2)] 
r² = 16 / 26 
r² = 8 / 13 
Therefore, the answer is C.

Question 15

A researcher predicts that scores in treatment A will be higher than scores in treatment B.Which of the following is the correct null hypothesis for a one-tailed test evaluating this prediction?

Accepted Answer

The answer of A researcher predicts that scores in treatment...

Question 16

Using an independent-measures t,the 90% confidence interval for the difference between two population means ranges from 19 to 23.Based on this confidence interval,how big is the difference between the two sample means?

Accepted Answer

Since the confidence interval for the difference between the two population means ranges from 19 to 23, we can assume that the point estimate of the difference between the two sample means is at the center of this interval. Therefore, the difference between the two sample means would be (23-19)/2 = 2 points away from the midpoint of the interval, which is 21. Thus, the best choice is C) 21 points.

Question 17

Two samples,each with n = 9 scores,produce an independent-measures t statistic of t = 2.00.If the effect size is measured using r²,what is the value of r²?

Accepted Answer

The formula for r² is t² / (t² + df), where df = n1 + n2 - 2. Here, df = 9 + 9 - 2 = 16. So, r² = 2² / (2² + 16) = 4 / 20.

Question 18

For the independent-measures t statistic,what is the effect of increasing the difference between sample means?

Accepted Answer

The independent-measures t statistic measures the difference between sample means, and increasing this difference will result in a larger t statistic value and a smaller p-value. This means that it is more likely to reject the null hypothesis, and the effect size (such as Cohen's d) will also increase as the difference between sample means becomes larger.

Question 19

The narrower the confidence interval,the more precise it is.With this in mind,which combination of factors will produce the most precise estimate of the difference between two population means?

Accepted Answer

Larger sample sizes and lower confidence levels both contribute to narrower (more precise) confidence intervals. Option C has the largest sample size (n = 50) and the lower confidence level (80%), making it the most precise estimate.

Question 20

An independent-measures research study uses two samples,each with n = 10 participants.If the data produce a t statistic of t = 2.095,which of the following is the correct decision for a two-tailed hypothesis test?

Accepted Answer

The answer of An independent-measures research study uses two samples,each...

Quiz 10: The t Test for Two Independent Samples