Chapter 8 QuestionA high school teacher has designed a new course intended to help students prepare for the mathematics section of the SAT. A sample of n 5 20 students is recruited for the course...

8.a) Ans.
To test whether the students who took the course score significantly higher than the general population, using a one-sample t-test. The null hypothesis is that there is no difference between the mean score of the sample and the mean score of the general population, while the alternative hypothesis is that the mean score of the sample is significantly higher than the mean score of the general population.
H0: μ = 500
Ha: μ > 500
The level of significance, α, is given as .01, and the sample size is 20. Since the population standard deviation, σ, is known, using a z-test.
First, need to standardize the sample mean to obtain the z-score:
z = (M - μ) / (σ / sqrt(n))
z = (562 - 500) / (100 / sqrt(20))
z = 3.16
The critical z-value for a one-tailed test with α = .01 is 2.33 (obtained from a standard normal distribution table or calculator). Since the calculated z-value of 3.16 is greater than the critical z-value of 2.33, we reject the null hypothesis.
Therefore, it can be conclude that the students who took the course score significantly higher than the general population on the SAT.
b.Ans)
To calculate Cohen's d, need to calculate the standardized mean difference between the sample mean and the population mean, in terms of the population standard deviation.
Cohen's d = (M - μ) / σ
Cohen's d = (562 - 500) / 100
Cohen's d = 0.62
The resulting Cohen's d of 0.62 indicates a medium effect size, which suggests that the students who took the course performed better than the general population on the SAT.
c.Ans)
The results of the hypothesis test indicated that students who took the course performed significantly better than the general population on the SAT (z = 3.16, p < .01). Furthermore, the effect size was moderate (Cohen's d = 0.62), indicating that the improvement was practically significant as well.
Q9.a)Ans.
To test whether cocaine influenced time perception in rats, using a two-sample t-test. The null hypothesis is that there is no difference between the mean duration of sound judged as long or short for rats under the influence of cocaine and the untreated population mean. The alternative hypothesis is that there is a difference between the mean duration of sound judged as long or short for rats under the influence of cocaine and the untreated population mean.
H0: μ = 4
Ha: μ ≠ 4
The level of significance, α, is given as .05, and the sample size is 16. Since the population variance is unknown, a t-test with 15 degrees of freedom.
First, calculate the t-value:
t = (M - μ) / (s / sqrt(n))
t = (3.78 - 4) / (sqrt(0.16) / sqrt(16))
t = -1.5
The critical t-value for a two-tailed test with α = .05 and 15 degrees of freedom is ±2.131 (obtained from a t-distribution table or calculator). Since the absolute value of the calculated t-value of 1.5 is less than the critical t-value of 2.131, it fail to reject the null hypothesis.
Therefore, research can conclude that there is not enough evidence to support the claim that cocaine influenced time perception in rats.
b)ans.
To construct a 95% confidence interval to estimate the population mean duration of sound judged as long or short for rats under the influence of cocaine,
CI = M ± t(α/2, df) * (s / sqrt(n))
Where M is the sample mean, s is the sample standard deviation, n is the sample size, t(α/2, df) is the critical t-value with α/2 level of significance and degrees of freedom (df) equals n - 1.
Substituting the given values,
CI = 3.78 ± t(0.025, 15) * (sqrt(0.16) / sqrt(16))
From the t-distribution table or calculator, t(0.025, 15) is approximately 2.131.
CI = 3.78 ± 2.131 * 0.04
CI = [3.70, 3.86]
Therefore, it can be 95% confident that the population mean duration of sound judged as long or short for rats under the influence of cocaine lies between 3.70 seconds and 3.86 seconds.
c)Ans.
Here are two different measures of effect size that can be calculated:
Cohen's d:
Cohen's d is a measure of standardized mean difference between two groups. It indicates the effect size in terms of the number of standard deviations by which the means of the two groups differ.
d = (M - μ) / s
where M is the sample mean, μ is the population mean (in this case, μ = 4), and s is the sample standard deviation.
Substituting the given values,
d = (3.78 - 4) / sqrt(0.16)
d = -1.25
Therefore, Cohen's d is -1.25, indicating a large effect size.
Eta-squared (η²):
Eta-squared is a measure of the proportion of variance in the dependent variable that can be explained by the independent variable.
Calculate the Eta-squared η² = SSbetween / SStotal
where SSbetween is the sum of squares between groups and SStotal is the total sum of squares. Since research have only one group (rats under the influence of cocaine
SSbetween = n * (M - μ)²
Substituting the given values
SSbetween = 16 * (3.78 - 4)²
SSbetween = 0.6912
To calculate SStotal, the sum of squares for the entire population. Since research don't have this information.
SStotal = (n - 1) * s²
Substituting the given values,
SStotal = 15 * 0.16
SStotal = 2.4
Therefore, η² = 0.6912 / 2.4 = 0.288, indicating a large effect size.
Q10.a)Ans.
Increasing the number of scores in each sample will generally increase the independent measures t statistic, which measures the difference between the means of two groups relative to the variability within each group. This is because as the sample size increases, the standard error of the mean decreases, which makes it easier to detect a significant difference between the groups. This, in turn, increases the likelihood of rejecting the null hypothesis.
The magnitude of measures of effect size, such as Cohen's d, may or may not be affected by increasing the sample size, depending on the size of the true effect. If the true effect size is small, increasing the sample size may not change the effect size estimate significantly. However, if the true effect size is large, increasing the sample size may increase the effect size estimate, indicating a stronger effect.
b)Ans.
Increasing the variance for each sample will generally decrease the independent measures t statistic, which measures the difference between the means of two groups relative to the variability within each group. This is because increasing the variance increases the variability within each group, making it harder to detect a significant difference between the groups. This, in turn, decreases the likelihood of rejecting the null hypothesis.
The magnitude of measures of effect size, such as Cohen's d, may or may not be affected by increasing the variance, depending on the size of the true effect. If the true effect size is small, increasing the variance may not change the effect size estimate significantly. However, if the true effect size is large, increasing the variance may decrease the effect size estimate, indicating a weaker effect. This is because increasing the variance reduces the signal-to-noise ratio, making it harder to detect a true effect.
Q11.a)Ans.
To test the hypothesis that gamification affected participants' motivation to work, using a paired samples t-test. The null hypothesis is that there is no difference between the mean motivation to work before and after gamification, while the alternative hypothesis is that there is a difference.
The mean difference between the before and after motivation scores:
Mean difference = (7-2) + (1-2) + (4-3) + (10-9) + (8-6) + (10-9) + (9-6) + (6-5) + (12-7) / 9
Mean difference = 5.22
Next, calculate the standard deviation of the differences:
s = sqrt(∑(X-M)^2 / (n-1))
s = sqrt(((7-2.22)^2 + (1-2.22)^2 + (4-2.22)^2 + (10-2.22)^2 + (8-2.22)^2 + (10-2.22)^2 + (9-2.22)^2 + (6-2.22)^2 + (12-2.22)^2) / 8)
s = 3.68
The t-statistic can then be calculated as:
t = (M - μ) / (s / sqrt(n))
t = (5.22 - 0) / (3.68 / sqrt(9))
t = 4.49
Using a two-tailed test with a significance level of .05 and degrees of freedom of 8, the critical t-value is ±2.306.
Since our calculated t-value of 4.49 is greater than the critical value of 2.306, research can reject the null hypothesis and conclude that gamification did have a significant effect on participants' motivation to work.
also calculate a measure of effect size, such as Cohen's d:
d = M / s
d = 5.22 / 3.68
d = 1.42
This indicates a large effect size, suggesting that the gamification program had a substantial impact on participants' motivation to work.
B)ans.
To calculate Cohen's d, first calculate the pooled standard deviation:
s_pooled = sqrt(((n1 - 1) * s1^2 + (n2 - 1) * s2^2) / (n1 + n2 - 2))
where n1 is the sample size before gamification, s1 is the standard deviation before gamification, n2 is the sample size after gamification, and s2 is the standard deviation after gamification.
n1 = n2 = 9
s1 = 2.563
s2 = 3.118
s_pooled = sqrt(((9 - 1) * 2.563^2 + (9 - 1) * 3.118^2) / (9 + 9 - 2)) = 2.835
Now Cohen's d:
Cohen's d = (M2 - M1) / s_pooled
where M1 is the mean motivation to work before gamification and M2 is the mean motivation to work after gamification.
M1 = 5.11
M2 = 6.78
Cohen's d = (6.78 - 5.11) /...