Childhood participation in sports, cultural groups, and youth groups appears to be related to improved self-esteem for adolescents. (McGee, Williams, Howden-Chapman, Martin & Kawachi, 2006). In representative study, a sample of n = 100 adolescents with a history of group participation is given a standardized self-esteem questionnaire. For the general population of adolescents scores on this questionnaire form a normal distribution with a mean of ?= 50and a standard deviation of ?=15. the sample group-participation adolescents had an average of M=53.8.
A) Does the sample provide enough evidence to conclude that self-esteem scores for those adolescents are significantly different from those of the general population? Use the two tailed test with a= .05
B)Comput cohens d to measure the size of the difference.
C) write a sentance describing the outcome of the hypothesis test and the measure of the effect size as it would appear in a research report.
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A) To determine if the self-esteem scores for the adolescents in the sample differ significantly from those of the general population, we can perform a two-tailed t-test. Given that the sample size is n = 100, the sample mean is M = 53.8, the population mean is μ = 50, and the population standard deviation is σ = 15, we can calculate the t-value using the following formula:
t = (M - μ) / (σ / √n)
= (53.8 - 50) / (15 / √100)
= 3.8 / (15 / 10)
= 3.8 / 1.5
= 2.53
With α = 0.05 and a two-tailed test, the critical t-value for a sample size of 100 is approximately ± 1.984. Since the calculated t-value (2.53) is larger than the critical t-value, we can conclude that the self-esteem scores for the adolescents in the sample are significantly different from those of the general population.
B) Cohen's d is a measure of effect size, which indicates the standardized difference between two means. To calculate Cohen's d, we can use the following formula:
d = (M1 - M2) / SDpooled
Where M1 is the sample mean, M2 is the population mean (50), and SDpooled is the pooled standard deviation. The pooled standard deviation is calculated using the following formula:
SDpooled = √((SD1^2 + SD2^2) / 2)
Given that SD1 is the standard deviation of the sample (σ = 15) and SD2 is the standard deviation of the population, we can calculate Cohen's d:
SDpooled = √((15^2 + 15^2) / 2)
= √(450 / 2)
= √225
= 15
d = (53.8 - 50) / 15
= 3.8 / 15
= 0.25333 (rounded to 5 decimal places)
Therefore, the Cohen's d value is approximately 0.25333.
C) The hypothesis test results indicate that the self-esteem scores for the adolescents who participated in group activities are significantly different from those of the general population (t(100) = 2.53, p < 0.05). Furthermore, the effect size, measured by Cohen's d, is approximately 0.25333, indicating a small difference in self-esteem scores between the sample and the general population.
A) To determine whether the sample of adolescents with group participation has significantly different self-esteem scores from the general population, we need to conduct a two-tailed t-test.
Step 1: Set up the null and alternative hypotheses:
Null hypothesis (H0): The mean self-esteem score for adolescents with group participation is equal to the mean self-esteem score for the general population (μ = 50).
Alternative hypothesis (Ha): The mean self-esteem score for adolescents with group participation is significantly different from the mean self-esteem score for the general population (μ ≠ 50).
Step 2: Calculate the t-value:
We can calculate the t-value using the formula:
t = (sample mean - population mean) / (standard deviation / √sample size)
In this case, the sample mean (M) is 53.8, the population mean (μ) is 50, the standard deviation (σ) is 15, and the sample size (n) is 100. Plugging these values into the formula, we get:
t = (53.8 - 50) / (15 / √100)
t = 3.8 / (15 / 10)
t = 3.8 / 1.5
t = 2.53
Step 3: Find the critical t-value:
With a two-tailed test and a significance level (α) of 0.05, we need to find the critical t-value that corresponds to an α/2 level of 0.025 in each tail. Looking up this value in a t-table with 99 degrees of freedom (n - 1), we find the critical t-value to be approximately ±1.984.
Step 4: Compare the calculated t-value with the critical t-value:
As the calculated t-value (2.53) is greater than the critical t-value (1.984), we can conclude that the sample provides enough evidence to reject the null hypothesis and conclude that self-esteem scores for adolescents with group participation are significantly different from those of the general population.
B) To measure the size of the difference, we can calculate Cohen's d.
Cohen's d formula:
d = (sample mean - population mean) / standard deviation
Using the values from the given information:
d = (53.8 - 50) / 15
d = 3.8 / 15
d = 0.2533 (rounded to four decimal places)
In a research report, the outcome of the hypothesis test can be described as:
"The t-test revealed a significant difference in self-esteem scores between adolescents with group participation (M = 53.8) and the general population (μ = 50); t(99) = 2.53, p < .05, two-tailed. Furthermore, Cohen's d indicated a small effect size (d = 0.2533) for this difference."