A researcher is evaluating the effectiveness of a new physical education program
ID: 3133067 • Letter: A
Question
A researcher is evaluating the effectiveness of a new physical education program for elementary school children. The program is designed to reduce competition and increase individual self-esteem. A sample of n = 16 children is selected and the children are placed in the new program. After 3 months, each child is given a standardized self-esteem test. For the general population of elementary school children, the scores on the self-esteem test form a normal distribution with = 40 and = 8.
If the researcher obtains a sample mean of M = 42, is this enough evidence to conclude that the program has a significant effect? Assume a two-tailed test with = .05. (Use 2 decimal places.)
z-critical = ±
z=
Conclusion
Reject the null hypothesis, the program has a significant effect.
Fail to reject the null hypothesis, the program has a significant effect.
Fail to reject the null hypothesis, the program does not have a significant effect.
Reject the null hypothesis, the program does not have a significant effect.
(b) If the sample mean is M = 44, is this enough to demonstrate a significant effect? Again, assume a two-tailed test with = .05. (Use 2 decimal places.)
z-critical = ±
z=
Conclusion
Reject the null hypothesis, the program has a significant effect.
Fail to reject the null hypothesis, the program has a significant effect.
Fail to reject the null hypothesis, the program does not have a significant effect.
Reject the null hypothesis, the program does not have a significant effect.
2. State College is evaluating a new English composition course for freshmen. A random sample of n = 25 freshmen is obtained and the students are placed in the course during their first semester. One year later, a writing sample is obtained for each student and the writing samples are graded using a standardized evaluation technique. The average score for the sample is M = 76. For the general population of college students, writing scores form a normal distribution with a mean of = 70.
(a) If the writing scores for the population have a standard deviation of = 17, does the sample provide enough evidence to conclude that the new composition course has a significant effect? Assume a two-tailed test with = 0.05. (Round your answers to two decimal places.)
z-critical = ±
z=
Conclusion
Fail to reject the null hypothesis, the course does not have a significant effect.
Reject the null hypothesis, the course does not have a significant effect.
Fail to reject the null hypothesis, the course does have a significant effect.
Reject the null hypothesis, the course does have a significant effect
(b) If the population standard deviation is = 13, is the sample sufficient to demonstrate a significant effect? Again, assume a two-tailed test with = 0.05. (Round your answers to two decimal places.)
-critical = ±
z=
Conclusion
Reject the null hypothesis, the course does not have a significant effect.
Fail to reject the null hypothesis, the course does have a significant effect.
Reject the null hypothesis, the course does have a significant effect.
Fail to reject the null hypothesis, the course does not have a significant effect.
3. To test the effectiveness of a treatment, a sample is selected from a normal population with a mean of = 40 and a standard deviation of = 12. After the treatment is administered to the individuals in the sample, the sample mean is found to be M = 46.
(a) If the sample consists of n = 4 individuals, is this result sufficient to conclude that there is a significant treatment effect? Use a two-tailed test with = .05. (Use 2 decimal places.)
-critical = ±
z=
Conclusion
Reject the null hypothesis, there is a significant treatment effect.
Reject the null hypothesis, there is not a significant treatment effect.
Fail to reject the null hypothesis, there is not a significant treatment effect.
Fail to reject the null hypothesis, there is a significant treatment effect.
If the sample consists of n = 36 individuals, is this result sufficient to conclude that there is a significant treatment effect? Use a two-tailed test with = .05. (Use 2 decimal places.)
-critical = ±
z=
Conclusion
Fail to reject the null hypothesis, there is a significant treatment effect.
Reject the null hypothesis, there is not a significant treatment effect.
Fail to reject the null hypothesis, there is not a significant treatment effect.
Reject the null hypothesis, there is a significant treatment effect.
(c) Compute Cohen's d to measure effect size for both tests (n = 4 and n = 36). (Use 2 decimal places.)
n = 4
n = 36
(d) Briefly describe how sample size influences the outcome of the hypothesis test. How does sample size influence measures of effect size?
A larger sample increases the likelihood of rejecting the null hypothesis, but has no influence on Cohen's d.
A larger sample increases the likelihood of rejecting the null hypothesis, but has a decreasing effect on Cohen's d.
A larger sample increases the likelihood of rejecting the null hypothesis and increases Cohen's d.
A larger sample reduces the likelihood of rejecting the null hypothesis, but has no influence on Cohen's d.
Explanation / Answer
1.
a)
Formulating the null and alternative hypotheses,
Ho: u = 40
Ha: u =/ 40
As we can see, this is a two tailed test.
Thus, getting the critical z, as alpha = 0.05 ,
alpha/2 = 0.025
zcrit = +/- 1.96 [ANSWER]
****************************
Getting the test statistic, as
X = sample mean = 42
uo = hypothesized mean = 40
n = sample size = 16
s = standard deviation = 8
Thus, z = (X - uo) * sqrt(n) / s = 1 [ANSWER, z]
*******************************
As |z| < 1.96,
OPTION C: Fail to reject the null hypothesis, the program does not have a significant effect. [ANSWER]
*****************************
B)
Formulating the null and alternative hypotheses,
Ho: u = 40
Ha: u =/ 40
As we can see, this is a two tailed test.
Thus, getting the critical z, as alpha = 0.05 ,
alpha/2 = 0.025
zcrit = +/- 1.96 [ANSWER]
*****************************************
Getting the test statistic, as
X = sample mean = 44
uo = hypothesized mean = 40
n = sample size = 16
s = standard deviation = 8
Thus, z = (X - uo) * sqrt(n) / s = 2 [ANSWER, Z]
*******************************************
As z > 1.96,we reject Ho:
OPTION A: Reject the null hypothesis, the program has a significant effect. [ANSWER]
*******************************************
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