Can someone please help with this problem to advise how to set up and in the book Essential of Statistics for the Behavioral Sciences.

A researcher is investigating the effectiveness of a new study-skills training program for elementary school children. A sample of n=25 third-grade children is selected to participate in the program and each child is given a standardized achievement test at the end of the year. For the regular population of third-grade children, scores on the test form a normal distribution with mean a u=150 and a standard deviation of o=25. The mean of the sample M=158.
a. Identify the independent and the dependent variables for this study.
b. Assuming a two-tailed test, state the null hypothesis in a sentence that includes the independent variable and the dependent variable.
c. Using symbols state the hypothesis (H0 and H1) for the two tailed test
d. Sketch the appropriate distribution and locate the critical region for a=.05
e. Calculate the test statistic (z-score) for the sample
f. What decision should be made about the null hypothesis, and what decision should be made about the effect of the program.

a. An independent variable is the potential stimulus or cause, usually directly manipulated by the experimenter, so it could also be called a manipulative variable.

A dependent variable is the response or measure of results.

b. The training program will not change the test scores.

c. Ho: training test scores = non-training test scores

H1: training test scores ≠ non-training test scores

d. Cannot sketch on these posts.

e. With that small a sample, a t-test might be more appropriate.

Z = (sample mean-population mean)/SEm

SEm = SD/√n

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.

f. From your results, you can make the decision.

To help you with this problem, let's go through each question step by step:

a. Identify the independent and dependent variables for this study.
In this study, the independent variable is the new study-skills training program. It is the variable that is manipulated by the researcher. The dependent variable, on the other hand, is the standardized achievement test score at the end of the year. It is the variable that is measured and affected by the independent variable.

b. Assuming a two-tailed test, state the null hypothesis in a sentence that includes the independent and dependent variables.
The null hypothesis (H0) states that there is no effect of the new study-skills training program on the standardized achievement test scores of third-grade children. In other words, the mean test score of children who participated in the program is equal to the mean test score of the regular population.

c. Using symbols, state the hypotheses (H0 and H1) for the two-tailed test.
H0: μ = 150 (Mean test score of program participants is equal to the mean test score of the regular population)
H1: μ ≠ 150 (Mean test score of program participants is not equal to the mean test score of the regular population)

d. Sketch the appropriate distribution and locate the critical region for a = 0.05.
To sketch the distribution, we would use the normal distribution with a mean of 150 and a standard deviation of 25 (representing the regular population of third-grade children). The critical region for a = 0.05 in a two-tailed test is split equally in both tails, with α/2 = 0.025 in each tail.

e. Calculate the test statistic (z-score) for the sample.
To calculate the z-score, we use the formula:
z = (M - μ) / (σ / √n)
where M is the sample mean (158), μ is the population mean (150), σ is the population standard deviation (25), and n is the sample size (25).
Plugging in the values, we get:
z = (158 - 150) / (25 / √25)
= 8 / 5
= 1.6

f. What decision should be made about the null hypothesis, and what decision should be made about the effect of the program.
To make a decision about the null hypothesis, we compare the test statistic (z-score) to the critical value(s) in the appropriate critical region. In this case, with a two-tailed test and α = 0.05, the critical z-values would be -1.96 and +1.96.

Since the calculated z-score of 1.6 falls outside the critical region (-1.96 to +1.96), we can reject the null hypothesis. This means that there is evidence to suggest that the new study-skills training program has a statistically significant effect on the standardized achievement test scores of third-grade children.

In terms of the effect of the program, based on the results, we can conclude that the program is likely to have a positive effect on the test scores, as the mean test score of the program participants (158) is significantly higher than the mean test score of the regular population (150).

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