APP6JMaloney problems 2. 4, 6, 10, 18, 22, 24 2 ) The value of the z score un a hypothesis test is influenced by a variety of factors. Assuming that all the other variables are held constant, explain how the value of Z is influenced by each of the following? Z= M – u / SD a) Increasing the difference between the sample mean and the original. The z score represents the distance of each X or score from the mean. If the distance between the sample mean and the population mean the z score will increase.

b) Increasing the population standard deviation.

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The standard deviation is the factor that is used to divide by in the equation. the bigger the SD, then the smaller the z score. c) Increasing the number of scores in the sample. Should bring the samples mean closer to the population mean so z score will get smaller. 4) If the alpha level is changed from . 05 to . 01 a) what happens to the boundaries for the critical region? It reduces the power of the test to prove the hypothesis.

You increase the chance of rejecting a true H b) what happens to the probability of a type 1 error? Type 1 error is falsely reporting a hypothesis,

Where you increase the chance that you will reject a true null hypothesis. 6) A researcher is investigating the effectiveness of a new study skills training program for elementary school childreen. A sample of n=25 third grade children is selected to participate in the program and each child is given a standardizrd achievement test at the end of year. For the regular population of third grade children, scores on the test form a normal distribution with a mean u = 150, and a standard deviation q = 25. The mean for the sample is M = 158. a) Identify the independent and the dependent variables in the study

Independent = third grade child Dependent = Score on test b) Assuming a two-tailed test, state null hypothesis that includes the independent & dependent variable. Ho: After the program the mean will still be 150 H1: After the program the mean will be different from 150 c) Using symbols, state the hypotheses (H and H) for the two tailed test. Ho: u=150 H1: ? ?150 d) Sketch the appropriate distribution, and locate the critical region for u=. 05 Put 150 instead of 50 for u e) Calculate the test statistic (z-score) for the sample qm = q / square root of number in sanple = 25 / sq root of 25 = 25 / 5 = 5

Z= M – u / qm = 158 – 150 / 5 = 8 / 5 = 1. 6 f) what decision should be made about the null hypothesis, & the effects of the program? – a statistical decision about the Null hypothesis. – and a conclusion about the outcome of the experiment. 10) State college is evaluating a new English composition course for freshman. A random sample of n=25 freshman is obtained and the students are placed in the course during their first semester. One year later, a writing sample is obtained for 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 from a normal distribution with a mean of u=70. a) If the writing scores for the population have a standard deviation of q=20, does the sample provide enough evidence to conclude that the new composition course has a significant effect? Assume a two-tailed test with alpha = . 05 I need to find the z score… q for the population = q for the sample / sq root of number in sample = Therefore = 20 / sq root of 25 = 20/5 = 4 is the q for population Z = M – u / 4 6 – 70 / 4 = 6 / 4 = 1. 5 is the z score No the sample does not, the z score was only 1. 5, you need at least 1. 96 (pos or neg) b) If the population standard deviation is q=10, is the sample sufficient to demonstrate a significant effect? Again, assume a two-tailed test with alpha=. 05 I need to find the z score… q for the population = q for the sample / sq root of number in sample = Therefore = 10 / sq root of 25 = 10/5 = 2 is the q for population Z = M – u / 2 76 – 70 / 2 = 6 / 2 = 3 is the z score Yes the sample does, the z score is 3, you needed at least 1. 6 (pos or neg) c) Briefly explains why you reached different conclusions for part (a) and part (b). The difference was the amount of deviation. 18) A sample of n = 16 individuals is selected from a normal population with a mean of u = 48 and a standard deviation of q = 12. After receiving a treatment, the sample mean is found to be M = 52 a) Compute Cohen’s d to evaluate the size of the treatment effect Cohen’s d = M treatment – M no treatment / standard deviation = 53 – 48 / 12 = 5 / 12 = 0. 416 b) If the sample size were n = 36, what value would be obtained for cohen’s d?

How does sample size influence the measure of effect size? q pop = sd / sq root of 36 = 12 / 6 = 2 Cohen’s d = 53 – 48 / 2 = 12 / 2 = 6 sample size increases Cohen’s d c) If the population standard deviation were q = 24, what value would be obtained for Cohen’s d? How does standard deviation influence the measure of effect size? 53 – 48 /24 = 5 / 24 = 0. 208 as standard deviation goes up, the measure of effect goes down d) If the sample mean were M = 56, what value would be obtained for Cohen’s d? How does the size of the mean difference influence the measure of effect size? 6 – 48 / 12 = 8 / 12 = . 666 the size of the mean difference decreases the the measure of effect size 22) Explain how the power of a hypothesis test is influenced by each of the following. Assume that all other factors are held constant. a) Increasing the alpha level from . 01 to . 05. Increases the chances of proving your hypothesis b) Changing from a one-tailed test to a two tailed test. Increases the chances of provong your hypothesis 24) A researcher is evaluating the influence of a treatment using a sample selected from a normally distributed population with a mean of u = 80 and a standard deviation f q = 20. The researcher expects a 12 – point treatment effect and plans to use a two-tailed hypothesis test with a alpha = . 05 a) compare the power of the text if the researcher uses a sample of n = 16 individuals. 20/sq root of 16 = 20 / 4 = 5 12 = X – 80 / 5, 12 x 5 = X – 80. 60 = X – 80, 60 / 80 = . 75 b) compare the power of the test if the researcher uses a sample of n = 25 individuals. 20/sq root of 25 = 20 / 5 = 4 12 = X – 80 / 4, 12 x 4 = X – 80, 48 = X – 80, 48 / 80 = . 6 Therefore the more scores, the less effect size