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.

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