PSY 315 Week 4 û Chapter problems
PSY 315 Week 4 û Chapter problems
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Date Created: 11/13/15
C h a p t e r P r o b l e m s - W e e k 4 P a g e | 1 Week 4 – Chapter problems PSY/315 Katherine Sainz University of Phoenix C h a p t e r P r o b l e m s - W e e k 4 P a g e | 2 11. List the five steps of hypothesis testing, and explain the procedure and logic of each. Step 1: Specify the Null Hypothesis and the Alternative Hypothesis In order to make a hypothesis test, the researcher needs to identify the problem in terms of a question that identifies the population of interest to the researcher, the parameters of the variable under investigation, and the hypothesized value of the parameters. The first step in hypothesis testing is to identify the null hypothesis (H ) an0 the alternative hypothesis (H ). 1 The null hypothesis is a statement of no effect, relationship, or difference between two or more groups or factors. The alternative hypothesis is the statement that there is an effect or difference. This is usually the hypothesis the researcher is interested in proving. The alternative hypothesis can be onesided or twosided. Researchers often use twosided tests even when their true hypothesis is onesided because it requires more evidence against the null hypothesis to accept the alternative hypothesis. If the research concerns whether one method of presenting pictorial stimuli leads to better recognition than another, the null hypothesis would most likely be that there is no difference between methods (H : μ μ = 0). 0 1 2 The alternative hypothesis would be H : μ ≠1μ .1 2 For example: Is the mean first salary of a newly graduated student equal to $30,000? The population of interest is all students who have just graduated. The parameter of interest is the mean and the variable salary is continuous. The hypothesized value of the parameter, the mean, is $30,000. Since the parameter is a population mean of a continuous variable, this C h a p t e r P r o b l e m s - W e e k 4 P a g e | 3 suggests a one sample test of a mean. Step 2: Set the Significance Level (a) One of the key concepts in hypothesis testing is that of significance level (or, equivalently the alpha ( ) level) which specifies the probability level for our evidence to be an unreasonable estimate. That is the estimate should not have taken its particular value unless some non chance factor(s) had operated to alter the nature of the sample such that it was no longer representative of the population of interest. The significance level is generally set at 0.05. This means that there is a 5% chance that you will accept your alternative hypothesis when your null hypothesis is actually true. Example: H0: µ = $30,000 HA: µ NE $30,000 alpha=.05 Test assumptions are 1) the population is normally distributed or sample size is approximately >=30 and 2) the sample we have used to collect the data was drawn randomly from the population. If these test assumptions have not been met, then data collection should be reevaluated or continued under caution. C h a p t e r P r o b l e m s - W e e k 4 P a g e | 4 Step 3: Calculate the Test Statistic and Corresponding PValue The third step is to calculate a statistic parallel to the parameter specified by the null hypothesis. Hypothesis testing generally uses a test statistic that compares groups or examines associations between variables. When describing a single sample without establishing relationships between variables, a confidence interval is commonly used. The p value describes the probability of obtaining a sample statistic as or more extreme by chance alone if your null hypothesis is true. This pvalue is determined based on the result of your test statistic. Your conclusions about the hypothesis are based on your pvalue and your significance level. Example: Suppose we randomly sampled 100 high school seniors and determined their salary of their first job. The sample mean salary, xbar, was $29,000 with a standard deviation of $6,000. Since sample size is >30, we don't have to worry about whether the population is normally distributed (Central Limit Theorem). The test statistic would be: z = $29,000 $30,000 = $1,000 = 1.667 $6,000/sqrt(100) $600 Step 4: Calculate probability of test statistic or rejection region. C h a p t e r P r o b l e m s - W e e k 4 P a g e | 5 The fourth step is to calculate the probability value pvalue which is the probability of the test statistic for both tails since this is a twotailed test. The probability value computed in this step is compared with the significance level selected in step 2. If the probability is less than or equal to the significance level, then the null hypothesis is rejected. If the probability is greater than the significance level then the null hypothesis is not rejected. When the null hypothesis is rejected, the outcome is said to be "statistically significant"; when the null hypothesis is not rejected then the outcome is said be "not statistically significant." If the outcome is statistically significant, then the null hypothesis is rejected in favor of the alternative hypothesis. Example: P(z> 1.667) =.048 + P(z< 1.667)=.048, the pvalue is . 048+.048=.096 Since this value is greater than alpha=.05 selected when we set up out hypotheses, we accept the null hypothesis, H0: µ = $30,000. If we wish to use a rejection region of alpha=.05 (.025 in each tail) to determine if we accept or reject the null hypothesis, the cutoff zscore would be 1.96 and 1.96. If our test statistic is >=1.96 or <= 1.96, then we would reject the null hypothesis at alpha=.05. We can say that C h a p t e r P r o b l e m s - W e e k 4 P a g e | 6 our test statistic (transformed into a zscore) is in the rejection region. In this example, our test statistic, z=1.667 for our test statistic does not fall in the rejection region (sometimes called the acceptance region), so we must accept the null hypothesis. Step 5: Drawing a Conclusion The fifth and final step is to describe the results and state correct statistical conclusions in an understandable way. Hypothesis testing is not set up so that you can absolutely prove a null hypothesis. Therefore, when you do not find evidence against the you can absolutely prove a null hypothesis. Thus, when you do not find evidence against the null hypothesis, you fail to reject the null hypothesis. When you do find strong enough evidence against the null hypothesis, you reject the null hypothesis. Your conclusions also translate into a statement about your alternative hypothesis Example: Accept the null hypothesis at alpha=.05 or pvalue of .096. Based on a sample mean of $25,000, the mean salary of a newly graduated student does not equal $30,000. Examples taken from: http://www.unc.edu/~jamisonm/fivesteps.htm 14. Based on the information given for each of the following studies, decide whether to reject the null hypothesis. For each, give (a) the Zscore cutoff (or cutoffs) on the comparison distribution at which the null hypothesis should be rejected, (b) the Z score on the comparison distribution for the sample score, and (c) your conclusion. Assume that all populations are normally distributed. C h a p t e r P r o b l e m s - W e e k 4 P a g e | 7 A: z–score = (7 – 5)/1 = 2 At .05 significance level for a onetailed test, the zscore value = +1.645 Since 2 > 1.645, we will reject the null hypothesis. B: z–score = (7 – 5)/1 = 2 At .05 significance level for twotailed test, the zscore value = ±1.96 Since 2 > +1.96, we will reject the null hypothesis. C: z –score = (7 – 5)/1 = 2 At .01 significance level for a onetailed test, the zscore value = +2.33 Since 2 < 2.33, we fail to reject the null hypothesis. D: z –score = (7 – 5)/1 = 2 At .01 significance level for a twotailed test, the zscore value = ±2.58 Since 2.58 <2< 2.58, we fail to reject the null hypothesis. 18. A researcher predicts that listening to music while solving math problems will make a particular brain area more active. To test this, a research participant has her brain scanned while listening to music and solving math problems, and the brain area of interest has a percentage signal change of 58. From many previous C h a p t e r P r o b l e m s - W e e k 4 P a g e | 8 studies with this same math problems procedure (but not listening to music), it is known that the signal change in this brain area is normally distributed with a mean of 35 and a standard deviation of 10. (a) Using the .01 level, what should the researcher conclude? Solve this problem explicitly using all five steps of hypothesis testing, and illustrate your answer with a sketch showing the comparison distribution, the cutoff (or cutoffs), and the score of the sample on this distribution. (b) Then explain your answer to someone who has never had a course in statistics (but who is familiar with mean, standard deviation, and Z scores). H0: μ = 35 H1: μ ≠ 35 α = 0.01 Critical Z values = ±2.5758 Z = (xxμ)/ (σ) = (5835)/ (10) = 2.3 Since 2.3 < 2.5758 we fail to reject H0 and therefore the researcher is unable to reject the .probability that the mean of signal change is 35. The difference is not significant at α = 0.01 The zscore that was established here lies outside the critical region. Critical region is a set of values of the test statistic or the zscore which rejects the possibility of the conclusion being .false In this case, we can't reject that possibility. Therefore the researcher shouldn't make any .conclusion C h a p t e r P r o b l e m s - W e e k 4 P a g e | 9
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