PSY 216. Assignment 9 Answers. Under what circumstances is a t statistic used instead of a z-score for a hypothesis test

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PSY 216 Assignment 9 Answers 1. Problem 1 from the text Under what circumstances is a t statistic used instead of a z-score for a hypothesis test The t statistic should be used when the population standard deviation, σ, is unknown and must be estimated from the sample variance, s 2. 2. Problem 8 from the text To evaluate the effect of a treatment, a sample is obtained from a population with a mean of μ = 75, and the treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M = 79.6 with a standard deviation of s = 12. a. If the sample consists of n = 16 individuals, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with α =.05? Step 1: State hypotheses and select α level H0: μ = 75 H1: μ 75 α =.05 Step 2: Locate the critical region df = n 1 = 16 1 = 15 Critical t(15), two-tailed, α =.05 = 2.131

Step 3: Compute the test statistic sm = (s 2 / n) = (12 2 / 16) = 3 = (79.6 75) / 3 = 1.533 Step 4: Make a decision Because the observed / calculated value of t (1.533) is not in the tails cut off by the critical t (2.131 the green area in the above diagrams), we fail to reject H0 and conclude that there is insufficient evidence to suggest that the treatment had an effect. b. If the sample consists of n = 36 individuals, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with α =.05? Step 1: State hypotheses and select α level H0: μ = 75 H1: μ 75 α =.05 Step 2: Locate the critical region df = n 1 = 36 1 = 35 Critical t(35), two-tailed, α =.05 = 2.030

Step 3: Compute the test statistic sm = (s 2 / n) = (12 2 / 36) = 2 = (79.6 75) / 2 = 2.300 Step 4: Make a decision Because the observed / calculated value of t (2.300) is in the tails cut off by the critical t (2.030 the green area in the above diagrams), we reject H0 and conclude that there is sufficient evidence to suggest that the treatment had an effect. c. Comparing your answer for parts a and b, how does the size of the sample influence the outcome of a hypothesis test? Increasing sample size increases the statistical power of the test. This makes it more likely that we will correctly reject a false H0. 3. Problem 14 from the text The librarian at the local elementary school claims that, on average, the books in the library are more than 20 years old. To test this claim, a student takes a sample of n = 30 books and records the publication date for each. The sample produces an average age of M = 23.8 years with a variance of s 2 = 67.5. Use this sample to conduct a one-tailed test with α =.01 to determine whether the average age of the library books is significantly greater than 20 years (μ > 20). Step 1: Hypotheses and α level H0: μ 20 H1: μ > 20 α = 0.01 Step 2: Critical region α =.01 One-tailed df = n 1 = 30 1 = 29 tcritical = 2.462 Step 3: Calculate tobserved sm = (s 2 / n) = (67.5 / 30) = 1.5

t = (23.8 20) / 1.5 t = 2.533 Step 4: Decide The observed t (2.533) is in the tail cut off by the critical t (2.462), therefore we reject H0. It is likely that the books are older than 20 years of age on average. 4. Problem 15 from the text For several years researchers have noticed that there appears to be a regular, year-by-year increases in the average IQ for the general population. This phenomenon is called the Flynn effect after the researcher who first reported it (Flynn, 1984, 1999), and it means that psychologists must continuously update IQ tests to keep the population mean at μ = 100. To evaluate the size of the effect, a researcher obtained a 10-year-old IQ test that was standardized to produce a mean IQ of μ = 100 for the population 10 years ago. The test was then given to a sample of n = 64 of today s 20-yearold adults. The average score for the sample was M = 107 with a standard deviation of s = 12. a. Based on the sample, is the average IQ for today s population significantly different from the average 10 years ago, when the test would have produced a mean of μ = 100? Use a two-tailed test with α =.01. Step 1: State hypotheses and select α level H0: μ = 100 H1: μ 100 α =.01 Step 2: Locate the critical region df = n 1 = 64 1 = 63 Critical t(63), two-tailed, α =.05 = 2.656

Step 3: Compute the test statistic sm = (s 2 / n) = (12 2 / 64) = 1.5 = (107 100) / 1.5 = 4.667 Step 4: Make a decision Because the observed / calculated value of t (4.667) is in the tails cut off by the critical t (2.656 the green area in the above diagrams), we reject H0 and conclude that there is sufficient evidence to suggest that the IQ for today s population is different than 10 years ago. b. Make an 80% confidence interval estimate of today s population mean IQ for the 10-yearold test. μ = M ± t sm Find the two-tailed critical t with α = 0.20 (0.20 = (1 0.80 (80% as a proportion))). tcritical(63) = 1.295. Lower bound of the 80% confidence interval: M t sm = 107 1.295 1.5 = 105.0575 Upper bound of the 80% confidence interval: M + t sm = 107 + 1.295 1.5 = 108.9425 5. Problem 18 from the text Other research examining the effects of preschool childcare has found that children who spent time in day care, especially high-quality day care, perform better on math and language tests than children who stay home with their mothers (Broberg, Wessels, Lamb, & Hwang, 1997). Typical results, for example, show that a sample of n = 25 children who attended day care before starting school had an average score of M = 87 with SS = 1536 on a standardized math test for which the population mean is μ = 81. Is this sample sufficient to conclude that the children with a history of preschool day care are significantly different from the general population? Use a two-tailed test with α =.01. Step 1: Hypotheses and α H0: μ = 81 H1: μ 81 α = 0.01

Step 2: Critical region α = 0.01 Two-tailed df = n 1 = 25 1 = 24 tcritical = 2.797 Step 3: Calculate tobserverd s 2 = SS / (n 1) = 1536 / (25 1) = 64 s = s 2 = 64 = 8 sm = (s 2 / n) = (64 / 25) = 1.6 t = (87 81) / 1.6 t = 3.75 Step 4: Decide The observed t (3.75) is in one of the tails cut off by the critical t (2.797), therefore we reject H0. It is likely that attending preschool changes math performance. a. Compute Cohen s d to measure the size of the preschool effect. estimated d = (M μ) / s = (87 81) / 8 = 0.75 b. Write a sentence showing how the outcome of the hypothesis test and the measure of effect size would appear in a research report. Infants who attended preschool scored an average of M = 87 on a standardized math test with SD = 8.00. Statistical analysis indicates that the score is significantly different than the population mean of 80, t(24) = 3.75, p.05, Cohen s d = 0.75. 6. Use SPSS with the class data set to answer the following question: Do students expect to earn at least a B- (more than 240 points) in a statistics class? Give H0, H1 and α. Sketch a t distribution and indicate the critical region(s) on the distribution and the critical t value(s). Is this a one-tailed or twotailed test? From the SPSS output, report the following: Statistic Value M 269.4 df 37 t 10.949 p.000 (remember that SPSS reports a two-tailed p value and its value must be divided by 2 to get a one-tailed p

value) What is the estimated value of Cohen s d? Is this a small, medium, or large effect? What is the value of r 2? Is this a small, medium, or large effect? Write the results of the statistical test in the appropriate format. Step 1: Hypothesis and α H0: μ 248 H1: μ > 248 α =.05 Step 2: Critical region α =.05 One-tailed df = n 1 = 38 1 = 37 tcritical = 1.687 (from a table of critical t values, one-tailed, df = 37, α =.05) Step 3: Calculate tobserved Load the class data set into SPSS. Click on Analyze Compare Means One Sample T Test. Move the Points Expected variable into the Test Variable(s) box on the right. In the Test Value box, type 240. Click OK.

One-Sample Statistics N Mean Std. Deviation Std. Error Mean Points expected 38 269.42 16.564 2.687 One-Sample Test Test Value = 240 95% Confidence Interval of the Difference t df Sig. (2-tailed) Mean Difference Lower Upper Points expected 10.949 37.000 29.421 23.98 34.87 Step 4: Decision Because this is a one-tailed test, we need to compare the observed t to the critical t. The observed t (10.949) is in the tail cut off by the critical t (1.812). Thus we reject H0. It is likely that students expect to earn a B- or better in statistics. Effect Sizes: Estimated Cohen s d: d = (M μ) / s = (269.42 240) / 16.564 = 1.78 This is a large effect. r 2 = t 2 / (t 2 + df) = 10.949 2 / (10.949 2 + 37) = 0.76 This is a large effect. Write the results: Students in previous sections of statistics expect to earn at least 269.4 out of 300 points in the class with a standard deviation of s =16.6. A one-sample t test indicates that this is significantly more than 240 points (a B-), t(37) = 10.949, p <.001, α =.05, r 2 =.76. When reporting the p value from SPSS (or similar software), one typically gives writes p = and gives the precise p value reported in the output. However, if the output reports the p value as equal to.000, then one writes p <.001.