EBK RESEARCH METHODS FOR THE BEHAVIORAL
6th Edition
ISBN: 8220106720271
Author: Forzano
Publisher: YUZU
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Chapter 15.3, Problem 7LO
To determine
the general purpose of hypothesis testing also how the test is helping the research goal.
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Chapter 15 Solutions
EBK RESEARCH METHODS FOR THE BEHAVIORAL
Ch. 15.1 - Define the two general categories of statistics,...Ch. 15.1 - Prob. 2LOCh. 15.2 - Prob. 3LOCh. 15.2 - Prob. 4LOCh. 15.2 - Prob. 5LOCh. 15.2 - Prob. 6LOCh. 15.3 - Prob. 7LOCh. 15.3 - Prob. 8LOCh. 15.3 - Explain how sample size and sample variance...Ch. 15.4 - Identify the characteristics that differentiate...
Ch. 15.5 - Describe the two basic concerns with the...Ch. 15.5 - Describe the basic concern with measuring...Ch. 15 - In addition to the key words, you should also be...Ch. 15 - Prob. 2ECh. 15 - A researcher who is interested in examining the...Ch. 15 - Construct a frequency distribution histogram or...Ch. 15 - Under what circumstances are the median and the...Ch. 15 - Describe a distribution of scores that has a mean...Ch. 15 - Prob. 7ECh. 15 - Describe the general concept of sampling error and...Ch. 15 - Briefly explain what is meant when a researcher...Ch. 15 - Prob. 10ECh. 15 - Describe the relationship between the alpha level...Ch. 15 - Explain how increasing the size of the sample can...Ch. 15 - Prob. 13ECh. 15 - Identify the appropriate hypothesis test for each...Ch. 15 - Prob. 15ECh. 15 - Prob. 16E
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- BUSINESS DISCUSSarrow_forwardA researcher wishes to estimate, with 90% confidence, the population proportion of adults who support labeling legislation for genetically modified organisms (GMOs). Her estimate must be accurate within 4% of the true proportion. (a) No preliminary estimate is available. Find the minimum sample size needed. (b) Find the minimum sample size needed, using a prior study that found that 65% of the respondents said they support labeling legislation for GMOs. (c) Compare the results from parts (a) and (b). ... (a) What is the minimum sample size needed assuming that no prior information is available? n = (Round up to the nearest whole number as needed.)arrow_forwardThe table available below shows the costs per mile (in cents) for a sample of automobiles. At a = 0.05, can you conclude that at least one mean cost per mile is different from the others? Click on the icon to view the data table. Let Hss, HMS, HLS, Hsuv and Hмy represent the mean costs per mile for small sedans, medium sedans, large sedans, SUV 4WDs, and minivans respectively. What are the hypotheses for this test? OA. Ho: Not all the means are equal. Ha Hss HMS HLS HSUV HMV B. Ho Hss HMS HLS HSUV = μMV Ha: Hss *HMS *HLS*HSUV * HMV C. Ho Hss HMS HLS HSUV =μMV = = H: Not all the means are equal. D. Ho Hss HMS HLS HSUV HMV Ha Hss HMS HLS =HSUV = HMVarrow_forward
- Question: A company launches two different marketing campaigns to promote the same product in two different regions. After one month, the company collects the sales data (in units sold) from both regions to compare the effectiveness of the campaigns. The company wants to determine whether there is a significant difference in the mean sales between the two regions. Perform a two sample T-test You can provide your answer by inserting a text box and the answer must include: Null hypothesis, Alternative hypothesis, Show answer (output table/summary table), and Conclusion based on the P value. (2 points = 0.5 x 4 Answers) Each of these is worth 0.5 points. However, showing the calculation is must. If calculation is missing, the whole answer won't get any credit.arrow_forwardBinomial Prob. Question: A new teaching method claims to improve student engagement. A survey reveals that 60% of students find this method engaging. If 15 students are randomly selected, what is the probability that: a) Exactly 9 students find the method engaging?b) At least 7 students find the method engaging? (2 points = 1 x 2 answers) Provide answers in the yellow cellsarrow_forwardIn a survey of 2273 adults, 739 say they believe in UFOS. Construct a 95% confidence interval for the population proportion of adults who believe in UFOs. A 95% confidence interval for the population proportion is ( ☐, ☐ ). (Round to three decimal places as needed.)arrow_forward
- Find the minimum sample size n needed to estimate μ for the given values of c, σ, and E. C=0.98, σ 6.7, and E = 2 Assume that a preliminary sample has at least 30 members. n = (Round up to the nearest whole number.)arrow_forwardIn a survey of 2193 adults in a recent year, 1233 say they have made a New Year's resolution. Construct 90% and 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. The 90% confidence interval for the population proportion p is (Round to three decimal places as needed.) J.D) .arrow_forwardLet p be the population proportion for the following condition. Find the point estimates for p and q. In a survey of 1143 adults from country A, 317 said that they were not confident that the food they eat in country A is safe. The point estimate for p, p, is (Round to three decimal places as needed.) ...arrow_forward
- (c) Because logistic regression predicts probabilities of outcomes, observations used to build a logistic regression model need not be independent. A. false: all observations must be independent B. true C. false: only observations with the same outcome need to be independent I ANSWERED: A. false: all observations must be independent. (This was marked wrong but I have no idea why. Isn't this a basic assumption of logistic regression)arrow_forwardBusiness discussarrow_forwardSpam filters are built on principles similar to those used in logistic regression. We fit a probability that each message is spam or not spam. We have several variables for each email. Here are a few: to_multiple=1 if there are multiple recipients, winner=1 if the word 'winner' appears in the subject line, format=1 if the email is poorly formatted, re_subj=1 if "re" appears in the subject line. A logistic model was fit to a dataset with the following output: Estimate SE Z Pr(>|Z|) (Intercept) -0.8161 0.086 -9.4895 0 to_multiple -2.5651 0.3052 -8.4047 0 winner 1.5801 0.3156 5.0067 0 format -0.1528 0.1136 -1.3451 0.1786 re_subj -2.8401 0.363 -7.824 0 (a) Write down the model using the coefficients from the model fit.log_odds(spam) = -0.8161 + -2.5651 + to_multiple + 1.5801 winner + -0.1528 format + -2.8401 re_subj(b) Suppose we have an observation where to_multiple=0, winner=1, format=0, and re_subj=0. What is the predicted probability that this message is spam?…arrow_forward
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