D. Surprisingly, a study found that children under stress reported toms of depression and anxiety but scored higher on a Lie Scale (a measure of tendency to give socially desirable responses). The population mean for the Lie Scale for nonstressed children is 3.8. A sample of 40 children under stress had a Lie Scale sample mean of 4.4, with a standard deviation of 2.5. Test the hypothesis that children under stress are more likely to give socially desirable answers. Write a complete conclusion in the context of the problem. Syt

D. Surprisingly, a study found that children under stress reported toms of depression and anxiety but scored higher on a Lie Scale (a measure of tendency to give socially desirable responses). The population mean for the Lie Scale for nonstressed children is 3.8. A sample of 40 children under stress had a Lie Scale sample mean of 4.4, with a standard deviation of 2.5. Test the hypothesis that children under stress are more likely to give socially desirable answers. Write a complete conclusion in the context of the problem. Syt

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Description: 20. Surprisingly, a study found that children under stress reported fewer symp-toms of depression and anxiety but scored higher on a Lie Scale (a measure oftendency to give socially desirable responses). The population mean for theLie Scale for nons

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Of interest is whether the percent of managers at High Tech Inc. that pay over $8 per day for lunch is the same as the percent of technicians. Sixty managers are surveyed and forty-six of them pay over $8 per day for lunch. One hundred forty-nine techniciansare surveyed and ninety-eight of them pay over $8 per day for lunch. The hypothesis test to use is???? independent group means, population standard deviations known independent group means, population standard deviations unknown matched or paired samples two proportions
Suppose that the average gregariousness score for the entire U.S. population is 70. The head research psychologist at Somerville University is interested in determining whether or not the university psychology students are statistically significantly more gregarious than the overall population. A random sample of 200 students from Somerville University shows a mean of 75 on the gregarious test with a sample standard deviation of 20. a. State the research hypothesis. b. State the null hypothesis. c. Identify the test statistic and provide the formula.
2. Does helping another person affect the helper's life satisfaction? To find out, a researcher randomly sampled 64 people and asked each of them to help one other person that day. At the end of the day, each participant completed a standardized measure of life satisfaction. The mean score for the sample was 51.60. In the general population, the mean score on this measure is 50 and the standard deviation is 6. Test the hypothesis that helping affects life satisfaction (a = .05, two-tailed). You don't have to outline all 5 steps of hypothesis testing, but specify the properties of the comparison distribution, the critical value of the test statistic for rejecting the null hypothesis, the obtained test statistic, and the final decision. What are the lower and upper confidence limits for the 95% confidence interval of the mean life satisfaction of people who have helped someone else that day? Compute the standardized effect size.
A professor believes that, for the introductory art history classes at his university, the mean test score of students in the evening classes is lower than the mean test score of students in the morning classes. He collects data from a random sample of 250 students in evening classes and finds that they have a mean test score of 76.8. He knows the population standard deviation for the evening classes to be 7.2 points. A random sample of 200 students from morning classes results in a mean test score of 77.8. He knows the population standard deviation for the morning classes to be 1.9 points. Test his claim with a 90% level of confidence. Let students in the evening classes be Population 1 and let students in the morning classes be Population 2. Step 1 of 3 : State the null and alternative hypotheses for the test. Fill in the blank below. H0: μ1=μ2 Ha: μ1__μ2
A researcher wants to know if men have a lower anxiety level than women in response to social rejection. The researcher brings 9 men and 16 women into a lab. He tells them all to write about a time they were socially rejected and he then measures their levels of anxiety using a 0-100 scale. The mean anxiety for men (A) was 81.67 with a standard deviation of 8.78. For women (B), he obtained a mean of 85.69 with a standard deviation of 4.8. A). What is the type of test you will use? (z-test, single-sample t-test, paired-samples t-test, or independent samples t-test) and why (what information provided in the problem) B). What are the hypotheses (Be Specific)
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IQ tests are approximately normally distributed with a mean of 100 for adults. A tech company CEO wants to conduct a hypothesis test to see if the average IQ of employees at high-tech firms in California is higher than average. What is Type I error in the context of this problem?
A survey was conducted where 27 students were asked which social media app is their favorite, Twitter or Tiktok. male female twitter 4 4 TikTok 4 15 create and interpret (in context) a one-proportion confidence interval. You may choose the level of confidence. conduct a Chi-Squared test for Association to determine if your data provide statistically significant evidence of an association between your two variables.
A marketing consultant is hired by a major restaurant chain wishing to investigate the preferences and spending patterns of lunch customers. The CEO of the chain hypothesized that the average customer spends at least $13.50 on lunch. A survey of 25 customers sampled at one of the restaurants found the average lunch bill per customer to be ?¯=$14.50 . Based on previous surveys, the restaurant informs the marketing manager that the standard deviation is ?=$3.50 . To address the CEO’s conjecture, the marketing manager carried out a hypothesis test of ?0:?=13.50 vs. ??:?>13.50 and obtained a ?‑value = 0.77. The marketing chooses a significance level of ?=0.10. If he uses this significance level throughout his work, how often will he reject a true null hypothesis? Group of answer choices a.He will reject 10% of all true null hypotheses. b. He will reject 1% of all true null hypotheses. c. He will reject 5% of all true null hypotheses. d. He will not reject 10% of all true null…
You are interested in seeing whether emotions impact decision making. You have three groups--a happy group, a sad group, and a neutral group. For the happy group there are 5 participants and the mean risky decision making score 5.0 with a standard deviation of 0.7. For the sad group there are 5 participants and the mean risky decision making score 5.4 with a standard deviation of 1.1. For the neutral group there are 5 participants and the mean risky decision making score 5.8 with a standard deviation of 2.8. The sum of squares between samples is equal to 1.6. The sum of squares within samples is equal to 38.0. You calculated the f-ratio above. The corresponding p-value for the f-ratio you calculated is p = 0.78. Is this finding significant?
A successful basketball player has a height of 6 feet 11 inches, or 211 cm. Based on statistics from a data set, his height converts to the z score of 5.17. How many standard deviations is his height above the mean?
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