The post-hoc test is used to ________. . Determine the standard error Test the statistical significance of pairs of means Determine whether ANOVA should be performed Test the statistical significance of the main effect

The post-hoc test is used to ________. . Determine the standard error Test the statistical significance of pairs of means Determine whether ANOVA should be performed Test the statistical significance of the main effect

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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use pic to answer, test is anova one way test 1. Did researchers use the correct test? a. yes b. no 2. Justify your answer, above. a. ANOVa can be used when you have three or more levels of an independent variable b. regression analysis would be prefereable because the independent variable is contiuous c. anovas primary strength is its ability to examine for difference in two groups d. you can use anove wiht three levels, but a t-test would be easier and give you the same results 3. How would you interpret the results of this oneway ANOVA? a. there is likely a difference in happiness in the "population" b. there is no mean differences in happiness across the three conditions c. there are mean differences but we do now know which conditions differ d. those who spend time reminiscing are signficantly more likely to be happy than those who dont
The total body bone mineral content is a measure of bone health. Two operators were assigned to collect measurements on the same individuals in a paired experiment. The results below were obtained. Test at the 5% significance level the null hypothesis that the two operators have the same mean against the alternative that the means are different. The value of the test statistic and the decision are
Test the claim that for the population of statistics final exams, the mean score is 80 using alternative hypothesis that the mean score is different from 80. Sample statistics include ?=22, ?⎯⎯⎯=81,, and ?=19. Use a significance level of ?=0.05.. (Assume normally distributed population.) The test statistic is:The positive critical value is: The negative critical value is: The conclusion is A. There is not sufficient evidence to reject the claim that the mean score is equal to 80.B. There is sufficient evidence to reject the claim that the mean score is equal to 80.
Test the claim that the mean GPA of night students is larger than the mean GPA of day students at the 0.025 significance level. The null and alternative hypothesis would be: Họ : UN = UD Họ: PN = PD Họ : PN = PD Ho : UN = HD Ho: PN = PD Ho: uN = HD H1 : UN > HD Hị : PN PD H1 : UN + HD H1:PN + PD H1: µN < HD The test is: two-tailed right-tailed left-tailed The sample consisted of 80 night students, with a sample mean GPA of 2.26 and a standard deviation of 0.03, and 80 day students, with a sample mean GPA of 2.23 and a standard deviation of 0.06. The test statistic is: (to 2 decimals) The p-value is: (to 2 decimals) Based on this we: Reject the null hypothesis O Fail to reject the null hypothesis
Construct a 95% confidence interval for the true difference in mean scores for these two groups. What does your interval say about the claim that there is no difference in mean scores?
Testing for a Difference in Pulse Rates Exercise 6.250 describes pulse rates collected from 10 students, once during a quiz and once during a lecture. We might expect mean pulse rates to increase under the stress of a quiz. Use the information in Exercise 6.250 and the data in Table 6.23 to conduct a matched pairs test and compute the sample statistic. Explain why a matched pairs test is appropriate.whether the data provides sufficient evidence to support this claim. Student Quiz Lecture Differences1 75 73 22 52 53 -13 52 47 54 80 88 -85 56 55 16 90 70 207 76 61 158 71 75 -49 70 61…
A real estate agent believes that the mean home price in the northern part of a county is higher than the mean price in the southern part of the county and would like to test the claim. A simple random sample of housing prices is taken from each region. The results are shown below. Mean Variance Observations Pooled Variance Hypothesized Mean Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail Confidence Level n = Ex: 9 Southern Northern 155.056 168.889 345.938 560.928 18 18 453.433 0 34 1.949 0.030 2.441 0.060 2.728 99% -3 -2 7 -1 0 Degrees of freedom: df = Point estimate for the southern part of the county: xs = Ex: 1.234 Point estimate for the northern part of the county: N = t 1 = р 2 = Ex: 1.234 3 Ex: 1.234
Read the following scenario and determine which test is the most appropriate. To test whether Internet use increases depression score, we measure the depression scores of a random sample of non-Internet-users, have them use the Internet for a specified time, then measure their depression scores again. O Inference for regression ANOVA O Two-Sample t-test O Matched-Pairs t-test O Chi-Squared test for independence
You'd like to test if marijuana use (no use, use) is associated with depression scores (O to 40) in a matched case-control based sample. One Sample Z test One Sample t test Two Sample Z test O Independent two Sample t test O Paired t test One-way ANOVA Chi Square test Pearson's Correlation Coefficient Simple Linear Regression
A real estate agent believes that the mean home price in the northern part of a county is higher than the mean price in the southern part of the county and would like to test the claim. A simple random sample of housing prices is taken from each region. The results are shown below. Mean Variance Observations Pooled Variance Hypothesized Mean Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail Confidence Level Southern Northern 155.056 168.889 345.938 560.928 18 18 453.433 0 34 1.949 0.030 2.441 0.060 2.728 99% n = Ex: 9 Degrees of freedom: df = -2 -1 0 Point estimate for the southern part of the county: s = Ex: 1.234 Point estimate for the northern part of the county: N = P = Ex: 1.234 2 t Ex: 1.234
Test the claim that the proportion of men who own cats is smaller than 60% at the .01 significance level. Based on a sample of 100 people, 59% owned catsThe test statistic is: (to 2 decimals)The critical value is: (to 2 decimals)
Would you favor spending more federal tax money on the arts? Of a random sample of n1 = 217 women, r1 = 52 responded yes. Another random sample of n2 = 175 men showed that r2 = 48 responded yes. Does this information indicate a difference (either way) between the population proportion of women and the population proportion of men who favor spending more federal tax dollars on the arts? Use ? = 0.05. (a) What is the level of significance? (b) What is the value of the sample test statistic? (Test the difference p1 − p2. Do not use rounded values. Round your final answer to two decimal places.)(c) Find (or estimate) the P-value. (Round your answer to four decimal places.)