You may need to use the appropriate appendix table or technology to answer this question.Consider the following hypothesis test.Ho: P 2 0.85H: P<0.85A sample of 170 items was selected. Compute the p-value and state your conclusion for each of the following sample results. Use a = 0.05.(a) P=0.79(b) p=0.82(c) P-0.80(d) p=0.89

The random variable X follows normal distribution. We have to find the p-value and write conclusion.…

Answered: You may need to use the appropriate…

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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Ida Find the P-value for the following values of the test statistic t, Sample size n, and alternate hypothesis H1. If you use table A. 3, you may specify that is between two values. t = -1.584, n = 19, H₁:μ ‡μO
Professor Nord stated that the mean score on the final exam from all the years he has been teaching is a 79%. Colby was in his most recent class, and his class’s mean score on the final exam was 82%. Colby decided to run a hypothesis test to determine if the mean score of his class was significantly greater than the mean score of the population. α = .01. What is the mean score of the population? What is the mean score of the sample? Is this test one-tailed or two-tailed? Why?
Imagine you are doing a Related-Samples t-test, and you are on Step Four of the hypothesis test. The study you are doing is testing whether there’s a significant difference between the anxiety of people before and after a final exam. The APA sentence is t(28) = 3.45, p < .05, d = 1.05. What is the correct set of conclusions that goes with this APA sentence? Group of answer choices Reject the H0, conclude that there IS a difference in anxiety, conclude based on effect size that the difference is small. Fail to Reject the H0, conclude that there IS a difference in anxiety. Reject the H0, conclude that there IS NOT a difference in anxiety, conclude based on effect size that the difference is small. Fail to Reject the H0, conclude that there IS NOT a difference in anxiety. Reject the H0, conclude that there IS a difference in anxiety, conclude based on effect size that the difference is large.
Please help me with this question.
State the Result: A hypothesis test was conducted at the alpha = 0.01 level of significance. The test resulted in a p-value of 0.044.
A news article that you read stated that 51% of voters prefer the Democratic candidate. You think that the actual percent is larger. 139 of the 262 voters that you surveyed said that they prefer the Democratic candidate. What can be concluded at the 0.10 level of significance? For this study, we should use Select an answer z-test for a population proportion t-test for a population mean The null and alternative hypotheses would be: Ho: ? μ p Select an answer = ≠ > < (please enter a decimal) H1: ? p μ Select an answer < = ≠ > (Please enter a decimal) The test statistic ? z t = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is ? > ≤ αα Based on this, we should Select an answer reject fail to reject accept the null hypothesis. Thus, the final conclusion is that ... The data suggest the population proportion is not significantly larger 51% at αα = 0.10, so there is sufficient…
Looking at youth depression scores from a general sample of random children in the US, we want to determine if our sample is significantly different than the national average of scores of children with depression in the US. The national average of scores on a measure of depression is 5 points. We have no reason to assume it will be higher or lower than the US national average. The sample: 11 9 12 3 3 4 13 1. What is the null and alternative hypotheses? Do not specify any directionality. 2. What did you discover (using 95% Cl)? Provide an interpretation.
econometrics
b c d
Please answer this with solution steps
A health manager wants to assess the relationship between number of doctor visits and patient feedbacks.Patient feedbacks are categorized between 1= very bad and 5= very goodin a scale. The SPSS analysis is given below. Write down the hypothesis for this case and specify the test used and comment on the results. Patient Feedback Visit to Doctor 5.00 2.00 4.00 1.00 4.00 2.00 5.00 3.00 4.00 3.00 3.00 4.00 3.00 4.00 4.00 4.00 2.00 5.00 2.00 5.00 3.00 5.00 1.00 4.00 One-Sample Kolmogorov-Smirnov Test Kisisel Saglik Degerlendirmesi Gecen Yil Icindeki Hekim Ziyaret Sayisi N 12 12 Normal Parametersa,b Mean 3.3333 3.5000 Std. Deviation 1.23091 1.31426 Most Extreme Differences Absolute .206 .232 Positive .127 .127 Negative -.206 -.232 Kolmogorov-Smirnov Z .713 .802 Asymp. Sig. (2-tailed) .689 .541 a. Test…
A sample of 100 yielded a test statistic of 2.1. For a level of significance of 0.05, the null hypothesis was rejected. What was the sample proportion for the sample of .1 100? What was the p-value for this test? 0.864

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