In reporting the results of a statistical analysis, a researcher reports that "p = .33". This implies that ________. a. the alternative hypothesis was accepted with an alpha level of .05 b. the alternative hypothesis was not accepted with an alpha level of .05 c. the null hypothesis was not rejected with an alpha level of .05 d. the null hypothesis was rejected with an alpha level of .05

In reporting the results of a statistical analysis, a researcher reports that "p = .33". This implies that ________. a. the alternative hypothesis was accepted with an alpha level of .05 b. the alternative hypothesis was not accepted with an alpha level of .05 c. the null hypothesis was not rejected with an alpha level of .05 d. the null hypothesis was rejected with an alpha level of .05

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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Bo is buying some nails for a project. They are interested in whether the boxes are significantly lighter than advertised. The company says the box of nails should weigh 10 lbs but Bo finds the average weight of 8 boxes is 9.8 lbs with a standard deviation of .4 lbs. Assume an alpha of .10.
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A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals. Data: 4; 2; 1; 3; 7; 2; 9; 4; 6; 6; 7; 0; 5; 6; 4; 2; 1; 3; 4; 2 a. At the α = 0.05 level can it be concluded that the sample mean is higher than 5.8 visits per year? α = Ho: Ha: T – Test inpt: Data Stats µ0: List: Freq:1 µ ≠ µ0 < µ0 > µ0 Calculate T – Test µ0: t = p= sx : n: Decision: Compare the p-value with alpha and give your decision. Conclusion: b. Find a 95 % confidence interval of the true mean years visits of a woman to her doctor. TInterval Inpt: Data Stats sx : n: C-Level: . Calculate T-Interval ( , ) sx : n: Conclusion: We are 95% Confident……
A Moving to the next question prevents changes to this answer. Question 15 An experiment where Preal = 0.10 is less likely to allow rejection of Ho if Ho is false than one where Preal = 0.80. O True O False A Moving to the next question prevents changes to this answer. 17F Sunny の % & * %24 4 %23 L. 8 3 E R Y P J K C al Σ
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Given the following data and using Jamovi, test (at an alpha of 0.05) whether type of music influences heart rate (measured in beats per minute, bpm). Participant Rock Classical Heavy Metal ROW MEANS = 1 70 64 90 74.67 2 78 69 85 77.33 3 80 72 67 73 4 82 65 100 82.33 5 81 67 120 89.33 6 80 58 98 78.67 GROUP MEANS = 78.5 65.83 93.33 Total mean = 79.22 What are the degrees of freedom for between persons? ________ What is the critical value (from table): ___________________
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If the t-statistic for a single sample t-test is 1.58, what can you conclude? a. The test is significant at an alpha level of .05. b. The sample size of the test was at least 30. c. The sample mean was larger than the population mean. d. The population standard deviation was smaller than the difference between the sample mean and the population mean.
Question 4 Suppose that, for some hypothesis test, a=0.05 and the P-value is computed to be 0.754. What is the appropriate conclusion? Reject the null hypothesis at a significance level of 0.05. It is not possible to make a conclusion based on the information provided. Fail to reject the null hypothesis at a significance level of 0.05. Accept the HO as the "P is low." Further testing is required before it can be determined if we should reject or fail to reject the null hypothesis at a significance level of 0.754.
1. If alpha is changed from .05 to .01, it is easier or harder to make a Type 1 error? 2.If alpha is changed from .05 to .01, it is easier or harder to make a Type 2 error? 3.If alpha is changed from .01 to .05, it is easier or harder to make a Type 2 error? 4.If alpha is changed from .01 to .05, it is easier or harder to make a Type 1 error?
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