A randomized clinical trial was designed to have power of 80% (i.e. Type IIerror probability of 20%) to detect a mean treatment effect of 8 > 0 at the 5% level . We assume that thetreatment effect values are normally distributed with the mean u and the standard deviation o = 108. Basedon the research question, we can set both null and alternative hypotheses as follows:Ho : µ = 0 vs. H1 : µ = 8.Find the sample size needed.

Answered: Image /qna-images/answer/78db14a5-cffa-493c-95f3-8dc536ae6e74.jpg

Answered: A randomized clinical trial was…

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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T/F: An R^2 value near 100% indicates that there is a strong causal relationship between the response and the predictor variables. T/F: Standard deviation of the sum of two independent random variables is the sum of their standard deviations.
The recommended daily dietary allowance for zinc among males older than age 50 years is 15 mg/day. An article reports the following summary data on intake for a sample of males age 65-74 years: n = 113, x = 12.4, and s = 6.63. Does this data indicate that average daily zinc intake in the population of all males age 65-74 falls below the recommended allowance? (Use a = 0.05.) State the appropriate null and alternative hypotheses. O Ho: H = 15 H: u > 15 O Ho: H = 15 H: u < 15 Ο H: μ15 H:us 15 Ο Η: μ- 15 H: u = 15 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = p-value = State the conclusion in the problem context. O Reject the null hypothesis. There is sufficient evidence that average daily zinc intake falls below 15 mg/day. O Do not reject the null hypothesis. There is sufficient evidence that average daily zinc intake falls below 15 mg/day. O Do not reject the null hypothesis. There…
You wish to test the following claim (HaHa) at a significance level of α=0.05 Ho:μ1=μ2 Ha:μ1≠μ2You believe both populations are normally distributed, but you do not know the standard deviations for either. However, assume that the variances of the two populations are equal. You obtain a sample of size n1=23 with a mean of ¯x1=74.5 and a standard deviation of SD1=20.4 from the first population. You obtain a sample of size n2=11with a mean of ¯x2=89.8and a standard deviation of SD2=7.3 from the second population.What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = The p-value is... less than (or equal to) αα greater than αα This test statistic leads to a decision to... reject the null accept the null fail to reject the null As such, the final conclusion is that... There is sufficient evidence to warrant rejection of…
Suppose that there are two borrowing strategies from a commercial bank as short-term and long-term. We have two small samples (n1=8 and n2=8) and sampled populations are normal. Standard deviation for the first sample is 200 and for the second one is 150. The researcher wants to determine whether the variation in the customers preferring short-term borrowing differs from the variation in the customers preferring long-term borrowing. (Use 0.10 significance level)
An insurance company is interested in studying the mean hospitalization time of patients who suffer strokes at a certain hospital. The hospital claims the mean length of stay is 22 days, but the insurance company believes it is less than claimed. A random sample of 9 patients showed a sample mean stay of 18 days with a sample standard deviation of 3 days. They would like to conduct the test using alpha = 0.1. %3D Calculate the test statistic for the test and enter your answer in the first blank. Round your answer to two decimals. Enter the critical value cutoff point for the rejection region in the second blank. Blank # 1 Blank # 2 Next Page Page 8 of 15 126 NOV 18 MacBook Pro
A random sample of 23 bags of apples (marked as 10 pounds each) received by a large grocery chain tested out as having a mean of 9.2 pounds with a variance of 2.56 pounds. Test whether the true mean of all bags is under 10 pounds. Set up hypotheses. Perform the appropriate test by showing your formula. Interpret the results using a Type I (alpha) error of .05. Also provide the p value here. Also construct a 95% confidence interval around your sample mean (X bar) that should contain the true mean (mu). Also interpret this interval.
Scores on an IQ test are normally distributed. A sample of 25 IQ scores had variance of 64. The developer of the test claims that the population standard deviation is 15. Do these data provide sufficient evidence to significance? Use α=0.05α=0.05 . Answer the following questions. (a) What is the parameter? . (input as "mean", "standard deviation", "variance" or "proportion") (b) Determine if the alternative hypothesis is "right", "left" or "two-tailed" test. (input as "right", "left" or "two-tailed") (c) Test by using a confidence interval. Round confidence interval to nearest thousandth. (e.g. 0.1345 would be entered as 0.135) between ( and ) (d) State conclusion.: There is (input as "sufficient" or "insufficient") to (input as "conclude" or "reject") to conclude that the population standard deviation of IQ test is 15.
After conducting your one-sample z-test, you find an obtained z-statistic of 1.08 and a critical z-statistic of 1.64. What do you conclude? Group of answer choices Fail to reject the null hypothesis; the sample mean and population mean are significantly different. Reject the null hypothesis; the sample mean and population mean are significantly different. Fail to reject the null hypothesis; the sample mean and population mean are not significantly different. The results of the test are inconclusive.
In a random sample of size 36 we observe a sample mean of 25 and a variance of 144. If we test the hypotheses below with a significance level of 0.05. Ho: µ = 20 HẠ :4 > 20 what will be the calculated test statistic?
My answer is as follows: T critical = 2.70 T value= 7.775 P< a (0.01) Therefore we reject the null. Is this correct?
In experimental situations, care is taken to control for as many factors as possible which might influence the results so as to increase the confidence of the causal relationship between the independent and dependent variables. In cases where results are of marginal statistical significance, say just less than 0.05, how might this level of control affect the ability to apply the results in real world settings?
You wish to test the following claim (HaHa) at a significance level of α=0.01α=0.01. Ho:μ1=μ2 Ha:μ1≠μ2You believe both populations are normally distributed, but you do not know the standard deviations for either. However, assume that the variances of the two populations are equal. You obtain a sample of size n1=20 with a mean of ¯x1=55.2 and a standard deviation of SD1=11.3SD1=11.3 from the first population. You obtain a sample of size n2=18 with a mean of ¯x2=48.3 and a standard deviation of SD2=7.2 from the second population.What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = The p-value is... less than (or equal to) αα greater than αα This test statistic leads to a decision to... reject the null accept the null fail to reject the null As such, the final conclusion is that... There is sufficient evidence to…

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