Say that we have a null hypothesis that X1,..., X16 that are Normally distributed withmean 0 and variance 1. Call the sample mean X. Consider the test wherep-value = 20 (4 X) -24 (4|X) – 1We will reject this if p-value < a.a) Show that the type I error rate of this test is a.b) Say that X1,..., X16 that are Normally distributed with mean 1 and variance 1.What is the type II error rate of this test when a = 0.05?•..

Answered: Say that we have a null hypothesis that…

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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Let x represent the number of mountain climbers killed each,year. The long-term variance of x is approximately o Suppose that for the past 12 years, the variance has been s = 118.3. = 136.2 Use a 1% level of significance to test the claim that the recent variance for number of mountain-climber deaths is less than 136.2 Find a 90% confidence interval for the population variance. (b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.) What are the degrees of freedom?
Suppose you needed to test the claim that the two samples described below come from populations with the same mean. Assume that the samples are independent simple random samples. Sample 1: n1 13, a1 = 24, 81 = 7.55 Sample 2: n2 = 3, x2 = 21.9, 82 = 8.92 Find: (a) The estimated degree of freedom is (b) The standardized test statistic is (use Sample 1 - Sample 2)
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Let x represent the average annual salary of college and university professors (in thousands of dollars) in the United States. For all colleges and universities in the United States, the population variance of x is approximately o² = 47.1. However, a random sample of 17 colleges and universities in Kansas showed that x has a sample variance s² = 82.7. Use a 5% level of significance 1 test the claim that the variance for colleges and universities in Kansas is greater than 47.1. (a) What is the level of significance? State the null and alternate hypotheses. Ho: o²=47.1; H: o> 47.1 Ho: o²=47.1; H: o2 0.100 0.050 a, we fail to reject the null hypothesis. Since the P-value > x, we reject the null hypothesis. Since the P-values ∞, we reject the null hypothesis. Since the P-values a, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application. At the 5% level of significance, there is insufficient evidence to conclude the variance of annual salaries…
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A study tested the claim that heights of Americans and heights of Austrians have different variances with s=7.48092 cm for Americans and 7.14756 cm for Austrians. The sample sizes are n1= 145 and n2= 157. When using the F test with these data, is it correct to reason that there is no need to check for normalcy because n1 is greater than 30, and n2 is greater than 30?
I need the answer as soon as possible
Independent simple random samples are taken to test the difference between the means of two populations whose variances are not known. The sample sizes are n1 = 12 and n2 = 14. Which is the correct distribution to use?
A certain IQ test is known to have a population mean of 100 and standard deviation of 15 in the general population. You want to test whether psychology majors have a different average IQ than the population as a whole. Assume the variance of IQ is the same for Psych majors as it is in the general population. Suppose that Psychology majors actually have an average IQ of 108. If you do a 2-tailed test at α= .05 with a sample of 56 Psychology majors, you will be able to reject the null hypothesis if the mean IQ of your sample is below [L] or above [H]. Find L and H values. Options listed below. [L] answer choices: 96.08, 98.00, 103.92, 104.08, 110.00, 111.92. [H] Answer choices: 96.08, 98.00, 103.92, 104.08, 110.00, 111.92.
To evaluate the effect of a treatment, a sample is obtained from a population with a mean of =40, and the treatment is administered to the individuals in the sample. After treatment, the sample mean is found to be M = 44.5 with a variance of s² = 36. If the sample consists of n = 16 individuals, are the data sufficient to conclude that the treatment has a significant effect using a two-tailed test with a = .05? O For these data, t= 1.50. This value is less than the critical value of 2.131, so we reject the null hypothesis and conclude that there is a significant treatment effect. O For these data, t = 3.00. This value is greater than the critical value of 2.131, so we reject the null hypothesis and conclude that there is a significant treatment effect. O For these data, t = 4.50. This value is greater than the critical value of 2.131, so we reject the null hypothesis and conclude that there is a significant treatment effect. O For these data, t = 1.50. This value is less than the…
2. The average zone of inhibition (in mm) for mouthwash L as tested by the medical technology students has been known to be 9mm. A random sample of 10 mouthwash L was tested and the test yielded an average zone of inhibition of 7.5mm with a variance of 25 mm. Is there enough reason to believe that the anti-bacterial property of the mouthwash has decreased? Test the hypothesis that the average zone of inhibition of the mouthwash is no less than 9mm using 0.05 level of significance. A. State the hypotheses. B. Determine the test statistic to use. C. Determine the level of significance, critical value, and the decision rule. D. Compute the value of the test statistic. E. Make a decision. F. Draw a conclusion.

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