A 95% confidence interval for u was given as -5.65 < u < 2.61. What would a testfor Ho : µ = 0 vs H1 : µ # 0 conclude?A. reject the null hypothesis at a =0.05 and all smaller aB. fail to reject the null hypothesis at a == 0.05 and all smaller aC. reject the null hypothesis at a = 0.05 and all larger aD. fail to reject the null hypothesis at a =- 0.05 and all larger

We have given that 95% confidence interval for α=0.05. ( -5.65 , 2.61 ) Null and alternative…

Answered: 5% confidence interval for u was given…

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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Construct the 90 % confidence interval for the difference −μ1μ2 when =x1479.30 , =x2323.56 , =s145.04 , =s222.40 , =n19 , and =n217 . Use tables to find the critical value and round the answers to at least two decimal places. A 90% confidence interval for the difference in the population means is
60% of students entering four-year colleges receive a degree within six years. Is this percent larger than for students who play intramural sports? 172 of the 249 students who played intramural sports received a degree within six years. What can be concluded at the level of significance of αα = 0.10? For this study, we should use (t-test, z-test) The null and alternative hypotheses would be: Ho: (symbol) (symbol) _____ (please enter a decimal) H1: (symbol) (symbol) _____ (Please enter a decimal) The test statistic (symbol) = _____ (please show your answer to 3 decimal places.) The p-value = _______ (Please show your answer to 4 decimal places.) The p-value is (symbol) αα Based on this, we should the null hypothesis. Thus, the final conclusion is that ... The data suggest the population proportion is not significantly larger than 60% at αα = 0.10, so there is sufficient evidence to conclude that the population proportion of students who…
#6.3 (p.344). Conduct a test of H0: μ1>= μ2-2.3 versus Ha: μ1< μ2-2.3 for the sample data summarized here. Use α=0.01 in reaching your conclusions.
An engineer who is studying the tensile strength of a steel alloy intended for use in golf club shafts knows that tensile strength is approximately normally distributed with o = 60 psi. A random sample of 12 specimens has a mean tensile strength of X = 3250 psi. (a) Test the hypothesis that mean strength is 3500 psi. Use a= 0.01. (b) What is the smallest level of significance at which you would be %3D willing to reject the null hypothesis? (c) Explain how you could answer the question in part (a) with a two-sided confidence interval on mean tensile strength.
You are performing a two-tailed matched-pairs test with 35 pairs of data.If α=0.1α=0.1, find the positive critical value, to two decimal places.
The mean blood cholesterol level for all men aged 20 to 34 years is μ=188μ=188 mg/dl. We suspect that the mean for cross-country runners is lower. We take a random sample from all cross-country runners and measure the cholesterol level of each runner. We set α=0.10α=0.10 and correctly calculate a p−valuep−value of 0.08. Suppose someone else has additional information unavailable to us and it is known that IN REALITY cross-country runners actually do have lower cholesterol levels. Based on our calculations, what conclusion did we reach and did we make an error? The mean blood cholesterol level for all men aged 20 to 34 years is μ=188μ=188mg/dl. We suspect that the mean for cross-country runners is lower. We take a random sample from all cross-country runners and measure the cholesterol level of each runner. We set α=0.10α=0.10and correctly calculate a p−valuep−valueof 0.08. Suppose someone else has additional information unavailable to us and it is known that IN REALITY…
I keep getting different answers for this so I'm checking
16% of all Americans suffer from sleep apnea. A researcher suspects that a lower percentage of those who live in the inner city have sleep apnea. Of the 398 people from the inner city surveyed, 52 of them suffered from sleep apnea. What can be concluded at the level of significance of αα = 0.01? For this study, we should use Z TEST FOR A POPULATION PROPORTION The null and alternative hypotheses would be: Ho: P=0.15 (please enter a decimal) H1: P <0.1 3. The test statistic _z__ = _______ (please show your answer to 3 decimal places.) 4. The p-value = __0.0144__ (Please show your answer to 4 decimal places.) 5. The p-value is __<__α 6. Based on this, we should ___FAIL TO REJECT___ the null hypothesis. Thus, the final conclusion is that ... The data suggest the populaton proportion is significantly smaller than 16% at αα = 0.01, so there is sufficient evidence to conclude that the population proportion of inner city residents who have sleep apnea is…
Only 15% of registered voters voted in the last election. Will voter participation increase for the upcoming election? Of the 363 randomly selected registered voters surveyed, 58 of them will vote in the upcoming election. What can be concluded at the αα = 0.01 level of significance? For this study, we should use Select an answer t-test for a population mean z-test for a population proportion The null and alternative hypotheses would be: H0:H0: ? μ p Select an answer > < ≠ = (please enter a decimal) H1:H1: ? μ p Select an answer = < ≠ > (Please enter a decimal) The test statistic ? t z = (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 accept reject fail to reject the null hypothesis. Thus, the final conclusion is that ... The data suggest the population proportion is not significantly higher than 15% at αα = 0.01, so there…
59% of students entering four-year colleges receive a degree within six years. Is this percent larger than for students who play intramural sports? 148 of the 232 students who played intramural sports received a degree within six years. What can be concluded at the level of significance of αα = 0.01? 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: ? p μ Select an answer ≠ > = < (please enter a decimal) H1: ? μ p 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 than 59% at αα =…
Only 16% of registered voters voted in the last election. Will voter participation increase for the upcoming election? Of the 327 randomly selected registered voters surveyed, 56 of them will vote in the upcoming election. What can be concluded at the αα = 0.05 level of significance? For this study, we should use: z-test for a population proportion or t-test for a population mean The null and alternative hypotheses would be: H0:H0: ? μ p Select an answer ≠ < = > (please enter a decimal) H1:H1: ? p μ Select an answer > = ≠ < (Please enter a decimal) The test statistic ? z or 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: accept, reject or fail to reject the null hypothesis. Thus, the final conclusion is that ... The data suggest the population proportion is not significantly higher than 16% at αα = 0.05, so there…
An aneroid sphygmomanometer is a mechanical device used to measure blood pressure. A simple random sample of these devices is tested for accuracy and the errors (mm Hg) are listed below. One of the devices is considered to be unacceptable if its error is more than 33 mm Hg. The criterion for concluding that the sample is from a population of unacceptable devices is sigmaσgreater than>1.51.5 mm Hg. Use a 0.050.05 significance level with the sample data to test the claim that the sample is from a population with a standard deviation greater than 1.51.5 mm Hg. Assume that the population is normally distributed. -7 -10 5 3 9 14 -12 -15 4 4 -3 What are the correct hypotheses for this test? Calculate the value of the test statistic. nothing (Round to two decimal places as needed.) Use technology to determine the P-value for the test statistic. The P-value is nothing. (Round to three decimal places as needed.) What is the conclusion at the 0.050.05…

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