Module 6Hypothesis TestingNameCourse and YearExercise 6.1,1. Suppose you are testing Ho = .45 versus Ha >.45 random sample of 310 peopleproduces a value of p = .465. Use= .05 to test this hypothesis.2. Instructions: Five - step Hypothesis Testing Procedure Used:A criminologist was interested whether there was disparity in sentencing based onthe race of the defendant. She selected at random 18 burglary convictions andcompared the prison terms given to the 10 whites and 8 blacks sampled. Thesentenced length s (in years) are shown for the white and blacks offenders. Using thedata, test the null hypothesis that white and blacks convicted burglary in thisjurisdiction do not differ with respect prison sentence length.Black8735 43547 5 5 6 4 3 2454WhiteNote: Follow all the steps and use two decimal places, round off if possible.Analysis of Variance (ANOVA)Exercise 6.2.1. Compute a one - way ANOVA on the following data. Use .05 level of significance3133413.6.351Determine the observed F value. Compare the observed F value with the criticaltable F value and decide whether to reject the null hypothesis.Note: Use other long bond paper.

Hypothesis testing is an inference process to draw the conclusions about the population parameter.…

Answered: Suppose you are testing Ho =.45 versus…

MATLAB: An Introduction with Applications

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ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

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Problem 1P

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Use a significance level of =α0.05 to test the claim that ≠μ39.2 . The sample data consists of 15 scores for which =x49 and =s6.9 .
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 = 116, x = 12.5, and s = 6.35. Does this data indicate that average daily zinc intake in the population of all males age 65−74 falls below the recommended allowance? (Use ? = 0.05.)State the appropriate null and alternative hypotheses. H0: ? = 15Ha: ? ≠ 15H0: ? = 15Ha: ? > 15 H0: ? = 15Ha: ? ≤ 15H0: ? = 15Ha: ? < 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. Do not reject the null hypothesis. There is not sufficient evidence that average daily zinc intake falls below 15 mg/day.Reject the null hypothesis. There is not sufficient evidence that average daily zinc intake falls below 15 mg/day. Do not reject…
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For the hypothesis test Ho: u = 14 against H1:µ < 14 and variance known, calculate the P-value for the following test statistic: Z0 = - 1.96. Round your answer to three decimal places (e.g. 98.765). P-value
11% of all Americans suffer from sleep apnea. A researcher suspects that a different percentage of those who live in the inner city have sleep apnea. Of the 344 people from the inner city surveyed, 34 of them suffered from sleep apnea. What can be concluded at the level of significance of αα = 0.01? The test statistic= (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.)
59
Use the AustStates data. Perform a repeated measures ANOVA using the 9 years as the paired variables and the states as the observations. Based on the Greenhouse-Geisser p-value of p=0.130, sphericity can be assumed for these data. The result of the analysis is F=9.563 on degrees of freedom, for a p-value of You the null hypothesis. 8 and 56, p<0.001, reject 8 and 56, p<0.001, fail to reject 1.0 and 7.3, p=0.016, reject 1.0 and 7.3, p=0.016, fail to reject
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Part 6 of 7 (f) Determine whether to reject H. Use the a=0.10 level of significance. Do not reject the null hypothesis H Part: 6/7 Part 7 of 7 (g) State a conclusion. There is not enough evidence to conclude that the mean lead amount is less than 15 micrograms per liter. X

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Suppose you are testing Ho =.45 versus Ha > .45 random sample of 310 people produces a value of p= .465. Use= .05 to test this hypothesis. %3D Instruot