### QuestionManufacturers are testing a die to make sure that it is fair (has a uniform distribution). They roll the die 72 times and record the outcomes.They conduct a chi-square Goodness-of-Fit hypothesis test at the 5% significance level.(a) The null and alternative hypotheses are:- \( H_0 \): The die has the uniform distribution.- \( H_a \): The die does not have the uniform distribution.(b) \( \chi^2_0 = 4.500. \)(c) \( \chi^2_{0.05} = 11.070. \)(d) What conclusions can be made? Select all that apply.#### Select all that apply:- [ ] We should reject \( H_0 \).- [x] We should not reject \( H_0 \).- [ ] At the 5% significance level, there is sufficient evidence to conclude that the die is not fair.- [x] At the 5% significance level, there is not enough evidence to conclude that the die is not fair.

Answered: Question Manufacturers are testing a…

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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You are conducting a study to see if the proportion of voters who prefer Candidate A is significantly more than 0.57. You use a significance level of a = 0.001. Ho:p = 0.57 H1:p > 0.57 You obtain a sample of size n = 331 in which there are 211 successes.
Test the claim that the proportion of people who own cats is significantly different than 80% at the 0.1 significance level. The alternative hypothesis would be: p 0.8 u > 0.8 µ < 0.8 p # 0.8 µ # 0.8 o The test is: left-tailed two-tailed right-tailed Based on a sample of 500 people, 75% owned cats The test statistic is: (to 2 decimals) The p-value is: (to 2 decimals)
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Algebra students in a local high school use a computer-based program to brush up on algebra skills to be assessed by a test. The program claims that 80% of students using the program will pass the test. After an early pilot of the new program, only 35 of 50 students passed the test. A 1-proportion z test is going to be conducted to see if there's evidence that the true passing rate is less than what was claimed. The null hypothesis is p = 0.80. (a) What is the alternative hypothesis? H₂: P = 0.80 H₂: p > 0.80 H₂ P= = 0.80 H₂: p< < 0.80 0000 (b) What is the value of z, the test statistic? (Round your answer to two decimal places.) Z = (c) What is the P-value? (Use a table or technology. Round your answer to four decimal places.)
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