---### Hypothesis Testing ExampleConsider the following scenario for hypothesis testing when a test statistic is provided:#### The test statistic of z = 2.92 is obtained when testing the claim that p ≠ 0.548.1. **Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.**2. **Find the P-value.**3. **Using a significance level of α = 0.05, should we reject H₀ or should we fail to reject H₀?**[Click here to view page 1 of the standard normal distribution table.](#) [Click here to view page 2 of the standard normal distribution table.](#)---#### Solution:##### a. Type of Hypothesis TestThis is a **two-tailed** test.##### b. Calculation of P-valueP-value = ☐ (Round to three decimal places as needed.)In a two-tailed test with z = 2.92, we look at the areas in both tails of the normal distribution.##### c. Decision MakingUsing a significance level of α = 0.05, compare the P-value with the significance level to determine whether to reject H₀ or fail to reject H₀.---### Additional Resources- [Standard Normal Distribution Table - Page 1](#)- [Standard Normal Distribution Table - Page 2](#)---

Given that Test statistic =2.92 Hypothesis is P=! 0.548

Answered: k The test statistic of z= 2.92 is…

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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reject Ho or should we fail to reject Ho? distribution table distribution table as needed.) support the clain ce to support the d ence to support the evidence to suppo 1 6 Y H N tv C FO & Standard Normal Distribution Table (Page 1) 7 9 MacBook Air U NEGATIVE z Scores J z -3.50 and lower -3.4 -3.3 -3.2 -3.1 -3.0 -2.9 -2.8 F7 e S74 M * Standard Normal (2) Distribution: Cumulative Area from the LEFT .00 .01 .02 .03 .04 .05 0001 8 0003 .0005 0007 0010 0013 .0019 0026 1 K DII FB 0003 .0005 .0007 .0009 .0013 0018 0025 V & a ( 6. 9 O 0003 0005 .0006 0009 0013 0018 0024 DD FO L ) O. V .0003 .0004 0006 .0009 0012 0017 0023 P F10 I command option .0003 .0004 .0006 .0008 .0012 0016 0023 - 0003 .0004 .0006 .0008 .0011 .0016 .0022 N fro { 4 [ ? 1 F11 +11 = 11 .06 0003 .0004 .0006 .0008 .0011 .0015 .0021 z F12 } 1 0 07 .0003 .0004 .0005 .0008 .0011 .0015 .0021 08 .0003 .0004 .0005 .0007 .0010 0014 0020 delete 38 return shi
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