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7q1r2 defects 2 5 0 1 3 1 3 1 1 1 1 5 0 1 4 1 5 1 4 1 3 1 2 2 2 1 3 1 2 1 0 2 3 1 0 0 2 1 1 1 1 0 6 3 0 A quality control engineer at a particular Icd screen manufacturer is studying the mean number of defects per screen. Based on historical evidence, the mean number of defects per screen was thought to be 2.58. There have recently been changes to the manufacturing process, and the engineer now feels that the mean number of defects per screen may be significantly smaller than 2.58. Using the number of defects on each of 50 sample screens shown below, conduct the appropriate hypothesis test using a 0.1 level of significance. Assignment 7q1 data a) What are the appropriate null and alternative hypotheses? 2.58 versus Ha: u < 2.58 Ho: Но: х = 2.58 versus Ha: x > 2.58 Но: и = 2.58 versus Ha: u # 2.58 Но: и %3D 2.58 versus Ha: И> 2.58 b) What is the test statistic? Give your answer to four decimal places. c) What is the P-value for the test? Give your answer to four decimal places. d) What is the appropriate conclusion? Reject the claim that the mean number of defects per screen is 2.58 because the P-value is larger than 0.1. Fail to reject the claim that the mean number of defects per screen is 2.58 because the P-value is larger than 0.1. Reject the claim that the mean number of defects per screen is 2.58 because the P-value is smaller than 0.1. Fail to reject the claim that the mean number of defects per screen is 2.58 because the P-value is smaller than 0.1

7q1r2 defects 2 5 0 1 3 1 3 1 1 1 1 5 0 1 4 1 5 1 4 1 3 1 2 2 2 1 3 1 2 1 0 2 3 1 0 0 2 1 1 1 1 0 6 3 0 A quality control engineer at a particular Icd screen manufacturer is studying the mean number of defects per screen. Based on historical evidence, the mean number of defects per screen was thought to be 2.58. There have recently been changes to the manufacturing process, and the engineer now feels that the mean number of defects per screen may be significantly smaller than 2.58. Using the number of defects on each of 50 sample screens shown below, conduct the appropriate hypothesis test using a 0.1 level of significance. Assignment 7q1 data a) What are the appropriate null and alternative hypotheses? 2.58 versus Ha: u < 2.58 Ho: Но: х = 2.58 versus Ha: x > 2.58 Но: и = 2.58 versus Ha: u # 2.58 Но: и %3D 2.58 versus Ha: И> 2.58 b) What is the test statistic? Give your answer to four decimal places. c) What is the P-value for the test? Give your answer to four decimal places. d) What is the appropriate conclusion? Reject the claim that the mean number of defects per screen is 2.58 because the P-value is larger than 0.1. Fail to reject the claim that the mean number of defects per screen is 2.58 because the P-value is larger than 0.1. Reject the claim that the mean number of defects per screen is 2.58 because the P-value is smaller than 0.1. Fail to reject the claim that the mean number of defects per screen is 2.58 because the P-value is smaller than 0.1

MATLAB: An Introduction with Applications
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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7q1r2 defects 2 5 0 1 3 1 3 1 1 1 1 5 0 1 4 1 5 1 4 1 3 1 2 2 2 1 3 1 2 1 0 2 3 1 0 0 2 1 1 1 1 0 6 3 0
Transcribed Image Text:7q1r2 defects 2 5 0 1 3 1 3 1 1 1 1 5 0 1 4 1 5 1 4 1 3 1 2 2 2 1 3 1 2 1 0 2 3 1 0 0 2 1 1 1 1 0 6 3 0
A quality control engineer at a particular Icd screen manufacturer is studying the mean number of defects per screen. Based on historical evidence, the mean number of defects per screen was thought to be 2.58. There have recently been changes to the manufacturing process, and the engineer now feels that the mean number of defects per screen may be significantly smaller than 2.58. Using the number of defects on each of 50 sample screens shown below, conduct the appropriate hypothesis test using a 0.1 level of significance. Assignment 7q1 data a) What are the appropriate null and alternative hypotheses? 2.58 versus Ha: u < 2.58 Ho: Но: х = 2.58 versus Ha: x > 2.58 Но: и = 2.58 versus Ha: u # 2.58 Но: и %3D 2.58 versus Ha: И> 2.58 b) What is the test statistic? Give your answer to four decimal places. c) What is the P-value for the test? Give your answer to four decimal places. d) What is the appropriate conclusion? Reject the claim that the mean number of defects per screen is 2.58 because the P-value is larger than 0.1. Fail to reject the claim that the mean number of defects per screen is 2.58 because the P-value is larger than 0.1. Reject the claim that the mean number of defects per screen is 2.58 because the P-value is smaller than 0.1. Fail to reject the claim that the mean number of defects per screen is 2.58 because the P-value is smaller than 0.1
Transcribed Image Text:A quality control engineer at a particular Icd screen manufacturer is studying the mean number of defects per screen. Based on historical evidence, the mean number of defects per screen was thought to be 2.58. There have recently been changes to the manufacturing process, and the engineer now feels that the mean number of defects per screen may be significantly smaller than 2.58. Using the number of defects on each of 50 sample screens shown below, conduct the appropriate hypothesis test using a 0.1 level of significance. Assignment 7q1 data a) What are the appropriate null and alternative hypotheses? 2.58 versus Ha: u < 2.58 Ho: Но: х = 2.58 versus Ha: x > 2.58 Но: и = 2.58 versus Ha: u # 2.58 Но: и %3D 2.58 versus Ha: И> 2.58 b) What is the test statistic? Give your answer to four decimal places. c) What is the P-value for the test? Give your answer to four decimal places. d) What is the appropriate conclusion? Reject the claim that the mean number of defects per screen is 2.58 because the P-value is larger than 0.1. Fail to reject the claim that the mean number of defects per screen is 2.58 because the P-value is larger than 0.1. Reject the claim that the mean number of defects per screen is 2.58 because the P-value is smaller than 0.1. Fail to reject the claim that the mean number of defects per screen is 2.58 because the P-value is smaller than 0.1
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