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3.2 ompute the p-value for this hypothesis test. Recall that i = 112. Hint: Compute P(X < E) (at ast as extreme). ve your answer to 4 decimal places with a leading zero if necessary. Do not use the continuity rrection - it is actually slightly different for this problem (of an average), and would not actually ect our answer. Asking P(X < 112) is the same as asking P(EX; < 200 · 112) = (EX;< 22400) (multiply both sides by 200). Applying the continuity correction would add 5 to this equation, and converting back to X gives P(X < 112 + 20) which is basically the me as the original. 0.5 Enter your answer here Save Answer

3.2 ompute the p-value for this hypothesis test. Recall that i = 112. Hint: Compute P(X < E) (at ast as extreme). ve your answer to 4 decimal places with a leading zero if necessary. Do not use the continuity rrection - it is actually slightly different for this problem (of an average), and would not actually ect our answer. Asking P(X < 112) is the same as asking P(EX; < 200 · 112) = (EX;< 22400) (multiply both sides by 200). Applying the continuity correction would add 5 to this equation, and converting back to X gives P(X < 112 + 20) which is basically the me as the original. 0.5 Enter your answer here Save Answer

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Q3.2 112. Hint: Compute P(X <) (at Compute the p-value for this hypothesis test. Recall that = least as extreme). Give your answer to 4 decimal places with a leading zero if necessary. Do not use the continuity correction - it is actually slightly different for this problem (of an average), and would not actually affect our answer. Asking P(X < 112) is the same as asking P(EX; < 200 · 112) = P(EXi < 22400) (multiply both sides by 200). Applying the continuity correction would add 0.5 to this equation, and converting back to X gives P(X < 112 + 200) which is basically the 0.5 same as the original. Enter your answer here Save Answer
Transcribed Image Text:Q3.2 112. Hint: Compute P(X <) (at Compute the p-value for this hypothesis test. Recall that = least as extreme). Give your answer to 4 decimal places with a leading zero if necessary. Do not use the continuity correction - it is actually slightly different for this problem (of an average), and would not actually affect our answer. Asking P(X < 112) is the same as asking P(EX; < 200 · 112) = P(EXi < 22400) (multiply both sides by 200). Applying the continuity correction would add 0.5 to this equation, and converting back to X gives P(X < 112 + 200) which is basically the 0.5 same as the original. Enter your answer here Save Answer
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