NBC claims that viewers spend a daily average of 185.1 minutes watching their content. An advertiser wants to verify this and conducts a poll of 15 random viewers who claim to watch NBC. The poll showed that this group spends a daily average of 198.1 minutes watching NBC with a standard deviation of 16.6 minutes. Use a 0.01 significance level to test the claim that the daily average amount of time NBC viewers watch NBC is at most 185.1 minutes, the average amount claimed by NBC.. Claim: Select an answer u ≥ 185.1 p < 185.1 p ≥ 185.1 u = 185.1 u > 185.1 u ≤ 185.1 p = 185.1 u ≠ 185.1 p ≠ 185.1 p > 185.1 u < 185.1 p ≤ 185.1 which corresponds to Select an answer H0: u = 185.1 H1: u < 185.1 H0: u ≤ 185.1 H0: u ≠ 185.1 H1: u ≠ 185.1 H0: p ≥ 185.1 H1: u > 185.1 Opposite: Select an answer u > 185.1 p ≥ 185.1 p ≤ 185.1 u = 185.1 u < 185.1 u ≤ 185.1 u ≥ 185.1 u ≠ 185.1 p = 185.1 p ≠ 185.1 p < 185.1 p > 185.1 which corresponds to Select an answer H0: p ≥ 185.1 H1: u > 185.1 H0: u = 185.1 H0: u ≠ 185.1 H1: u ≠ 185.1 H0: u ≤ 185.1 H1: u < 185.1 The test is: Select an answer left-tailed two-tailed right-tailed The test statistic is: (to 3 decimals) The Critical Value is: (to 3 decimals) Based on this we: Select an answer Reject the null hypothesis Fail to reject the null hypothesis Conclusion There Select an answer does not does appear to be enough evidence to support the claim that the daily average amount of time NBC viewers watch NBC is at most 185.1 minutes, the average amount claimed by NBC..

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Description: NBC claims that viewers spend a daily average of 185.1 minutes watching their content. An advertiser wants to verify this and conducts a poll of 15 random viewers who claim to watch NBC. The poll showed that this group spends a daily average of 198.1

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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NBC claims that viewers spend a daily average of 185.1 minutes watching their content. An advertiser wants to verify this and conducts a poll of 15 random viewers who claim to watch NBC. The poll showed that this group spends a daily average of 198.1 minutes watching NBC with a standard deviation of 16.6 minutes. Use a 0.01 significance level to test the claim that the daily average amount of time NBC viewers watch NBC is at most 185.1 minutes, the average amount claimed by NBC..

Claim: Select an answer u ≥ 185.1 p < 185.1 p ≥ 185.1 u = 185.1 u > 185.1 u ≤ 185.1 p = 185.1 u ≠ 185.1 p ≠ 185.1 p > 185.1 u < 185.1 p ≤ 185.1  which corresponds to Select an answer H0: u = 185.1 H1: u < 185.1 H0: u ≤ 185.1 H0: u ≠ 185.1 H1: u ≠ 185.1 H0: p ≥ 185.1 H1: u > 185.1

Opposite: Select an answer u > 185.1 p ≥ 185.1 p ≤ 185.1 u = 185.1 u < 185.1 u ≤ 185.1 u ≥ 185.1 u ≠ 185.1 p = 185.1 p ≠ 185.1 p < 185.1 p > 185.1  which corresponds to Select an answer H0: p ≥ 185.1 H1: u > 185.1 H0: u = 185.1 H0: u ≠ 185.1 H1: u ≠ 185.1 H0: u ≤ 185.1 H1: u < 185.1

The test is: Select an answer left-tailed two-tailed right-tailed

The test statistic is:  (to 3 decimals)

The Critical Value is:  (to 3 decimals)

Based on this we: Select an answer Reject the null hypothesis Fail to reject the null hypothesis

Conclusion There Select an answer does not does  appear to be enough evidence to support the claim that the daily average amount of time NBC viewers watch NBC is at most 185.1 minutes, the average amount claimed by NBC..

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