A dietician read in a survey that 60.41% of adults in the U.S. do not eat breakfast at least 3 days a week. She believes that the proportion that skip breakfast 3 days a week is different than 0.6041. To verify her claim, she selects a random sample of 77 adults and asks them how many days a week they skip breakfast. 52 of them report that they skip breakfast at least 3 days a week. Test her claim at αα = 0.10. The correct hypotheses would be: H0:p≤0.6041H0:p≤0.6041 HA:p>0.6041HA:p>0.6041 (claim) H0:p≥0.6041H0:p≥0.6041 HA:p<0.6041HA:p<0.6041 (claim) H0:p=0.6041H0:p=0.6041 HA:p≠0.6041HA:p≠0.6041 (claim) Since the level of significance is 0.10 the critical value is 1.645 and -1.645 The test statistic is: (round to 3 places) The p-value is: (round to 3 places) The decision can be made to: reject H0H0 do not reject H0H0 The final conclusion is that: There is enough evidence to reject the claim that the proportion that skip breakfast 3 days a week is different than 0.6041. There is not enough evidence to reject the claim that the proportion that skip breakfast 3 days a week is different than 0.6041. There is enough evidence to support the claim that the proportion that skip breakfast 3 days a week is different than 0.6041. There is not enough evidence to support the claim that the proportion that skip breakfast 3 days a week is different than 0.6041.

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Description: A dietician read in a survey that 60.41% of adults in the U.S. do not eat breakfast at least 3 days a week. She believes that the proportion that skip breakfast 3 days a week is different than 0.6041. To verify her claim, she selects a random sample

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

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A dietician read in a survey that 60.41% of adults in the U.S. do not eat breakfast at least 3 days a week. She believes that the proportion that skip breakfast 3 days a week is different than 0.6041. To verify her claim, she selects a random sample of 77 adults and asks them how many days a week they skip breakfast. 52 of them report that they skip breakfast at least 3 days a week. Test her claim at αα = 0.10.

The correct hypotheses would be:

H0:p≤0.6041H0:p≤0.6041
HA:p>0.6041HA:p>0.6041 (claim)
H0:p≥0.6041H0:p≥0.6041
HA:p<0.6041HA:p<0.6041 (claim)
H0:p=0.6041H0:p=0.6041
HA:p≠0.6041HA:p≠0.6041 (claim)

Since the level of significance is 0.10 the critical value is 1.645 and -1.645

The test statistic is: (round to 3 places)

The p-value is: (round to 3 places)

The decision can be made to:

reject H0H0
do not reject H0H0

The final conclusion is that:

There is enough evidence to reject the claim that the proportion that skip breakfast 3 days a week is different than 0.6041.
There is not enough evidence to reject the claim that the proportion that skip breakfast 3 days a week is different than 0.6041.
There is enough evidence to support the claim that the proportion that skip breakfast 3 days a week is different than 0.6041.
There is not enough evidence to support the claim that the proportion that skip breakfast 3 days a week is different than 0.6041.

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