A dietician read in a survey that 70.5% 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 from 0.705. To verify her claim, she selects a random sample of 71 adults and asks them how many days a week they skip breakfast. 44 of them report that they skip breakfast at least 3 days a week. Test her claim at α = 0.10.The correct hypotheses would be:- \( H_0: p = 0.705 \)- \( H_A: p \neq 0.705 \) (claim)Since the level of significance is 0.10, the critical value is 1.645 and -1.645.The test statistic is: [textbox] (round to 3 places)The p-value is: [textbox] (round to 3 places)The decision can be made to:- ○ reject \( H_0 \)- ○ do not reject \( H_0 \)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 0.705.- ○ There is not enough evidence to reject the claim that the proportion that skip breakfast 3 days a week is different than 0.705.- ○ There is enough evidence to support the claim that the proportion that skip breakfast 3 days a week is different than 0.705.- ○ There is not enough evidence to support the claim that the proportion that skip breakfast 3 days a week is different than 0.705.

Name: A dietician read in a survey that 70.5% of adults in the U.S. do not
Uploaded: 2024-03-20T07:06:41.000Z
Channel: Bartleby
Description: A dietician read in a survey that 70.5% 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 from 0.705. To verify her claim, she selects a random sample o