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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 than 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 a = 0.10.

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 than 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 a = 0.10.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.6: Summarizing Categorical Data
Problem 10CYU
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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 than 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 a = 0.10. The correct hypotheses would be: О Но:р< 0.705 HA:P > 0.705 (claim) Но: р > 0.705 HA:p < 0.705 (claim) О Но: р — 0.705 HA:P + 0.705 (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: O reject Ho O do not reject Ho The final conclusion is that: O There is enough evidence to reject the claim that the proportion that skip breakfast 3 days a week is different than 0.705. O There is not enough evidence to reject the claim that the proportion that skip breakfast 3 days a week is different than 0.705. O There is enough evidence to support the claim that the proportion that skip breakfast 3 days a week is different than 0.705. O There is not enough evidence to support the claim that the proportion that skip breakfast 3 days a week is different than 0.705.
Transcribed Image Text: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 than 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 a = 0.10. The correct hypotheses would be: О Но:р< 0.705 HA:P > 0.705 (claim) Но: р > 0.705 HA:p < 0.705 (claim) О Но: р — 0.705 HA:P + 0.705 (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: O reject Ho O do not reject Ho The final conclusion is that: O There is enough evidence to reject the claim that the proportion that skip breakfast 3 days a week is different than 0.705. O There is not enough evidence to reject the claim that the proportion that skip breakfast 3 days a week is different than 0.705. O There is enough evidence to support the claim that the proportion that skip breakfast 3 days a week is different than 0.705. O There is not enough evidence to support the claim that the proportion that skip breakfast 3 days a week is different than 0.705.
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