26 A dietician read in a survey that 76.4% of adults in the U.S. do not eat breakfast at least 2 days a week. Shebelieves that a smaller proportion skip breakfast 2 days a week. To verify her claim, she selects a randomsample of 75 adults and asks them how many days a week they skip breakfast. 55 of them report that theyskip breakfast at least 2 days a week. Test her claim at a = 0.01.The correct hypotheses would be:О Но:р < 0.764HA:p > 0.764 (claim)Но:р > 0.764HA:p < 0.764 (claim)O Ho:p = 0.764HA:P + 0.764 (claim)Since the level of significance is 0.01 the critical value is -2.326The test statistic is:(round to 3 places)The p-value is:(round to 3 places)The decision can be made to:O reject HoO do not reject HoThe final conclusion is that:There is enough evidence to reject the claim that a smaller proportion skip breakfast 2 days a week.There is not enough evidence to reject the claim that a smaller proportion skip breakfast 2 days aweek.There is enough evidence to support the claim that a smaller proportion skip breakfast 2 days aweek.O There is not enough evidence to support the claim that a smaller proportion skip breakfast 2 days aweek.

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Description: 26 A dietician read in a survey that 76.4% of adults in the U.S. do not eat breakfast at least 2 days a week. Shebelieves that a smaller proportion skip breakfast 2 days a week. To verify her claim, she selects a randomsample of 75 adults and asks th

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A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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A dietician read in a survey that 76.4% of adults in the U.S. do not eat breakfast at least 2 days a week. She
believes that a smaller proportion skip breakfast 2 days a week. To verify her claim, she selects a random
sample of 75 adults and asks them how many days a week they skip breakfast. 55 of them report that they
skip breakfast at least 2 days a week. Test her claim at a = 0.01.
The correct hypotheses would be:
О Но:р < 0.764
HA:p > 0.764 (claim)
Но:р > 0.764
HA:p < 0.764 (claim)
O Ho:p = 0.764
HA:P + 0.764 (claim)
Since the level of significance is 0.01 the critical value is -2.326
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:
There is enough evidence to reject the claim that a smaller proportion skip breakfast 2 days a week.
There is not enough evidence to reject the claim that a smaller proportion skip breakfast 2 days a
week.
There is enough evidence to support the claim that a smaller proportion skip breakfast 2 days a
week.
O There is not enough evidence to support the claim that a smaller proportion skip breakfast 2 days a
week.

Expert Solution

Step 1

The claim of the test is that smaller proportion than 0.764 of skip breakfast 2 days a week. Let p be the population proportion of adults who do not eat breakfast at least 2 days a week.

Null hypothesis:

$H_{0} : p \geq 0.764$

Alternative hypothesis:

$H_{A} : p < 0.764$

Correct answer:

$H_{0} : p \geq 0.764 H_{A} : p < 0.764$

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