Question

A dietician read in a survey that 82.04% 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 77 adults and asks them how many days a week they skip breakfast. 49 of them report that they skip breakfast at least 2 days a week. Test her claim at αα = 0.01.

The correct hypotheses would be:

• H0:p≤0.8204H0:p≤0.8204
  HA:p>0.8204HA:p>0.8204 (claim)
• H0:p≥0.8204H0:p≥0.8204
  HA:p<0.8204HA:p<0.8204 (claim)
• H0:p=0.8204H0:p=0.8204
  HA:p≠0.8204HA:p≠0.8204 (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:

• reject H0H0
• do not reject H0H0

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.
• There is not enough evidence to support the claim that a smaller proportion skip breakfast 2 days a week.

Answer 1

The correct hypothesis would be:

H0: p ≥ 0.8204 against HA : p < 0.8204 (claim)

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Here  

  

   (sample size)

   (number of adults skip breakfast at least 2 days a week)

   (significance level)

   (critical value)

Sample proportion is,

  

The test statistic is:

    

  

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The p-value = 0

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• Since Z = -4.207 < Z= -2.326, we can conclude that the null hypothesis is rejected.
• Using p-value approach, since p-value < ,we can conclude that the null hypothesis is rejected.

The decision can be made to: Reject H0

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The final conclusion is that:

There is enough evidence to support the claim that a smaller proportion skip breakfast 2 days a week.

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