Question

A dietician read in a survey that 88.44% of adults in the U.S. do not eat...

A dietician read in a survey that 88.44% of adults in the U.S. do not eat breakfast at least 3 days a week. She believes that a larger proportion skip breakfast 3 days a week. To verify her claim, she selects a random sample of 65 adults and asks them how many days a week they skip breakfast. 38 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≤0.8844H0:p≤0.8844
    HA:p>0.8844HA:p>0.8844 (claim)
  • H0:p≥0.8844H0:p≥0.8844
    HA:p<0.8844HA:p<0.8844 (claim)
  • H0:p=0.8844H0:p=0.8844
    HA:p≠0.8844HA:p≠0.8844 (claim)



Since the level of significance is 0.10 the critical value is 1.282

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 larger proportion skip breakfast 3 days a week.
  • There is not enough evidence to reject the claim that a larger proportion skip breakfast 3 days a week.
  • There is enough evidence to support the claim that a larger proportion skip breakfast 3 days a week.
  • There is not enough evidence to support the claim that a larger proportion skip breakfast 3 days a week.

Homework Answers

Answer #1

given data and some calculations are:-

sample size (n) = 65

sample proportion () = 38/65 = 0.5846

The correct hypotheses would be:

the critical value is :-



The test statistic is: -



The p-value is:-



The decision can be made to:

do not reject H0

[ p value = 1.000 > 0.10 (alpha)..so we fail to reject the null hypothesis.]

The final conclusion is that:

There is not enough evidence to support the claim that a larger proportion skip breakfast 3 days a week.

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