SOLUTION: A dietician read in a survey that 68.9% of adults in the U.S. do not eat breakfast at least 2 days a week. She believes that a larger proportion skip breakfast 2 days a week. To ve

Question 1185131: A dietician read in a survey that 68.9% of adults in the U.S. do not eat breakfast at least 2 days a week. She believes that a larger proportion skip breakfast 2 days a week. To verify her claim, she selects a random sample of 71 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
α= 0.05.
The test statistic is:______
The p-value is: _____
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Assuming randomness and independence,
p hat=49/71=0.690
Ho: p >0.690
Ha: p < = 0.690
alpha=0.05 p{reject Ho|Ho true}
test stat is a z
reject for z> 1.645
z=(0.690-0.689)/sqrt (0.689*0.311/71)
=0.001/0.0549=+0.018
fail to reject Ho, insufficient evidence to support the claim.
p-value is 0.493

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