Question 748124: A dietitian read in a survey that at least 55% of adults do not eat breakfast at least 3 days a week. To verify this, she selected a random sample of 80 adults and asked them how many days a week they skipped breakfast. A total of 50% responded that they skipped breakfast at least 3 days a week. At a=0.01, test the claim
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A dietitian read in a survey that at least 55% of adults do not eat breakfast at least 3 days a week. To verify this, she selected a random sample of 80 adults and asked them how many days a week they skipped breakfast. A total of 50% responded that they skipped breakfast at least 3 days a week. At a=0.01, test the claim
-----------------------------
Ho: p <= 0.55
Ha: p < 0.55 (claim)
-----
z(0.50) = (0.50-0.55)/sqrt[0.55*0.45/80] = -0.8989
--------------
P-value = P(z < -0.8989) = normalcdf(-100,-8989) = 0.1844
-----
Since the p-value is greater than 1%, fail to reject Ho.
The test results do not support the claim.
Cheers,
Stan H.
================
|
|
|
