1. Hippocrates magazine states that 37 percent of all Americans take multiple vi
ID: 3226495 • Letter: 1
Question
1. Hippocrates magazine states that 37 percent of all Americans take multiple vitamins regularly. Suppose a researcher surveyed 750 people to test this claim and found that 299 did regularly take a multiple vitamin. Is this sufficient evidence to conclude that the actual percentage is different from 37% at the 5% significance level?
Select the [p-value, Decision to Reject (RH0) or Failure to Reject (FRH0)].
a) [p-value = 0.104, RH0]
b) [p-value = 0.104, FRH0]
c) [p-value = 0.052, FRH0]
d) [p-value = 0.154, FRH0]
e) [p-value = 0.052, RH0]
2. Quart cartons of milk should contain at least 32 ounces. A sample of 20 cartons contained the following amounts in ounces. Does sufficient evidence exist to conclude the mean amount of milk in cartons is less than 32 ounces at the 5% significance level?
The data is: (28.9, 31.6, 31.5, 31.7, 31.7, 28.9, 32.2, 31.1, 31.3, 32.7, 28.3, 32.4, 31.8, 32.1, 31.6, 28.8, 27.8, 28.2, 31.3, 32.6)
Select the [p-value, Decision to Reject (RH0) or Failure to Reject (FRH0)].
a) [p-value = 0.998, FRH0]
b) [p-value = 0.002, FRH0]
c) [p-value = 0.002, RH0]
d) [p-value = 0.998, RH0]
e) [p-value = 0.001, RH0]
Please answer with clear steps
Explanation / Answer
(1)
Data:
n = 750
p = 0.37
p' = 0.398666667
Hypotheses:
Ho: p = 0.37
Ha: p 0.37
Decision Rule:
= 0.05
Lower Critical z- score = -1.9600
Upper Critical z- score = 1.9600
Reject Ho if |z| > 1.9600
Test Statistic:
SE = {p (1 - p)/n} = (0.37 * (1 - 0.37)/750) = 0.0176
z = (p'- p)/SE = (0.398666666666667 - 0.37)/0.0176295206968312 = 1.6261
p- value = 0.104
Option B.
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