A nutritionist claims that the mean tuna consumption by a person is 3.93.9 pound
ID: 3135026 • Letter: A
Question
A nutritionist claims that the mean tuna consumption by a person is 3.93.9 pounds per year. A sample of 9090 people shows that the mean tuna consumption by a person is 3.83.8 pounds per year. Assume the population standard deviation is 1.181.18 pounds. At alphaequals=0.030.03, can you reject the claim?
(a) Identify the null hypothesis and alternative hypothesis.
A. Upper H 0H0: muequals=3.93.9 Upper H Subscript aHa: munot equals3.93.9
B. Upper H 0H0: mugreater than>3.93.9 Upper H Subscript aHa: muless than or equals3.93.9
C. Upper H 0H0: mugreater than>3.83.8 Upper H Subscript aHa: muless than or equals3.83.8
D. Upper H 0H0: muless than or equals3.83.8 Upper H Subscript aHa: mugreater than>3.83.8
E. Upper H 0H0: muless than or equals3.93.9 Upper H Subscript aHa: mugreater than>3.93.9
F. Upper H 0H0: munot equals3.83.8 Upper H Subscript aHa: muequals=3.83.8
(b) Identify the standardized test statistic. zequals=nothingm (Round to two decimal places as needed.)
(c) Find the P-value. nothingm (Round to three decimal places as needed.)
(d) Decide whether to reject or fail to reject the null hypothesis.
A. RejectReject Upper H 0H0. There isis sufficient evidence to reject the claim that mean tuna consumption is equal to 3.93.9 pounds.
B. RejectReject Upper H 0H0. There is notis not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.93.9 pounds.
C. Fail to rejectFail to reject Upper H 0H0. There is notis not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.93.9 pounds.
D. Fail to rejectFail to reject Upper H 0H0. There isis sufficient evidence to reject the claim that mean tuna consumption is equal to 3.93.9 pounds.
Explanation / Answer
A)
Formulating the null and alternative hypotheses,
Ho: u = 3.9
Ha: u =/ 3.9 [ANSWER, OPTION A]
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B)
Getting the test statistic, as
X = sample mean = 3.8
uo = hypothesized mean = 3.9
n = sample size = 90
s = standard deviation = 1.18
Thus, z = (X - uo) * sqrt(n) / s = -0.803968897 [ANSWER]
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c)
As we can see, this is a two tailed test.
Thus, the p value is
p = 0.421414938 [ANSWER]
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d)
As P > 0.05, we FAIL TO REJECT THE NULL HYPOTHESIS.
Hence,
OPTION C: C. Fail to reject H0. There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.9 pounds. [ANSWER, C]
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