Question

A nutritionist claims that the mean tuna consumption by a person is 3.5 pounds per year. A sample of 70 people shows that the mean tuna consumption by a person is 3.3 pounds per year. Assume the population standard deviation is 1.22 pounds. At alphaequals0.02​, can you reject the​ claim?

Answer 1

Let μ = mean tuna consumption (in pounds per year)

H0: μ = 3.1

HA: μ ≠ 3.1

Sample mean = 2.9

Standard deviation = 0.94

Standard error of mean = s / √ n

Standard error of mean = 0.94 / √ 60

SE = 0.94/7.746

Standard error of mean 0.1214

z = (xbar- μ ) / SE

z = (2.9-3.1) / 0.1214

z = -1.6481

p-value = P( z < -1.6481)+P( z > 1.6481) = 2 (0.0495) = 0.099

Fail to reject null hypothesis since 0.099 >0.08

The nutritionist's claim that the mean tuna consumption by a person in the U.S is 3.1 pounds per year is not rejected.

Answer 2

If the sample size is 70, population standard deviation is 1.22 and α=0.002 then we can reject the nutritionist's claim of mean tuna consumption by a person is 3.5 pounds,

What is hypothesis?

Hypothesis is the statement that is required to be prove.It is of two types one is null hypothesis and alternate hypothesis.

How to check hypothesis?

First of all we are required to make hypothesis:

$H_{0}$:μ≠3.5

$H_{1}$:μ=3.5

n=70

σ=1.22

Confidence level=1-0.02=0.98=98%

Z=(X-μ)/σ

=(3.5-3.3)/1.22

=0.2/1.22

=0.1639

Z value of 985 confidence level =2.326

2.326>0.1639

So we will accept null hypothesis and it means that mean is not equal to 3.5.

Hence if the sample size is 70, population standard deviation is 1.22 and α=0.002 then we can reject the nutritionist's claim of mean tuna consumption by a person is 3.5 pounds,

Learn more about hypothesis at https://brainly.com/question/11555274

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