Anutritionist 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?

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.

if im undersanding the question correctly, its organized from 10

step-by-step explanation:

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)