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.

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is symmetric about the line y = 120,000.

step-by-step explanation: