Question

A Nutritionist claims that the mean tuna consumption by a person is 3.8 pound per year .A sample of 50 people shows that the mean tuna consumption by a person is 3.6 pounds per year .Assume the population standard deviation is 1.08 pounds.At alpha=0.09 can you reject the claim?

Answer 1

To test the claim about mean tuna consumption, we use a Z-test with an alpha of 0.09, a population standard deviation of 1.08, a sample mean of 3.6, and a sample size of 50. If the calculated Z-value is in the rejection region, we can reject the null hypothesis.

Step-by-step explanation:

To test the nutritionist's claim that the mean tuna consumption by a person is 3.8 pounds per year, we will perform a hypothesis test using the given sample data. The null hypothesis (Hば) is that the true mean tuna consumption is 3.8 pounds (μ = 3.8), and the alternative hypothesis (H぀) is that the true mean is not 3.8 pounds (μ ≠ 3.8). Given an alpha level of 0.09, a population standard deviation of 1.08 pounds, a sample mean of 3.6 pounds, and a sample size of 50, we will use a Z-test to determine if the null hypothesis can be rejected.

The Z-value is calculated using the formula:

Z = (Xbar - μ) / (σ / √ n)

Where Xbar is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. Plugging in the values:

Z = (3.6 - 3.8) / (1.08 / √ 50)

After performing the calculation, the Z-score will be compared to the critical Z-value for a two-tailed test at alpha = 0.09. If the calculated Z-value is within the rejection region, the null hypothesis can be rejected.