A nutritionist claims that the mean tuna consumption by a person is 3.5 pounds per year. A sample of 90 people shows that the mean tuna consumption by a person is 3.3 pounds per year. Assume the population standard deviation is 1.02 pounds. At a = 0.08, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. Ο Α. H : μS 3.5 Ha: u> 3.5 О В. Но: > 3.5 Ha: us3.5 С. Но: и#3.3 Ha: u = 3.3 D. Ho: μ= 3.5 Ο Ε. Hρ: μ5 3.3 Ha: u> 3.3 O F. Ho: u> 3.3 Ha: us3.3 Ha: u#3.5 (b) Identify the standardized test statistic. z= (Round to two decimal places as needed.)

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### Hypothesis Testing for Mean Tuna Consumption

A nutritionist claims that the mean tuna consumption by a person is 3.5 pounds per year. A sample of 90 people shows a mean tuna consumption of 3.3 pounds per year. It is given that the population standard deviation is 1.02 pounds. At a significance level of α = 0.08, can you reject the claim?

#### (a) Identify the Null Hypothesis and Alternative Hypothesis

The given options for hypotheses are:

- **A**. \( H_0: \mu \leq 3.5 \); \( H_a: \mu > 3.5 \)
- **B**. \( H_0: \mu > 3.5 \); \( H_a: \mu \leq 3.5 \)
- **C**. \( H_0: \mu \neq 3.3 \); \( H_a: \mu = 3.3 \)
- **D**. \( H_0: \mu = 3.5 \); \( H_a: \mu \neq 3.5 \) (Correct)
- **E**. \( H_0: \mu \leq 3.3 \); \( H_a: \mu > 3.3 \)
- **F**. \( H_0: \mu > 3.3 \); \( H_a: \mu \leq 3.3 \)

The correct hypothesis statement is:

- **Null Hypothesis (\( H_0 \))**: \( \mu = 3.5 \)
- **Alternative Hypothesis (\( H_a \))**: \( \mu \neq 3.5 \)

#### (b) Identify the Standardized Test Statistic

To calculate the z-score, the formula is:

\[
z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}
\]

Where:
- \( \bar{x} \) = Sample mean = 3.3
- \( \mu \) = Population mean under null hypothesis = 3.5
- \( \sigma \) = Population standard deviation = 1.02
- \( n \) = Sample size = 90

Calculate the z-score and fill in the provided box, rounding to two decimal places.

\[ z = \frac{3.3 -
Transcribed Image Text:### Hypothesis Testing for Mean Tuna Consumption A nutritionist claims that the mean tuna consumption by a person is 3.5 pounds per year. A sample of 90 people shows a mean tuna consumption of 3.3 pounds per year. It is given that the population standard deviation is 1.02 pounds. At a significance level of α = 0.08, can you reject the claim? #### (a) Identify the Null Hypothesis and Alternative Hypothesis The given options for hypotheses are: - **A**. \( H_0: \mu \leq 3.5 \); \( H_a: \mu > 3.5 \) - **B**. \( H_0: \mu > 3.5 \); \( H_a: \mu \leq 3.5 \) - **C**. \( H_0: \mu \neq 3.3 \); \( H_a: \mu = 3.3 \) - **D**. \( H_0: \mu = 3.5 \); \( H_a: \mu \neq 3.5 \) (Correct) - **E**. \( H_0: \mu \leq 3.3 \); \( H_a: \mu > 3.3 \) - **F**. \( H_0: \mu > 3.3 \); \( H_a: \mu \leq 3.3 \) The correct hypothesis statement is: - **Null Hypothesis (\( H_0 \))**: \( \mu = 3.5 \) - **Alternative Hypothesis (\( H_a \))**: \( \mu \neq 3.5 \) #### (b) Identify the Standardized Test Statistic To calculate the z-score, the formula is: \[ z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \] Where: - \( \bar{x} \) = Sample mean = 3.3 - \( \mu \) = Population mean under null hypothesis = 3.5 - \( \sigma \) = Population standard deviation = 1.02 - \( n \) = Sample size = 90 Calculate the z-score and fill in the provided box, rounding to two decimal places. \[ z = \frac{3.3 -
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