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Let’s go through the problem step-by-step.

### Problem Summary
- A nutritionist claims that the mean tuna consumption by a person is **3.5 pounds per year**.
- A sample of **90 people** has a mean consumption of **3.4 pounds**.
- The population standard deviation is **1.17 pounds**.
- The significance level \(\alpha\) is **0.04**.
- We are asked if we can reject the claim based on the data.

### Solution Steps

#### Part (a): Null and Alternative Hypotheses
The null and alternative hypotheses should address whether the mean tuna consumption is equal to the claimed 3.5 pounds.

The correct choice, as indicated, is:
- **\( H_0: \mu = 3.5 \)**
- **\( H_a: \mu \neq 3.5 \)**

This setup is a **two-tailed test** because we are checking if the mean is significantly different (not just greater or less) from 3.5 pounds.

#### Part (b): Standardized Test Statistic
To calculate the test statistic \( z \), use the formula:

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

where:
- \( \bar{x} = 3.4 \) (sample mean),
- \( \mu = 3.5 \) (claimed population mean),
- \( \sigma = 1.17 \) (population standard deviation),
- \( n = 90 \) (sample size).

Substitute the values:

\[
z = \frac{3.4 - 3.5}{\frac{1.17}{\sqrt{90}}}
\]

First, calculate the standard error \( \frac{\sigma}{\sqrt{n}} \):

\[
\frac{1.17}{\sqrt{90}} \approx \frac{1.17}{9.49} \approx 0.1233
\]

Then, calculate \( z \):

\[
z = \frac{3.4 - 3.5}{0.1233} \approx \frac{-0.1}{0.1233} \approx -0.81
\]

### Answer for Part (b)
The standardized test statistic \( z \) is approximately **-0.81**, rounded to two decimal places.

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Would you like to continue with interpreting the results based on the significance level, or have any questions on this part?

### Related Questions:
1. What does the value of the \( z \)-score tell us about the sample mean relative to the claimed mean?
2. How do you determine if the result is statistically significant at \( \alpha = 0.04 \)?
3. Why is it necessary to use the population standard deviation in this calculation?
4. What are the critical values for a two-tailed test at \( \alpha = 0.04 \)?
5. What would happen to the \( z \)-score if the sample mean were lower than 3.4?

### Tip:
In hypothesis testing, a larger sample size generally makes the test more sensitive, making it easier to detect smaller differences from the claimed mean.

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 3.4 pounds per year. Assume the population standard deviation is 1.17 pounds. At α = 0.04, can you reject the claim?

This solution explains how to test a nutritionist's claim about mean tuna consumption using a two-tailed z-test. Given a sample mean of 3.4 pounds, population standard deviation of 1.17 pounds, and sample size of 90, we determine if the claim of 3.5 pounds per year can be rejected at a 0.04 significance level.

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