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To test the hypothesis, we need to follow these steps:

### 1. **Set up the Hypotheses**:
   - Null Hypothesis (\(H_0\)): \(\mu = 88\) calories.
   - Alternative Hypothesis (\(H_a\)): \(\mu \neq 88\) calories (two-tailed test).

### 2. **Given Values**:
   - Sample mean (\(\bar{x}\)) = 90 calories.
   - Population mean (\(\mu\)) = 88 calories.
   - Standard deviation (\(\sigma\)) = 10 calories.
   - Sample size (\(n\)) = 150.
   - Significance level (\(\alpha\)) = 0.05.

### 3. **Z-Value Calculation**:
   The formula to calculate the z-value is:
   \[
   z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}
   \]
   Substituting the values:
   \[
   z = \frac{90 - 88}{\frac{10}{\sqrt{150}}} = \frac{2}{\frac{10}{12.247}} = \frac{2}{0.8165} \approx 2.45
   \]

### 4. **Find the Critical Z-Value**:
   For a two-tailed test at a 0.05 significance level, you will split the significance level between both tails (i.e., 0.025 in each tail).

   From the z-table or using the inverse normal table given, you will find that a z-value of 1.96 corresponds to a probability of 0.975, which is the critical value for a two-tailed test at the 0.05 level.

### 5. **Compare the Calculated Z-Value with the Critical Z-Value**:
   - Calculated \(z\)-value = 2.45.
   - Critical \(z\)-value = ±1.96.

   Since \(2.45 > 1.96\), the calculated z-value falls in the rejection region.

### 6. **Conclusion**:
   Since the calculated z-value exceeds the critical z-value, we reject the null hypothesis (\(H_0\)). This means there is sufficient evidence to conclude that the average number of calories is different from 88 at the 0.05 significance level.

Would you like a more detailed explanation, or do you have any questions?

### Related Questions:
1. How does the significance level affect the decision in hypothesis testing?
2. What are the implications of a two-tailed test versus a one-tailed test?
3. How do you interpret the p-value in relation to the z-value?
4. Why do we use the standard normal distribution for hypothesis testing?
5. What would happen if the sample size were smaller or larger in this scenario?

### Tip:
Always check whether the test is one-tailed or two-tailed before finding the critical value from the z-table.

A dietician claims that the average number of calories in a serve of taro and fish soup is 88 with a standard deviation of 10. Test the null hypothesis H₀: μ = 88 calories against the alternative hypothesis Hₐ: μ ≠ 88 calories if a random sample of 150 serves yields an average of 90 calories. Use a 0.05 level of significance and state your conclusions clearly.

Learn how to test a dietician's claim about the average calories in taro and fish soup using hypothesis testing. This problem involves calculating the z-value, determining critical values, and making conclusions at a 0.05 significance level. Suitable for students in Grades 11-12.

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