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This problem involves hypothesis testing for a single sample mean with a known population standard deviation. Let's go through it step by step.

### Problem Summary
1. **Population Mean (\(\mu\))**: 75 insect parts per 100 grams.
2. **Sample Mean (\(\bar{x}\))**: 76.7 insect parts per 100 grams.
3. **Population Standard Deviation (\(\sigma\))**: 10.
4. **Sample Size (\(n\))**: 48.
5. **Significance Level**: 5% (0.05).

### Step 1: Define Hypotheses
- **Null Hypothesis (\(H_0\))**: The mean insect parts per 100 grams is 75.
- **Alternative Hypothesis (\(H_a\))**: The mean insect parts per 100 grams is greater than 75.

So:
- \( H_0: \mu = 75 \)
- \( H_a: \mu > 75 \)

### Step 2: Calculate the Test Statistic \( z \)
The formula for the \( z \)-score in a single-sample test is:
\[
z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}
\]
Plugging in the values:
\[
z = \frac{76.7 - 75}{10 / \sqrt{48}} \approx \frac{1.7}{1.4434} \approx 1.1781
\]

### Step 3: Find the p-value
Since this is a one-tailed test, we look up the \( z \)-score of 1.1781 in the standard normal distribution table or use a calculator to find the p-value.

For \( z = 1.1781 \), the p-value is approximately 0.119. 

### Step 4: Decision Rule
At a significance level of 0.05:
- If the p-value is less than 0.05, reject \( H_0 \).
- If the p-value is greater than 0.05, fail to reject \( H_0 \).

Since 0.119 > 0.05, we **fail to reject \( H_0 \)**. This means there is insufficient evidence to conclude that the mean number of insect parts exceeds 75 per 100 grams.

### Step 5: Type I Error
A **Type I error** in this context would occur if we **incorrectly reject \( H_0 \)**, concluding that the flour is unfit for consumption when it is actually fit.

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Would you like a more detailed breakdown of each step, or do you have any questions?

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### Related Questions
1. How is the sample size affecting the confidence of this test?
2. What would happen if we used a 1% significance level instead of 5%?
3. How does the z-score change if the sample mean were 78 instead of 76.7?
4. What would a Type II error represent in this scenario?
5. How would this test differ if the population standard deviation was unknown?

### Tip
Always compare your calculated p-value to the significance level to determine whether to reject or fail to reject the null hypothesis.

The FDA takes a random sample of 48 bags of flour to evaluate the mean insect parts per 100 grams. The known population mean is 75 insect parts per 100 grams, and the standard deviation is 10. The sample mean is found to be 76.7 insect parts per 100 grams. Test the claim at a 5% significance level to see if the entire shipment of wheat is not fit for human consumption.

Learn how to conduct a hypothesis test to determine if a sample of flour is unfit for human consumption based on insect parts per 100 grams. This example uses a z-score and a 5% significance level with known population parameters to test the claim. Suitable for high school and college-level statistics.

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