To calculate the 95% confidence interval for the mean weight of the items produced, we use the formula for the confidence interval when the population standard deviation is known:

$CI = \bar{x} \pm Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$

Where:

• $\bar{x}$ is the sample mean (75 grams),
• $Z_{\alpha/2}$ is the Z-score for the 95% confidence level (which is 1.96),
• $\sigma$ is the population standard deviation (5 grams),
• $n$ is the sample size (70).

Step 1: Compute the standard error

The standard error is given by:

$SE = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{70}} \approx \frac{5}{8.3666} \approx 0.5977$

Step 2: Multiply by the Z-score

Now, calculate the margin of error:

$ME = Z_{\alpha/2} \times SE = 1.96 \times 0.5977 \approx 1.1709$

Step 3: Calculate the confidence interval

Now, we can calculate the lower and upper bounds of the confidence interval:

• Lower bound: $75 - 1.1709 = 73.8291$
• Upper bound: $75 + 1.1709 = 76.1709$

Final Answer:

The 95% confidence interval for the mean weight is approximately:

$\boxed{73.8291 \text{ to } 76.1709}$

Would you like any further details or clarifications?

Related Questions:

1. How would the confidence interval change if the sample size increased to 100 items?
2. How can you calculate the margin of error for different confidence levels?
3. What happens to the confidence interval if the standard deviation of the population is unknown?
4. How do you interpret a 95% confidence interval in the context of this problem?
5. How does the confidence interval relate to hypothesis testing?

Tip:

The larger the sample size, the narrower the confidence interval, as the standard error decreases with an increase in $n$.