To construct a 95% confidence interval for the true population mean weight, we use the formula:

$\text{Confidence Interval} = \bar{x} \pm z^* \left( \frac{\sigma}{\sqrt{n}} \right)$

Where:

• $\bar{x}$ is the sample mean.
• $z^*$ is the z-score corresponding to the desired confidence level (1.96 for 95% confidence).
• $\sigma$ is the population standard deviation.
• $n$ is the sample size.

Given:

• Sample mean, $\bar{x} = 72$ ounces.
• Population standard deviation, $\sigma = 10.2$ ounces.
• Sample size, $n = 50$.

Step 1: Calculate the standard error (SE)

$SE = \frac{\sigma}{\sqrt{n}} = \frac{10.2}{\sqrt{50}} \approx \frac{10.2}{7.07} \approx 1.44$

Step 2: Find the margin of error (ME)

$ME = z^* \times SE = 1.96 \times 1.44 \approx 2.82$

Step 3: Construct the confidence interval

$\text{Lower limit} = 72 - 2.82 = 69.18$ $\text{Upper limit} = 72 + 2.82 = 74.82$

The 95% confidence interval for the true population mean dog weight is $69.18$ to $74.82$ ounces.

Would you like more details on the calculations, or have any questions?

Here are five related questions to expand on this topic:

1. How would the confidence interval change if the sample size were increased?
2. What would happen to the interval if a 99% confidence level were used instead?
3. How is the confidence interval affected by the population standard deviation?
4. Why do we use the z-score of 1.96 for a 95% confidence interval?
5. What is the difference between using the t-distribution and the z-distribution for constructing confidence intervals?

Tip: The width of a confidence interval decreases as the sample size increases, which leads to a more precise estimate of the population mean.