You measure 49 watermelons' weights, and find they have a mean weight of 60 ounces. Assume the population standard deviation is 5 ounces. Based on this, construct a 99% confidence interval for the true population mean watermelon weight.

Question

Give your answers as decimals, to two places

Answer 1

To construct a 99% confidence interval for the true population mean watermelon weight, we can use the formula for the confidence interval based on the population standard deviation:

$$CI=\overline{x}\pm z^{\ast }\left(\frac{\sigma }{\sqrt{n}}\right)$$

Where:

• $\overline{x}$ is the sample mean.
• $z^{\ast }$ is the z-value that corresponds to the desired confidence level (99% in this case).
• $\sigma$ is the population standard deviation.
• $n$ is the sample size.

Step 1: Identify the values

• Sample mean ($\overline{x}$) = 60 ounces
• Population standard deviation ($\sigma$) = 5 ounces
• Sample size ($n$) = 49

Step 2: Find the z-value for 99% confidence level

For a 99% confidence level, the z-value (from the z-table) is approximately:

$$z^{\ast }\approx 2.576$$

Step 3: Calculate the standard error of the mean (SEM)

The standard error of the mean is calculated as follows:

$$SE=\frac{\sigma }{\sqrt{n}}=\frac{5}{\sqrt{49}}=\frac{5}{7}\approx 0.7143$$

Step 4: Calculate the margin of error

The margin of error (ME) is given by:

$$ME=z^{\ast }\cdot SE=2.576\cdot 0.7143\approx 1.8387$$

Step 5: Calculate the confidence interval

Now we can calculate the confidence interval:

$$CI=\overline{x}\pm ME=60\pm 1.8387$$

Calculating the lower and upper bounds:

• Lower bound: $60-1.8387\approx 58.16$
• Upper bound: $60+1.8387\approx 61.84$

Final Result

Thus, the 99% confidence interval for the true population mean watermelon weight is:

$$\boxed{{\displaystyle (58.16,61.84)}}$$