To construct a 99% confidence interval for the true population mean watermelon weight, we can use the formula for the confidence interval when the population standard deviation is known:
\[ \text{Confidence Interval} = \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) \]
where:
- \(\bar{x} = 60\) ounces (sample mean)
- \(\sigma = 5\) ounces (population standard deviation)
- \(n = 49\) (sample size)
- \(z\) is the z-score corresponding to the desired confidence level.
For a 99% confidence interval, we look up the z-score for 99% in standard normal distribution tables, or it can be found using a calculator. The critical z-value \(z\) for a 99% confidence level is approximately \(2.576\).
Now, we can calculate the standard error (SE):
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{49}} = \frac{5}{7} \approx 0.7142857 \]
Now we can calculate the margin of error (ME):
\[ ME = z \times SE = 2.576 \times 0.7142857 \approx 1.838856 \]
Now, we construct the confidence interval:
\[ \text{Lower limit} = \bar{x} - ME = 60 - 1.838856 \approx 58.161144 \] \[ \text{Upper limit} = \bar{x} + ME = 60 + 1.838856 \approx 61.838856 \]
Rounding these values to two decimal places, we find:
- Lower limit: 58.16 ounces
- Upper limit: 61.84 ounces
Thus, the 99% confidence interval for the true population mean watermelon weight is:
(58.16, 61.84) ounces.
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