A random sample of 36 trout from this pond was caught and weighed, and they showed a mean weight of 560g with standard deviation of 84g. What would the upper limit of the required confidence interval be?

Answer 1

To compute the upper limit of the confidence interval for the mean weight of trout in the pond, we will use the following steps:

1. Determine the sample mean and standard deviation:

   • Sample Mean (\( \bar{x} \)) = 560g
   • Sample Standard Deviation (s) = 84g
   • Sample size (n) = 36

2. Choose the confidence level: We need to specify a confidence level. Common levels are 90%, 95%, and 99%. For this example, let's assume a 95% confidence level.

3. Find the critical value (z-score): For a 95% confidence level, the critical value (z) can be found using a z-table or calculator:

   • The critical value for a 95% confidence level is approximately 1.96.

4. Calculate the standard error (SE): \[ SE = \frac{s}{\sqrt{n}} = \frac{84}{\sqrt{36}} = \frac{84}{6} = 14 \]

5. Calculate the margin of error (ME): \[ ME = z \times SE = 1.96 \times 14 = 27.44 \]

6. Calculate the upper limit of the confidence interval: \[ \text{Upper limit} = \bar{x} + ME = 560 + 27.44 = 587.44 \]

Thus, the upper limit of the 95% confidence interval for the mean weight of trout is approximately 587.44 grams.