Question

The average number of mosquitos caught in 36 mosquito traps was 800 per trap. The standard deviation is 100 mosquitos. What is the 99% confidence level for the true mean

Answer 1

uh i think it's 25%

Explanation:

Answer 2

The 99% confidence interval for the true mean number of mosquitoes caught per trap is approximately:
`\[ (757.0667, 842.9333) \]`

To calculate the 99% confidence interval for the true mean of the number of mosquitoes caught per trap, we can use the formula for the confidence interval:

`\[ \text{Confidence Interval} = \text{Mean} \pm \left( \text{Critical Value} * \frac{\text{Standard Deviation}}{\sqrt{\text{Sample Size}}} \right) \]`

The critical value depends on the desired confidence level and the distribution of the data. For a normal distribution, the critical value for a 99% confidence level is approximately 2.576.

Given:

- Mean (\(\bar{X}\)) = 800 mosquitoes per trap,

- Standard Deviation (\(S\)) = 100 mosquitoes,

- Sample Size (\(n\)) = 36,

- Critical Value (\(Z\)) for 99% confidence level ≈ 2.576.

Substitute these values into the formula:

`\[ \text{Confidence Interval} = 800 \pm \left( 2.576 * (100)/(√(36)) \right) \]`

`\[ \text{Confidence Interval} = 800 \pm \left( 2.576 * (100)/(6) \right) \]`

`\[ \text{Confidence Interval} = 800 \pm 42.9333 \]`

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

`\[ \text{Lower Bound} = 800 - 42.9333 \]`

`\[ \text{Upper Bound} = 800 + 42.9333 \]`

Therefore, the 99% confidence interval for the true mean number of mosquitoes caught per trap is approximately:

`\[ (757.0667, 842.9333) \]`