Question

(d) Find the sample size necessary for an 80% confidence level with a maximal margin of error $E = 0.11$ for the mean weights of the hummingbirds. Recall that the formula for finding the sample size $n$ is $$n = \left( \frac{z_\alpha \sigma}{E} \right)^2$$ where $E$ is the specified maximal margin of error. Use the formula above to find the sample size $n$. $$n = \left( \frac{1.28 \cdot 0.35}{0.11} \right)^2$$

Name: d find the sample size necessary for an 80 confidence level with a maximal margin of error e 011 for the mean weights of the hummingbirds recall that the formula for finding the sample size 17513
Uploaded: 2022-12-09T01:11:24-08:00
Duration: 1 min 50 s
Channel: Christopher Dzorkpata
Description: d find the sample size necessary for an 80 confidence level with a maximal margin of error e 011 for the mean weights of the hummingbirds recall that the formula for finding the sample size 17513

          (d) Find the sample size necessary for an 80% confidence level with a maximal margin of error $E = 0.11$ for the mean weights of the hummingbirds.
Recall that the formula for finding the sample size $n$ is
$$n = \left( \frac{z_\alpha \sigma}{E} \right)^2$$
where $E$ is the specified maximal margin of error.
Use the formula above to find the sample size $n$.
$$n = \left( \frac{1.28 \cdot 0.35}{0.11} \right)^2$$

d find the sample size necessary for an 80 confidence level with a maximal margin of error e 011 for the mean weights of the hummingbirds recall that the formula for finding the sample size 17513

Added by Rita W.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Christopher Dzorkpata

12/09/2022

Step 1

Step 1: We are given the formula for the sample size $n$: $$n = \left( \frac{z_\alpha \sigma}{E} \right)^2$$ where $z_\alpha$ is the critical value from the normal distribution for the desired confidence level, $\sigma$ is the population standard deviation, and $E$ Show more…

Show all steps

Thanks for your feedback!

(d) Find the sample size necessary for an 80% confidence level with a maximal margin of error $E = 0.11$ for the mean weights of the hummingbirds. Recall that the formula for finding the sample size $n$ is $$n = \left( \frac{z_\alpha \sigma}{E} \right)^2$$ where $E$ is the specified maximal margin of error. Use the formula above to find the sample size $n$. $$n = \left( \frac{1.28 \cdot 0.35}{0.11} \right)^2$$