Question 1185509
Here's how to calculate the 80% confidence interval and margin of error:

1. **Find the critical value (z_c):**

For an 80% confidence level, the alpha (α) is 1 - 0.80 = 0.20. Since confidence intervals are two-tailed, we divide alpha by 2: 0.20 / 2 = 0.10.

We want to find the z-score that corresponds to an area of 0.10 in *each* tail of the standard normal distribution. This means the area to the *left* of the z-score is 1 - 0.10 = 0.90.  Using a standard normal (z) table or calculator, look up the z-score corresponding to a cumulative area of 0.90.  You'll find that z_c ≈ 1.28.

So, z_c = 1.28

2. **Calculate the margin of error (E):**

The margin of error is calculated as:

E = z_c * (σ / √n)

Where:

* z_c is the critical value (1.28)
* σ is the population standard deviation (0.36 gram)
* n is the sample size (11 hummingbirds)

E = 1.28 * (0.36 / √11)
E ≈ 1.28 * (0.36 / 3.3166)
E ≈ 1.28 * 0.1085
E ≈ 0.14 gram (rounded to two decimal places)

3. **Calculate the confidence interval:**

The confidence interval is calculated as:

(x̄ - E, x̄ + E)

Where x̄ is the sample mean (3.15 grams).

Lower Limit = x̄ - E = 3.15 - 0.14 = 3.01 grams
Upper Limit = x̄ + E = 3.15 + 0.14 = 3.29 grams

Therefore:

* z_c = 1.28
* Lower limit = 3.01 grams
* Upper limit = 3.29 grams
* Margin of error = 0.14 grams