The manager of the aluminum recycling division of Environmental Services wants a survey which will tell him how many households in the city of Seattle, Washington, will voluntarily wash out, store, and then transport all of their aluminum cans to a central recycling center located in the downtown area and only open on Sunday mornings.

A random survey of 500 households determines that 20% of the households would do so and that each participating household expects to recycle about 100 cans monthly with a standard deviation of 30 cans. (a) What is the 95% confidence interval regarding the percentage of households expected to recycle? (b) What is the 95% confidence interval regarding the number of cans to be recycled?

To answer part (a) of the question, we need to calculate the 95% confidence interval for the percentage of households expected to recycle.

Here's how you would do it:

1. Calculate the standard error (SE) using the formula:
SE = √((p * (1 - p)) / n)
where p is the percentage of households in the sample that would recycle (20% or 0.2 in this case) and n is the sample size (500 in this case).

SE = √((0.2 * (1 - 0.2)) / 500)
SE = √(0.16 / 500)
SE = √0.00032
SE ≈ 0.0179

2. Determine the z-value for a 95% confidence level. The z-value for a 95% confidence level is approximately 1.96.

3. Calculate the margin of error (ME) by multiplying the standard error by the z-value:
ME = z * SE
ME = 1.96 * 0.0179
ME ≈ 0.035

4. Calculate the lower and upper bounds of the confidence interval:
Lower bound = p - ME
Lower bound = 0.2 - 0.035
Lower bound ≈ 0.165

Upper bound = p + ME
Upper bound = 0.2 + 0.035
Upper bound ≈ 0.235

Therefore, the 95% confidence interval for the percentage of households expected to recycle is approximately 16.5% to 23.5%.

To answer part (b) of the question, we need to calculate the 95% confidence interval for the number of cans to be recycled.

Here's how you would do it:

1. Calculate the standard error (SE) using the formula:
SE = (standard deviation / √n)
where the standard deviation is 30 cans and n is the sample size (500 in this case).

SE = 30 / √500
SE = 30 / 22.36
SE ≈ 1.34

2. Determine the z-value for a 95% confidence level. The z-value for a 95% confidence level is approximately 1.96.

3. Calculate the margin of error (ME) by multiplying the standard error by the z-value:
ME = z * SE
ME = 1.96 * 1.34
ME ≈ 2.63

4. Calculate the lower and upper bounds of the confidence interval:
Lower bound = sample mean - ME
Lower bound = 100 - 2.63
Lower bound ≈ 97.37

Upper bound = sample mean + ME
Upper bound = 100 + 2.63
Upper bound ≈ 102.63

Therefore, the 95% confidence interval for the number of cans to be recycled is approximately 97.37 cans to 102.63 cans.