Question

1. Garbage Production. Based on a sample of 62 households, the mean weight of discarded plastic is 1.93 pounds and the standard deviation is 1.08 pounds (data from the Garbage Project at the University of Arizona). Use a single value to estimate the mean weight of discarded plastic for all households. Also, find the 90% confidence interval.

Answer 1

The answer is below

Explanation:

Given that:

Mean (μ) = 1.93, standard deviation (σ) = 1.08 pounds, sample (n) = 62.

the mean weight of discarded plastic for all household is given by:

`\mu_x=\mu = 1.93\ pounds`

The standard deviation of discarded plastic for all household is given by:

`\sigma_x=(\sigma)/(√(n) ) =(1.08)/(√(62) )=0.137\ pounds`

The confidence (c) = 90% = 0.9

α = 1 - c = 1 - 0.9 = 0.1

α/2= 0.1/2 = 0.05

The z score of 0.05 corresponds to the z score of 0.45 (0.5 - 0.05) which is 1.645. i.e.
`z_(\alpha)/(2)=z_(0.05)=1.645`

The margin of error (E) =
`z_(0.05)(\sigma)/(√(n) )=1.645*(1.03)/(√(62) )= 0.2256`

The confidence interval =
`\mu \pm E=1.93 \pm 0.2256=(1.7044,2.1556)`

We are 90% confidence that the value is between 1,7044 and 2.1556