Answer:
The answer is below
Step-by-step 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:
[tex]\mu_x=\mu = 1.93\ pounds[/tex]
The standard deviation of discarded plastic for all household is given by:
[tex]\sigma_x=\frac{\sigma}{\sqrt{n} } =\frac{1.08}{\sqrt{62} }=0.137\ pounds[/tex]
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. [tex]z_\frac{\alpha}{2}=z_{0.05}=1.645[/tex]
The margin of error (E) = [tex]z_{0.05}\frac{\sigma}{\sqrt{n} }=1.645*\frac{1.03}{\sqrt{62} }= 0.2256[/tex]
The confidence interval = [tex]\mu \pm E=1.93 \pm 0.2256=(1.7044,2.1556)[/tex]
We are 90% confidence that the value is between 1,7044 and 2.1556
