Answer:
The 99% confidence interval for the mean paper products recycled per person per day for the population of Dallas is
[tex]0.7454 < \mu < 1.1546[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is n = 20
The sample mean is [tex]\= x = 0.95 \ pounds[/tex]
The standard deviation is [tex]\sigma = 0.32 \ pounds[/tex]
Generally given that the sample size is small , n< 30 we will be making use of t distribution table
Generally the degree of freedom is mathematically represented as
[tex]df = 20 - 1[/tex]
=> [tex]df = 19[/tex]
From the question we are told the confidence level is 99% , hence the level of significance is
[tex]\alpha = (100 - 99 ) \%[/tex]
=> [tex]\alpha = 0.01[/tex]
Generally from the t distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] at a degree of freedom of [tex]df = 19[/tex] is
[tex]t_{\frac{\alpha }{2}, 19 } = 2.86[/tex]
Generally the margin of error is mathematically represented as
[tex]E =t_{\frac{\alpha }{2}, 19 } * \frac{\sigma }{\sqrt{n} }[/tex]
=> [tex]E = 2.86 * \frac{0.32}{\sqrt{20} }[/tex]
=> [tex]E = 0.2046[/tex]
Generally 99% confidence interval is mathematically represented as
[tex]\= x -E < \mu < \=x +E[/tex]
=> [tex]0.95 -0.2046 < \mu < 0.95 + 0.2046[/tex]
=> [tex]0.7454 < \mu < 1.1546[/tex]
