Respuesta :
Answer:
A 95% confidence interval for the mean credit card debt of all college students in Illinois is [$316.06, $375.94] .
Step-by-step explanation:
We are given that in a sample of 50 college students in Illinois, the mean credit card debt was $346. Suppose that we also have reason to believe that the population standard deviation of credit card debts for this group is $108.
Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;
                P.Q. =  [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]  ~ N(0,1)
where, [tex]\bar X[/tex] = sample mean credit card debt = $346
      [tex]\sigma[/tex] = population standard deviation = $108
      n = sample of college students = 50
      [tex]\mu[/tex] = population mean credit card debt
Here for constructing a 95% confidence interval we have used a One-sample z-test statistics because we know about population standard deviation.
So, a 95% confidence interval for the population mean, [tex]\mu[/tex] is;
P(-1.96 < N(0,1) < 1.96) = 0.95 Â {As the critical value of z at 2.5%
                         level of significance are -1.96 & 1.96}   P(-1.96 < [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 1.96) = 0.95
P( [tex]-1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}} }[/tex] < [tex]1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.95
P( [tex]\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] , [tex]\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }[/tex] ]
                    = [ [tex]\$346-1.96 \times {\frac{\$108}{\sqrt{50} } }[/tex] , [tex]\$346+1.96 \times {\frac{\$108}{\sqrt{50} } }[/tex] ]
                    = [$316.06, $375.94]
Therefore, a 95% confidence interval for the mean credit card debt of all college students in Illinois is [$316.06, $375.94] .
