In a random sample of 64 audited estate tax​ returns, it was determined that the mean amount of additional tax owed was ​$3489 with a standard deviation of ​$2595.Construct and interpret a​ 90% confidence interval for the mean additional amount of tax owed for estate tax returns.1) The lower bound is ​$?2) The upper bound is $?

Answer with explanation:Formula for confidence interval for population mean ( if population standard deviation is unknown ) : [tex]\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}[/tex], where n= sample size [tex]\overline{x}[/tex] = sample means= sample standard deviation[tex]t_{\alpha/2}[/tex] = two-tailed t-value for significance level of ([tex]\alpha[/tex]).Let x denotes the amount of additional tax owed .We assume that amount of additional tax owed is normally distributed .As per given , we haven=  64 Degree of freedom : df = 63    [ df= n-1][tex]\overline{x}=\$3489[/tex] s= $2595[tex]\alph=1-0.90=0.1[/tex]Using t-distribution table ,[tex]t_{\alpha/2, df}= t_{0.05, 63}= 1.6694[/tex] Then , 90% confidence interval for the mean additional amount of tax owed for estate tax returns would be :[tex]\$3489\pm (1.6694)\dfrac{2595}{\sqrt{64}}\\\\\$3489\pm\$541.51\\\\ =($3489-\$541.51, $3489+\$541.51)=(\$2947.49,\ $4030.51) [/tex]1) The lower bound is ​$2947.49  .2) The upper bound is $4030.51  .Interpretation : We are 90% confident that the true population mean amount of additional tax owed lies between $2947.49 and  $4030.51.