Question

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 1

Answer with explanation:

Formula for confidence interval for population mean ( if population standard deviation is unknown ) :

$\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}$

, where n= sample size

$\overline{x}$ = sample mean

s= sample standard deviation

$t_{\alpha/2}$ = two-tailed t-value for significance level of ($\alpha$).

Let x denotes the amount of additional tax owed .

We assume that amount of additional tax owed is normally distributed .

As per given , we have

n=  64

Degree of freedom : df = 63    [ df= n-1]

$\overline{x}=\ 3489$

s= $2595

$\alph=1-0.90=0.1$

Using t-distribution table ,

$t_{\alpha/2, df}= t_{0.05, 63}= 1.6694$

Then , 90% confidence interval for the mean additional amount of tax owed for estate tax returns would be :

$\ 3489\pm (1.6694)\dfrac{2595}{\sqrt{64}}\\\\\ 3489\pm\ 541.51\\\\ =( 3489-\ 541.51, 3489+\ 541.51)=(\ 2947.49,\ 4030.51)$

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.