In a random sample of 81 audited estate tax returns, it was determined that the mean amount of additional tax owed was $3408 with a standard deviation of $2565. construct and interpret a 90% confidence interval for the mean additional amount of tax owed for estate tax returns. the lower bound is $ nothing. (round to the nearest cent as needed.) the upper bound is $ nothing. (round to the nearest cent as needed.) interpret a 90% confidence interval for the mean additional amount of tax owed for estate tax returns. choose the correct answer below.

a. one can be 90% confident that the mean additional tax owed is less than the lower bound
b. one can be 90% confident that the mean additional tax owed is between the lower and upper bounds
c. one can be 90% confident that the mean additional tax owed is greater than the upper bound.

Using the t-distribution, it is found that:

b. One can be 90% confident that the mean additional tax owed is between the lower and upper bounds

We are given the standard deviation for the sample, which is why the t-distribution is used to solve this question.

The information given is:

Sample mean of .

Sample standard deviation of .

Sample size of .

The confidence interval is:

The critical value, using a t-distribution calculator, for a two-tailed 90% confidence interval, with 81 - 1 = 80 df, is t = 1.6641.

Then, the bounds of the interval are:

The lower bound is $2,933.73.

The upper bound is $3,822.27.

The interpretation is that we are 90% sure that the true mean is in the interval, that is, option b.

A similar problem is given at link

B. One can be 90% confident that the mean additional tax owed is between the lower and upper bounds.

Step-by-step explanation:

Given:

n= 81

Solution:

A confidence interval, in statistics, refers to the probability that a population parameter will fall between two set values for a certain proportion of times. Confidence intervals measure the degree of uncertainty or certainty in a sampling method. A confidence interval can take any number of probabilities, with the most common being a 95% or 99% confidence level.

Confidence interval =

To Find the z  value:

Degree of freedom = n-1

=>81- 1

=> 80

Significance level = 1- confidence level

=>

=>

=>

=>0.05

using this value In T- Distribution table we get

z =  1.645

Substituting the values  we have,

confidence interval =

confidence interval =

confidence interval =

confidence interval =

confidence interval= (2939.18, 3876.83)

idk

step-by-step explanation:

payback

the answer is 5/14.

hope this .