Arandom sample of 300 citibank visa cardholder accounts indicated a sample mean debt of 1,220 with a sample standard deviation of 840. construct a 95 percent confidence interval estimate of the average debt of all cardholders.

(1124.5619, 1315.4381)

Step-by-step explanation:

The confidence interval for population mean when populatin standard deviation is unknown :-

, where = Sample mean

=Sample standard deviation

t* = Critical t-value.

Given : n= 300

Degree of freedom : df = n-1 = 299

Confidence interval = 95%

Significance level :

Using t-distribution table ,

The critical value for 95% Confidence interval for significance level 0.05 and df = 299 :

Then, a 95% confidence interval estimate of the average debt of all cardholders will be :-

Hence, a 95% confidence interval estimate of the average debt of all cardholders is (1124.5619, 1315.4381) .

total amount of money that he should pay if he has not used his duty free exemption is $98

step-by-step explanation:

given a united states citizen returning to the states declares the following items at the customs office: 3 shirts at $8.50 each 2 dresses at $27.50 each 1 pair of gold cuff links at $17.50 per pair. we have to find how much he pay

cost of 3 shirts if one cost at $8.50 is

cost of 2 dresses if one cost at $27.50 is  

cost of 1 pair of gold cuff at $17.50

so, total amount of money that he should pay if he has not used his duty free exemption is   $25.5+$55+$17.50=$98