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) .

d≈7.07in

step-by-step explanation:

sinn is the total amount of students now and m were the students before, you subtract them both (n-m) and divide that by m. then, since to get a percentage from a fraction you have to multiply by 100, your answer should look like this: n-m/m*100

step-by-step explanation: