20emmanuelg1030
03.12.2019 •
Mathematics
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.
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Ответ:
(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) .
Ответ:
??
Step-by-step explanation: