Mathematics, 17.12.2019 06:31 rosyposy43
Alocal bank needs information concerning the savings account balances of its customers. a random sample of 15 accounts was checked. the mean balance was $686.75 with a standard deviation of $256.20. find a 98% confidence interval for the true mean. assume that the account balances are normally distributed. a) ($532.86, $840.64) c) (s513.14, $860.36) b) (s544.87, $828.63) d) (s326.21, $437.90)
Answers: 3
with a standard deviation of s = 256.20. The confidence interval is given by 
where 
is the 100
th quantile of the t distribution with n-1=15-1=14 degrees of freedom. As we want a 
% = 98% confidence interval, we have that 
and the confidence interval is 
where 
is the 1st quantile of the t distribution with 14 df, i.e., 
. Then, we have 
and the 98% confidence interval is given by (513.1379, 860.3621)Another question on Mathematics
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Alocal bank needs information concerning the savings account balances of its customers. a random sam...
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