A savings and loan association needs information concerning the checking account balances of its local customers. A random sample of 14 accounts was checked and yielded a mean balance of $664.14 and a standard deviation of $297.29.

a) Find a 98% confidence interval for the true mean checking account balance for local customers.
b) Find a 95% confidence interval for the standard deviation.

a) 98% confidence interval for the true mean checking account balance for local customers.

(453.586 , 874.693)

b)   95% confidence interval for the standard deviation.

(214.91 , 441.53)

Step-by-step explanation:

Given a size of sample 'n' =14

given mean of the sample x⁻ = $664.14

standard deviation of the sample 'S' = $297.29.

a)

98% of confidence intervals

The degrees of freedom γ=n-1 =14-1 =13

t₁₃ = 2.650 at 98% of confidence level of signification.

on calculation, we get

(664.14-210.553 , 664.14+210.553)

(453.586 , 874.693)

98% confidence interval for the true mean checking account balance for local customers.

(453.586 , 874.693)

95% of confidence intervals

The degrees of freedom γ=n-1 =14-1 =13

X^2_{0.05,13} =22.36     (check table)

X^2_{0.95,13} = 5.892    (check table)

(214.91 , 441.53)

95% confidence interval for the standard deviation.

(214.91 , 441.53)

i believe

step-by-step explanation:

from my conclusion c

= 5

step-by-step explanation:

to find the 15th term substitute n = 15 into the nth term formula

= (- 3 × 15 ) + 50 = - 45 + 50 = 5