Question

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 a nd a standard deviation of $297.29. use this data to find a 98% confidence interval for the true mean checking account balance for all local customers. $455.65 < ? < $872.63 $492.52 < ? < $835.76 $453.56 < ? < $874.72 $493.71 < ? < $834.57

Answer 1

In this case, since the standard deviation is known and the sample size is less than 30, we will use the t-distribution. The formula for calculating the confidence interval is:

Confidence Interval = X ± t * s / sqrt(n)

Where,

X = sample mean = $664.14

t = t score (taken from standard distribution tables)

s = standard deviation = $297.29

n = sample size = 14

 

At Degrees of Freedom = n – 1 = 13 and 98% Confidence interval, z = 2.65

Substituting the values in the equation:

Confidence Interval = 664.14 ± 2.65 * 297.29 / sqrt(14)

Confidence Interval = 664.14 ± 210.55

Confidence Interval = 453.59 to 874.69

ANSWER: $453.56 < CI < $874.72