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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

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barnuts

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
abiolataiwo2015

The 98% confidence interval for the true mean checking account balance for all local customers is: $453.59 < CI < $874.69.

Confidence interval

Confidence Interval = X ± z × s / √(n)

Where:

Sample mean = $664.14

z score for 98% confidence interval=2.65

Standard deviation = $297.29

Sample size = 14

Let plug in the formula

Confidence Interval = 664.14 ± 2.65 × 297.29 / √(14)

Confidence Interval = 664.14 ± 2.65 × 297.29 /3.741657

Confidence Interval = 664.14 ± 210.55

Confidence Interval =(664.14 -210.55);  (664.14 + 210.55)

Confidence Interval = $453.59 < CI < $874.69

Therefore the 98% confidence interval for the true mean checking account balance for all local customers is: $453.59 < CI < $874.69.

Learn more about Confidence interval here:https://brainly.com/question/15071962

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