A savings and loan association needs information concerning the checking account

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 and a standart deviation of $297.29.

a)For a 95% confidence level, find the margin of error E and construct a confidence interval for the true population mean checking account balance of all local customers. Assume the population has a normal distribution.

b)Write a brief statement that correctly interprets the confidence interval obtained in part"a"

c)The savings and loan association claims it's local customers have, on average, a checking account balance of $800. What does the confidence interval suggest about this claim?

Explanation / Answer

a)

Margin of Error = t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
Mean(x)=664.14
Standard deviation( sd )=297.29
Sample Size(n)=14
Margin of Error = t a/2 * 297.29/ Sqrt ( 14)
= 2.1604 * (79.454)
= 171.653

CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=664.14
Standard deviation( sd )=297.29
Sample Size(n)=14
Confidence Interval = [ 664.14 ± t a/2 ( 297.29/ Sqrt ( 14) ) ]
= [ 664.14 - 2.1604 * (79.454) , 664.14 + 2.1604 * (79.454) ]
= [ 492.487,835.793 ]

b)
We are 95% confident the mean lies in the interval inpart b
c)
It can be, on average, a checking account balance of $800. reason being
is it lies in

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