Let's say the sample size is 12 - that would give us degrees of freedom of 12-1=11.

Now, let's say we need to compute the 95% confidence limits. In the t-table, you'll find the value in the row=11, but in the column = .025 - why? Because in a two-sided confidence interval, you need to split the 5% between both sides of the distribution. So the t-value in this case would be 2.201.

info given:
n=250
acct error=reported-actual
total error is material if at least $5,000
average error for 250 (n)= $0.25
stnd deviation= $193.05

A) Find the estimated total error based on your sample, and compare it to the material amount.

is this the average error for the sample size times the number of accounts
$0.25*12,000=3,000

B) Identify the null and research hypotheses for a one-sided test of the population mean error per account and explain why a one-sided test is appropriate here.

Ho: total error is not material
Ha: total error is material
One sided becasue it is material if at least $5,000

C) Find the appropriate one sided 95% Confidence interval statement for the population mean error per account.

For this part, I did
Xbar-t(stnd error) which came out to .25-(1.645)(12.21)=-19.84---which I know is a wrong answer becasue it just doesn't make sense.

D) Find the t statistic.

The t statistic formula is t-stat=Xbar-mu(o)/stnd error

E) Which hypothesis is accepted as a result of a one-sided test at the 5% level?

B) the hypotheses are "verbally" correct, but need to be put into numerical form

C) why don't you think this makes sense? (hint: could it be possible that the confidence interval includes an actual error in the negative direction; do all accounts have to have errors in the positive direction?)

C) why don't you think this makes sense? (hint: could it be possible that the confidence interval includes an actual error in the negative direction; do all accounts have to have errors in the positive direction?)

I don't think it makes senses becasue the number is so low compared to the other information pertaining to the problem

average error for the accounts = 0.25
std dev for the errors = 193.05
n = 250

Now, picture a normal distribution curve, centered at 0.25. If you draw a bunch of random samples of 250 accounts and compute the mean each time, the values at each of the standard deviations of those means (or the standard errors of the means) would be:

Well thanks, for all your help, I really needed it. Those were my last problems, but I may be back--tomorrow we start regressions. I also will be back around exam time, I'm sure I'll need guidance then.

can anyone please answer this question for me....i have this one question left n my assignment and if i m not able to do this i woul get failed ....so need help

Q: from past records of personal loans the sanguisunga bank has determined that 10% of borrowers default on their loans. it also found that, of those who default, 32% are unemployed while only 2% of those who do not default are unemployed.

a: what percentage of unemployed borrowers defualt?
b: what proportion of borrowers is unemployed?
c; what proportion of employed borrowers does not default?