A credit union needs information concerning the checking account balances of its customers. A random sample of 14 accounts was checked and yielded a mean balance of $664 and a standard deviation of $297. Assume the population has a normal distribution. You will be finding a confidence interval for the true mean checking account balance for all customers of the credit union.
A.) What distribution will be used. Z or T?
B.) Construct a 90% confidence interval. Remember to write the formula before substituting values. Round confidence interval limits to the nearest dollar.
C.) Complete the sentence summarizing the results: We are 90% confident the true mean checking account balance of all the credit union customers is between ______________ and ___________.
I'm really not just looking for the answer. We may have a question just like this on the final so I'd like to know how to do it as well. Any and all help would be greatly appreciated.
A.) What distribution will be used. Z or T?
B.) Construct a 90% confidence interval. Remember to write the formula before substituting values. Round confidence interval limits to the nearest dollar.
C.) Complete the sentence summarizing the results: We are 90% confident the true mean checking account balance of all the credit union customers is between ______________ and ___________.
I'm really not just looking for the answer. We may have a question just like this on the final so I'd like to know how to do it as well. Any and all help would be greatly appreciated.
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First to extract the mathematical information away from the "word problem" part.
A sample of 14 had sample mean 664 and sample std. dev. 297
Find a confidence interval for the true mean checking account balance.
This is a T test because the population standard deviation isn't known.
"In testing the null hypothesis that the population mean is equal to a specified value μ0, one uses the statistic
t = (x - μo) / (s / sqrt(n))" (Wikipedia).
We don't have a particular μo to work from so that means we have to work backwards from the 90% confidence level.
This means I don't really comprehend what B wants, unless I just did it already (put the dollars in and round as it requests, see below)
It is 2-tailed and presumably centered at 664, so we need to know the t-level that corresponds to the 95% p-level. There are n-1=13 degrees of freedom.
There are calculators online but they seem to have drawbacks.
One has a Java applet and I don't comprehend whether those have a lot of permissions or need to be trusted or signed etc. (I think they are OK but I'm not certain.) http://www.stat.tamu.edu/~west/applets/t…
That is best because it shows what it's actually doing, and it let me enter in 0.05 for the left tail and 13 degrees of freedom, and gave me the value, and didn't have ads.
Others have stupid ads etc
One of them wants you to paste in all your data, which is cool if you have the data
t level: +/- 1.770933
Then 1.770933 = |664 - μo| / (297 / sqrt(14))
1.770933 (297/sqrt(14)) = |664 - μo| = 140.6
With 90% confidence the mean is between 664+104.6 and 664-104.6.
A sample of 14 had sample mean 664 and sample std. dev. 297
Find a confidence interval for the true mean checking account balance.
This is a T test because the population standard deviation isn't known.
"In testing the null hypothesis that the population mean is equal to a specified value μ0, one uses the statistic
t = (x - μo) / (s / sqrt(n))" (Wikipedia).
We don't have a particular μo to work from so that means we have to work backwards from the 90% confidence level.
This means I don't really comprehend what B wants, unless I just did it already (put the dollars in and round as it requests, see below)
It is 2-tailed and presumably centered at 664, so we need to know the t-level that corresponds to the 95% p-level. There are n-1=13 degrees of freedom.
There are calculators online but they seem to have drawbacks.
One has a Java applet and I don't comprehend whether those have a lot of permissions or need to be trusted or signed etc. (I think they are OK but I'm not certain.) http://www.stat.tamu.edu/~west/applets/t…
That is best because it shows what it's actually doing, and it let me enter in 0.05 for the left tail and 13 degrees of freedom, and gave me the value, and didn't have ads.
Others have stupid ads etc
One of them wants you to paste in all your data, which is cool if you have the data
t level: +/- 1.770933
Then 1.770933 = |664 - μo| / (297 / sqrt(14))
1.770933 (297/sqrt(14)) = |664 - μo| = 140.6
With 90% confidence the mean is between 664+104.6 and 664-104.6.