Statistic 43?

43. The Warren County Telephone Company claims in its annual report that 鈥渢he typical customer
spends $60 per month on local and long distance service.鈥?A sample of 12 subscribers
revealed the following amounts spent last month.
$64 $66 $64 $66 $59 $62 $67 $61 $64 $58 $54 $66
a. What is the point estimate of the population mean?
b. Develop a 90 percent confidence interval for the population mean.
c. Is the company鈥檚 claim that the 鈥渢ypical customer鈥?spends $60 per month reasonable?
Justify your answer.

Confidence intervals are used to find a region in which we are 100 * ( 1 - 伪 )% confident the true value of the parameter is in the interval.
In order for the Confidence Interval to be valid you must have data from a normal distribution, at least if you are using the method here. If you do not have normal data then this type of confidence interval is not valid.

To clear up the notation I will use here. "t" is the test statistic and "t_(n-1)" is a Student t random variable with n - 1 degrees of freedom, e.g. a Student t random variable with 18 degrees of freedom is denoted as t_18.

For small sample confidence intervals about the mean you have:
xBar 卤 t * sx / sqrt(n)

where xBar is the sample mean
t is the t - score with n - 1 degrees of freedom such that 伪% of the data in the tails, i.e., P( |t_(n-1)| > t) = 伪
sx is the sample standard deviation
n is the sample size

For small sample confidence bounds we have the same estimators save the t-score.
A 100 * (1 - 伪 )% Confidence Lower Bound is:
xBar - t * sx / sqrt(n)
where t is such that P( t_(n-1) < t ) = 伪

A 100 * (1 - 伪 )% Confidence Upper Bound is:
xBar + t * sx / sqrt(n)
where t is such that P( t_(n-1) > t ) = 伪

The sample mean xbar = 62.58333
The sample standard deviation sx = 3.941812
The sample size n = 12

The t score for a 0.9 confidence interval is the t score such that 0.05 is in each tail.
t = 1.795885
The confidence interval is:

( xbar - t * sx / sqrt( n ) , xbar + t * sx / sqrt( n ) )
( 60.53979 , 64.62688 )

this CI indicates that the Telephone Company is low balling the typical customer. $60 is not in the CI and we conclude that that estitmate is to low.

hi,
a)point estimate of the population mean = alpha/2(standard deviation / root n)
b ) 90% cofidence interval = mu +/ 1.65 standard deviation/root n