~Class Meeting #18-Estimating an unknown Population mean (s known)



~Note: This is not realistic since the population standard deviation, s, is known. The population mean m is used in the computation of s.
Therefore, there is no need to estimate it. The next lesson deals with a much more realistic situation of an unknown s.

~Note: Assumptions:
1) the population standard deviation, s, is known   
2) the population is normally distributed and/or the sample size n is 30 or greater, either will do.

~Note: We know, from a previous lesson, that the mean of the sample means equals the population mean, regardless of sample size.

~However, each individual mean in the sample would not. However, a sample mean is the best point estimate of the population mean m. (a value that estimates a population parameter)

~Note: We would like to use the sample mean to estimate the population mean at a given level of confidence. The margin of error E could calculated once the interval is found or it could be found by formula (see below).

~Note: The techniques we will use are the same as the ones applied to estimating population proportions (Class Meeting #17).

~Note: For a given sample mean, x bar,  the confidence interval is written as x bar- E <m<x bar + E  or  (x bar- E , x bar + E), where E is the margin of error.

~Note: The margin of error E can also be calculated by:  
E = (Zalpha/2) times the population standard deviation/sqr(n), where Zalpha/2 is the critical z score based on the desired  confidence level.
                                                                                                                          
~Note: You will have to know this formula for E, since it is used to determine a minimum sample size for a given E & confidence level.
(i.e., you do this by substituting for E, Zalpha/2, and the know population standard deviation, then solving for n) (remember to roundup).

~Example 1:  A manufacturer of auto tires needs to estimate the mean life of a new line of tires. From past experience, it is known that the population standard deviation for tire life is approximately 2500 miles and tire life is normally distributed. Here are the results of a sample indicating tire life for 16 new line tires.

40,133   37,494   39,446   42,294   39,433   40,403   41,559   37,176   38,309   41,224   35,012   39,322   40,572   38,544   40,882   39,704

~Obtain a 90% confidence interval for the true mean of this new line of tires & give the margin of error E.

~Method 1: (by using formulas). Confidence level = 1-alpha, so, Zalpha/2 = Z.05 = 1.645, n=16, s= 2500,  Sample mean=∑x/n = 39,469, then use
(sample mean- (Zalpha/2) s/sqr(n) ,  sample mean+ (Za/2)s/sqr(n) ) = (39,469-1,028, 39,469+1,028) = (38,441, 40,497), where E = 1,028
                                                 
~Method 2: (all by calculator). Clear L1, enter 16 data points in L1, STAT, Tests, Zinterval(7), Data, s, L1, Freq:1, C-I .90, Calculate.

~The display will give you the confidence interval. To get E, subtract the lower limit from the sample mean or subtract the sample mean from the upper limit. This is definitely the way to do it.

~Example 2:  Find the 95% confidence interval & margin of error for the true mean monthly fuel expenditure, if a sample of 30 gives a mean of 58.56 (assume the standard deviation of the population is 20.65).
Ans: (51.17, 65.95), E=7.39

~What  sample size is required for E= .50 with a 95% confidence level?  Must use  E = (Zalpha/2) [s/sqr(n)] --->  .50 = (1.96)(20.65)/sqr(n)
                                                                                                                                                                 
Solve for n:   sqr(n)= (1.96)(20.65)/.50 = 80.948--> n = 6552.58 = 6,553 (roundup)

~Note: For finite populations of size N where n is > or =.05N, replace s/sqr(n) by s/sqr(n) times sqr[(N-n)/N-1)].
Confidence intervals and margins of error will change.