ond column, and total them. This column contains full information concerning the sam-
pling distribution of the mean. Thus, since no duplications occur^P(X = 20) = 1/10; P(X
= 23) = 1/10; etc. Compute the mean of the possible values of X 9 or ^x = 300/10 = 30.
Record the deviations from 30 in Table 27, square the deviations, and total the results.
Compute the variance of the possible values of X, or (Tx = 330/10 = 33. Then (TX V33 =
5.74. These results are consistent with those in step 2.
ESTIMATION OF POPULATION MEAN
ON BASIS OF SAMPLE MEAN
A firm produces rods, and their lengths vary slightly because of unavoidable differences
in manufacture. Assume that the lengths are normally distributed. One hundred rods were
selected at random, and they were found to have a mean length of 1.856 m and a standard
deviation of 0.074 m. Estimate the mean length of all rods manufactured by this firm, us-
ing a 95 percent confidence level.
Calculation Procedure:- Compute the z value corresponding to the given
confidence level
Let X= length of a rod. The population consists of all rods manufactured by the firm, and
it may be considered infinite. There are two types of estimates: a point estimate, which as-
signs a specific value to X 9 and an interval estimate, which states that X lies within a given
interval. This interval is called a confidence interval, its boundaries are called the confi-
dence limits, and the probability that the estimate is correct is called the confidence level,
or confidence coefficient. Statistical inference can supply only interval estimates.
The central-limit theorem states: (a) If the population is extremely large and the prob-
ability distribution of X is normal, then the sampling distribution of the sample mean X is
also normal; (b) if the population is extremely large but the probability distribution of X is
not normal, the sampling distribution of X is approximately normal if the sample size is
30 or more. Thus, in the present case, the sampling distribution of X is considered to be
normal. _
Figure 23 is the sampling distribution diagram of the sample mean X. Let M= area un-
der curve from B to^ C. If a sample is drawn at random, there is a probability M that the
true sample mean Xj lies within the interval BC, or P(JJLX ~
2
Px
<
^j
<
Nox
z
i°"jr)
M.
This equation can be transformed to P(Xj - zt^x < ^x < Xj + Z 1 Crx) = M 9 Eq. a.
In the present case, Xj = 1.856 m, s = 0.074 m, n = 100, and M= 0.95. Then area under
curve from A to C= (0.50)(0.95) = 0.475. From the table of areas under the normal curve,
if A(Z 1 ) = 0.475, zt =1.96.
- Set up expressions for the mean and standard deviation
of the sampling distribution of the mean
By Eq. 14, fjbx = fa where JJL is the population mean to be estimated. Use the standard de-
viation s of the sample as an estimate of the standard deviation cr of the population. Then
o- = 0.074 m, and by Eq. 1 Sa 9 ax = 0.074/VKJO = 0.0074 m. - Estimate the mean length of the rods
Refer to Eq. a, and compute zpz = (1.96)(0.0074) = 0.015 m. Compute the confidence
limits in Eq. a, or 1.856 - 0.015 = 1.841 m and 1.856 + 0.015 = 1.871 m. Equation a