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CHAPTER 7: ESTIMATION
(Skip sections 7.5 and 7.6)
“Worldwide demand for cars will not exceed 1 million (primarily because of the
limitation of available chauffeurs).”
- Mercedes Benz, 1900
Point estimate of the mean is X .
Interval estimate of the mean for large samples (n > 30) is X + Z s / n .
Desirable properties of estimators:
Unbiasedness: Expected value of the estimator equals parameter.
Consistency: The difference between the estimator and the parameter decreases as the
sample size increases.
Example 1: A manufacturer developed a new method of assembling an electronic
component. The average assembly time using the current method is 20 minutes. The
manufacturer trained 45 randomly selected employees in the new method. Then each
employee assembled one component and the time taken was recorded. The average time
was 15.82 minutes and the standard deviation of the sample of 45 was 1.13. Build a 95%
confidence interval for the average assembly time to determine if the new assembly
method takes less time than the current method.
Level of confidence is 100(1-)% and the sample size affects the width of the interval.
Note: The level of confidence in most media polls is 95% (19 out of 20).
The t-distribution is continuous, symmetric about zero and bell-shaped.
The parameter of the t-distribution is the degrees of freedom (see Appendix E, Table 3).
Interval estimate of the mean for small samples is X  t s / n.
Example 2: As part of the budgeting process for next year, the manager of an electric
generating plant must estimate the coal she will require for this year. Last year, the plant
almost ran out, so she is reluctant to budget for the same amount, however she feels that
past usage will allow her to estimate the number of tons of coal to order. A random
sample of 10 operating weeks chosen over the past 5 years, yielded a mean usage of
11400 tons a week with a sample standard deviation of 700 tons a week. What would a
99% confidence interval be for the average number of tons used per week?
Estimating a population proportion
Point estimate of the proportion is ps.
Interval estimate of the proportion is ps + Z  ps (1- ps) / n.
Example 3: A survey of brand name and private label jeans in retail stores found a high
percentage with incorrect waist and/or inseam measurements on the label. In fact, only
18 of 240 pairs of men's pre-washed jeans sold in Minneapolis stores came to within halfinch of all their label measurements. Find a 90% confidence interval for the true
proportion of jeans that have inseam and waist measurements that fall within half an inch
of their labeled measurements.
Sample Size Determination:
n = {z  / e}2
A sample of this size makes us 100(1-)% confident that the sample mean is within B
units of the population mean.
The error bound e is half the width of the interval and is also referred to as the margin of
error (in %).
Note: The standard deviation can be approximated using the range if the population is
approximately normal.
Example 4: National Motors has equipped their ZX-900 cars with a new disc brake
system. The stopping distance is the number of feet required to bring the automobile to a
complete stop from a speed of 35 mph. One of the ZX-900’s major competitors is
advertising a mean stopping time of 60 feet. National would like to claim that the ZX900 achieves a shorter stopping distance. A few stopping distances were recorded and
the shortest distance was 44 feet, while the longest was 76 feet. Determine the sample
size needed to make National Motors 95% confident that the average stopping distance in
their sample is within 1 foot of the true mean stopping distance.
Sample size required: n = p(1-p)(z / e)2
A sample of this size makes us 100(1-)% confident that the sample proportion is within
B units of the population proportion.
Polls usually estimate proportions to within 3% (1% is too expensive and 5% is too
wide).
Example 5: A new product is to be test marketed in Toledo, Ohio using TV, radio and
newspaper ads. Two months after the campaign began, 475 out of 601 randomly selected
consumers were aware of the product. Find the sample size needed to be 95% confident
that the proportion of consumers that are aware of the product is within 0.02 of the true
proportion.
Homework: # 72.