Download 2015 Lesson 7: Estimation and Confidence Intervals

LESSON 7
LESSON 7: ESTIMATION AND CONFIDENCE INTERVALS
It is usually not feasible to inspect the entire population; means of small samples have more
dispersion or scatter than means of large samples. The Central Limit Theorem states that if all
samples of a particular size are selected from any population, the sampling distribution of the
sample mean is approximately a normal distribution. This approximation improves with larger
samples. Point estimates are sample measures of central tendency such as mean, median
and mode. Sample measures of dispersion such as variance and standard deviation are also
point estimates.
Confidence intervals state the range within which a population parameter probably lies. The
specified probability is called the level of confidence. A 95 percent confidence interval says
that we have a 95 percent level of confidence that a similarly constructed interval will include
the parameter being estimated.
A random sample of 500 restaurants reveals a sample mean yearly income of waitresses of
$ 35,743. The standard deviation of the population is 1092. What are the confidence limits
at the 95% level of confidence?
Confidence Interval - Mean
Mean=35743
z=1.96
s =1092
n=500
Confidence Limits
Lower
Upper
35647
35839
(Distribution Plot using Minitab 16 )
We are 95% confident that the interval from 35647 to 35839 includes the true value of the
mean. Solution:
$35,743 - [(1.96)(1092)]/(500)^.5 < $35,743 <
$35,743 + [(1.96)(1092)]/(500)^.5
Confidence Interval-Mean
Mean= 35743
z=
1.96
StdDev= 1092
n=
500
Confidence limits
Lower
Upper
35647 35839
Confidence Interval-Proportion
Propor= 0.8
z=
1.96
n=
5000
Confidence Limits
Lower
Upper
0.794
0.806
E= 0.006= margin of error
.0.80 -E =.794
0.80 + E =.806
For the Confidence Interval for the Proportion, the Solution is:
Sample proportion - margin of error < Population Prop. < Sample Prop. + margin of error
0.80 - 1.96 *[.8 * .2 / 5000]^.5
<
0.80
0.80 -
<
0.80
0.006
0.794 <0.800< 0.806
< 0.80 + 1.96*[.8 * .2 /5000]^.5
<
0.80 + 0.006
A survey reported that 70% of those responding to a national survey of college freshmen
were interested in taking summer courses. Using the sample size of 49, calculate a 95%
confidence interval for the proportion of college students who are interested in taking
summer courses.
E=z*(p*q/n)^.5 = 1.96((.70)(.30)/49)^.5 = .124 = 12.4%
Confidence interval
.700 - .124 < .700 < .700 + .124
.576 < .700 <
.824
2
A researcher wants to determine the proportion of pro-peace students at Ocean County
College. He has no idea what the sample proportion will be. How large a sample is required
to be 95% sure that the sample proportion is off by no more than 4%?
Z= 1.96 p=0.5 = q since we do not know what the sample proportion will be. E=.04
n= 1.96^2 * 0.5 * 0.5 / 0.04^2 = 600.25
n=601 samples.