Download Finding Margin of Error and Confidence Intervals

Avon High School
Section:
12.4
ACE COLLEGE ALGEBRA II - NOTES


Z-Scores
Margin of Error & Confidence Intervals
Mr. Record: Room ALC-129
Semester 2 - Day 55
Z-Scores
You can compare samples to determine if the difference in mean or proportion for a large population, based
on a given confidence level, is significant. If a population is large and there are at least 30 data points in a
sample, then the means and proportions can be compared using a normal distribution.
An important measure for normally distributed data is the z-score, which indicates the number of standard
deviations a value lies above or below the mean of a population. When finding the z-score of a data point of
x
a population, the formula is z 
, where x is a data point,  is the mean of a population, and  is the

xx
standard deviation of the population. For a sample, the z-score of a data point is found by using z 
s
Where x is the mean of the sample and s is the standard deviation.
Example 1
Finding Z – Scores
In a given population, the weights of newborns are normally distributed about the mean 3250 g. The
standard deviation of the population is 500 g.
a. What is the z-score of a newborn weighing 2500 g?
b. What is the z-score of a newborn weighing 4500 g?
c. What is probability that a newborn weights between 2270 g and 4230 g? Use z-scores of the weights and
the normal curve.
To compare the mean of a sample with the mean of a population, you use the formula below.
Comparing the Mean of a Sample to the Mean of a Population
x
,
z

n
where x is the mean of the sample,  is the mean of the population,
of the population, and n is the sample size.
 is the standard deviation
A z-score that is between  1.96 and 1.96 means that at a 95% confidence level, there is not a significant
difference between the sample mean and the population mean. A z-score of less than  1.96 or greater than
1.96 indicates that at a 95% confidence level, the differences between the sample mean and the population
mean is significant and the differences are not simply due to chance.
Example 2
Comparing the Mean of a Sample to the Mean of a Population
A company that develops fertilizers wants to know whether either of the two new fertilizers they have in
development shows a significant difference in the growth of plants based on a 95% confidence level. The
company has data on the growth of bean plants without fertilizers. For a growth period of one month, the
population of the beans grown without fertilizers have a mean of 20 cm with a standard deviation of 1 cm.
a. A company researcher chooses 30 plants at random and uses fertilizer A for one month. The researcher
finds that after using fertilizer A, the mean of the bean plants’ growth is 20.4 cm. What is the z-score for
the mean of the sample treated with fertilizer A compared to the population of beans planted without
fertilizer?
b. Does fertilizer A meet a 95% confidence level for having growth that is significantly different from
growth without fertilizer?
c. To test fertilizer B, the researcher chooses 35 plants at random. The bean plants’ growth averages 20.3
cm. What is the z-score for the mean of the sample treated with fertilizer B compared to the population
of bean plants without fertilizer.
d. Does fertilizer B meet a 95% confidence level for having growth that is significantly different from
growth without fertilizer?
e. Based on these data, would you advise the company to market fertilizer A or fertilizer B? Explain.
Margin of Error
When investigating any sample of data, there is always going to be some margin of error. The margin of
error can help you find the interval in which the mean of the population is likely to be in. The margin of
error is based on the size of the sample and, in addition, the confidence interval desired.
A 95% confidence interval means that the probability is 95% tha the true population mean is within a range
of values called a confidence interval. It also means that when you select many different large samples
from the same population, 95% of the confidence intervals will actually contain the population mean.
The normal distribution to the left shows that 95% of values
are between -1.96 and 1.96 standard deviations from the
mean. To find the margin of error based on the mean of a
large set of data at a 95% confidence level, you must use
the formula below.
Margin of Error for Normally Distributed Data
Margin of Error (ME)  1.96 
s
n
Where s is the standard deviation of the sample data, and
n is the number of values in the sample.
The confidence interval for the population mean
Example 3
 is x  ME    x  ME , where x is the sample mean.
Finding Margin of Error and Confidence Intervals
A grocery store manager wanted to determine the wait times for
customers in the express lines. He times customers chosen at random
and compiled the data in the chart to the right.
a. What is the mean and standard deviation of the sample? Round to the
nearest tenth of a minute.
b. At a 95% confidence level, what is the appropriate margin of error? Round to the nearest tenth of a
minute.
c. What is the confidence interval for a 95% confidence level?
d. What is the meaning of the interval in terms of wait times for customers?
You can also find the margin of error and the confidence interval for a sample proportion. A sample
x
proportion p , is the ratio , where x is the number of times an event occurs in a sample of size n.
n
Example 4
Finding Sample Portions
Find the sample portion for each situation. Write the ratios as percents rounded to the nearest tenth
of a percent.
a. In a poll of 1085 voters selected randomly, 564 favor Candidate A.
b. A coin is tossed 40 times, and it comes up heads 25 times.
To find the margin of error for a sample proportion at a 95% confidence level, use the formula below.
Margin of Error for a Sample Proportion
Margin of Error (ME)  1.96 
where

p 1 p

n
p is the sample proportion and n is the sample size.
The confidence interval for the population mean
Example 5
 is p  ME  p  p  ME .
Finding Margin of Error of a Sample Proportion
Find (i) the sample proportion, (ii) the margin of error, and (iii) the 95% confidence interval for the
population proportion for each sample below.
a. In a survey of 530 randomly selected high school students, 280 preferred watching football to watching
basketball.
b. In a simple random sample of 500 people, 342 responded using social networkng sites on the internet.
Comparing the Proportion of a Sample to the Proportion of a
Population
To compare the proportion of a sample with the proportion of a population, you use the formula below.
Comparing the Proportion of a Sample to the Proportion of a Population
p p
,
z
p 1  p 
n
where
p is the sample proportion, p is the population proportion, and n is the sample size.
For a 95% confidence level, a z-score between  1.96 and 1.96 means that difference between the sample
and the population is not significant. A z-score of less than  1.96 or greater than 1.96 means that the
difference in the proportions is significant, and not simply due to chance.
Example 7
Comparing the Proportion of a Sample to the Proportion of a Population
Suppose your teacher accidently gives you a true-false calculus test, instead of an Algebra 2 test.
There are 40 true-false questions, and you and your classmates answer the questions randomly. The
probability of getting an answer correct is ½. You can use ½ as the population proportion.
a. Develop a simulation for answering the 40 questions. Record the results.
b. What is the z-score for the sample based on your trials?
c. Compare your results with the results obtained by the rest of the class. At a 95% confidence level, what
interval represents the percentage of the questions a student would get correct simply by guessing?