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Chapter 6
Estimates and Sample Sizes with One Sample
Section 6.2 – Estimating a Population Proportion
Requirements for Using a Normal Distribution as an
Approximation to a Binomial Distribution
1. The sample must be a simple random sample.
2. The conditions for the Binomial Distribution are satisfied.
3. Both np and nq are greater than or equal to 5.
Notation
p
p̂
(can be in proportion, percent or probability form)
q̂
Terms:
Point Estimate –
Confidence Interval –
Confidence Level –
α
1–α
Critical Values
zα/2 is the symbol used to denote P Z  z α/ 2    / 2
Therefore, P  z α/ 2  Z  z α/ 2   1  
Margin of Error
Sample Size for Estimating Proportion p
Examples
1. Find the critical values for 90%, 95%, 98%, and 99%.
2. Use the confidence interval (0.444, 0.484) to find the point estimate and the margin of error.
3. A private consulting firm is given a governmental grant to study underreporting of income
on federal income tax returns. The firm takes a random sample of 160 taxpayers and
determines that 64 of them underreported their income on their tax return.
a) What is the population proportion p in this setting?
b) What is the sample proportion of taxpayers underreporting their income on their tax
return?
c) What is the standard deviation for this proportion?
d) Construct a 95% confidence interval for this population proportion.
e) Interpret the resulting interval from part (d) in words that a statistically naïve reader
would understand.
4. In a random sample of 100 students at a particular college, 60 indicated that they favored
having the option of receiving pass-fail grades for elective courses. Obtain a 95% confidence
interval for the proportion of the population of students who favor pass-fail grades for
elective courses. Does this confidence interval contain the value p = .5? Explain why this
particular value might be of interest.
5. Suppose you want a margin of error of 0.060 with a confidence interval of 99% and that you
know from a previous study that pˆ  .60 . What sample size should you use?
Section 6.3 – Estimating a Population Mean: σ Known
Requirements for Estimating μ when σ is known
1. The sample must be a simple random sample.
2. The value of σ is known
3. Either the population is already normal or n ≥ 30 so the Central Limit Theorem can be
applied.
Terms:
Point Estimate –
Confidence Interval –
Confidence Level –
Critical Values
Margin of Error
Sample Size for Estimating μ
Examples
1. Look at exercises 5-8. (pg 282)
2. The National Assessment of Educational Progress includes a short test of quantitative skills,
covering mainly basic arithmetic and the ability to apply it to realistic problems. This test
was given to 1000 women of ages 21 to 25 years. Their mean quantitative score was 275.
Suppose that you know that σ is 58. Give a 95% confidence interval for the mean score in
the population of all young women ages 21 to 25 years. Interpret the resulting interval in
words that a statistically naïve reader would understand.
3. The U.S. Department of Agriculture wants to determine the average number of eggs that
children under 14 years of age consume each year. A random sample of 900 such children is
obtained. In this sample, the average number of eggs consumed was x = 92. Assume that
σ = 16.
a) Find a 95% confidence interval for the mean.
b) Suppose the department was hoping that µ = 95. What would you tell them based on
your interval?
4. A professor wants to know the average age of the night students at a certain college. The
variance (σ2) is known to be 36. A random sample of 400 night students yields a sample
mean of 27 years.
a) Find a 90% confidence interval for the mean.
b) What is the width for a 99% confidence interval for the mean?
5. Starting annual salaries for college graduates with business administration degrees are
believed to have a standard deviation of approximately $1,700. How large a sample should
be taken if we want to be 95% confident that the maximum sampling error is $500?
Section 6.4 – Estimating a Population Mean: σ Not Known
Requirements for Estimating μ when σ is not known
1. The sample must be a simple random sample.
2. Either the population is already normal or n ≥ 30 so the Central Limit Theorem can be
applied.
Terms:
Point Estimate –
Student t distribution
The use of the t distribution is based on the following argument. If the population is
normally distributed with mean  and variance  2, the sample mean is normally
distributed (assuming the CLT applies) with mean  and standard deviation
standardized Z score, Z 
X 
/ n

n
and the
, is a standard normal variable. If  is unknown and
replaced by the sample standard deviation s in the Z score formula, the Z score no longer
follows the standard normal distribution. If we replace the true standard deviation by its
X 
estimate s, we obtain what is called the t score, where t =
s/ n
This t score follows what is called the Student t distribution.
The statistic t can be thought of as an estimated standardized normal variable because the
estimated standard deviation s is used instead of  . The larger the sample becomes, the
more the t becomes a normal.
Characteristics of the t Distribution
1. The Student t distribution is different for different sample sizes.
2. It is symmetric about 0 and ranges from –  to .
3. The distribution is bell shaped and has approx. the same appearance as the standard
normal.
4. The mean is 0.
5. The t distribution depends on a parameter  (nu) called the degrees of freedom of the
distribution. When constructing confidence intervals for the population mean, the
appropriate number of degrees of freedom is  = (n –1), where n is the sample size.

6. The variance of the t distribution is
for  >2
 2
7. The variance is always > 1.
8. As  increases, the variance of the t distribution approaches 1 and the shape approaches
that of the standard normal distribution.
9. Because the variance of t exceeds 1, the t distribution is slightly flatter in the middle
than the standard normal distribution and has thicker tails.
To use the table, we must know the degrees of freedom and how much area we want in the
right tail of the curve.
The symbol t/2, denotes the value of t such that the area to its right is /2 and t has 
degrees of freedom. The value t/2, satisfies the equation P(t > t/2,) = /2 where the
random variable t has the t distribution with  degrees of freedom (d.f.).
Conditions for Using the Student t distribution
1. σ is unknown
2. Either the population has a distribution that is essentially normal or n ≥ 30.
Confidence Interval –
Margin of Error
Examples
1. Look at exercises 1-8. (pgs 294-295)
2. A large university is considering giving each faculty member a telephone answering machine
rather than having calls forwarded to a secretary when the professor is not in his or her office.
A budget committee wants to estimate the average amount of time that secretaries spend
handling phone calls for absent faculty members. Because getting the information is
relatively expensive, the sample size must be kept small. A random sample of 18 secretaries
is observed. The results are x = 33 minutes per day and s 2 = 320. Assuming the population
distribution is normal, find a 99% confidence interval for the mean number of minutes per
day that secretaries handle phone calls for absent professors.
3. A company has a new process for manufacturing large artificial sapphires. The production of
each gem is expensive, so the number available for examination is limited. In a trial run, 12
sapphires are produced. The mean weight for these 12 gems is x = 6.75 carats, and the
sample standard deviation is s = 0.33 carats. Assuming the population distribution is
normal, find a 95% confidence interval for the mean weight for this type of artificial
sapphire.
4. The life in hours of a 75-watt light bulb is known to be normally distributed with a standard
deviation of σ = 25 hours. A random sample of 20 bulbs has a mean life of x =1014 hours.
Construct a 95% confidence interval for the mean life.
Section 6.5 – Estimating a Population Variance
Requirements for Estimating σ or σ
2
1. The sample must be a simple random sample.
2. The population must have normally distributed values (even if the sample is large).
Terms:
Chi-Square distribution
1.
2.
3.
4.
This distribution is skewed to the right.
Its values are non-negative. (zero or positive)
The degrees of freedom are d.f. = n – 1.
Each chi-square distribution changes as the degrees of freedom changes. As the degrees
of freedom become larger, the distribution becomes more normal.
5. The chi-square table (A-4) gives the area to the RIGHT of the critical value.
The best point estimate for σ 2 is s 2.
The common point estimate for σ is s, even though it is a biased estimator.
2 
n 1 s 2
2
Confidence Interval:
n 1s 2
 R2
 2 
n 1s 2
 L2
Determining Sample Size:
Use Table 6-2 on page 305.
Examples:
Problems 1, 5, and 9 from the book.
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