1) The sample shows the mean weight of men is 172.55 lbs with a standard deviation of 26 lbs. 2) A 95% confidence interval for the population mean weight is estimated to be between 164.49 lbs and 180.61 lbs. 3) This suggests that the outdated estimate of 166.3 lbs used for safety capacities is likely an underestimate, and updating to the point estimate of 172.55 lbs could help prevent overloading issues.

1. Elementary Statistics
Chapter 7:
Estimating Parameters
and Determining
Sample Sizes
7.2 Estimating a
Population Mean
1

2. 7.1 Estimating a Population Proportion
7.2 Estimating a Population Mean
7.3 Estimating a Population Standard Deviation or Variance
7.4 Bootstrapping: Using Technology for Estimates
2
Chapter 7:
Estimating Parameters and Determining Sample Sizes
Objectives:
• Find the confidence interval for a proportion.
• Determine the minimum sample size for finding a confidence interval for a proportion.
• Find the confidence interval for the mean when  is known.
• Determine the minimum sample size for finding a confidence interval for the mean.
• Find the confidence interval for the mean when  is unknown.
• Find a confidence interval for a variance and a standard deviation.

3. A point estimate is a specific numerical value estimate of a parameter. (A point estimate is a single
value (or point) used to approximate a population parameter. )
Properties of a good estimator:
1. The estimator should be an unbiased estimator. That is, the expected value or the mean of the
estimates obtained from samples of a given size is equal to the parameter being estimated.
2. The estimator should be consistent. For a consistent estimator, as sample size increases, the value
of the estimator approaches the value of the parameter estimated.
3. The estimator should be a relatively efficient estimator; that is, of all the statistics that can be
used to estimate a parameter, the relatively efficient estimator has the smallest variance.
• The sample proportion 𝑝 is the best point estimate of the population proportion p.
• p =
𝑋
𝑁
= population proportion 𝑝 =
𝑥
𝑛
(read p “hat”) = sample proportion, 𝑞 = 1 − 𝑝
Confidence Interval: We can use a sample proportion to construct a confidence interval estimate of
the true value of a population proportion.
Sample Size: We should know how to find the sample size necessary to estimate a population
proportion.
Recall: 7.1 Estimating a Population Proportion
3

4. Point estimate of p: 𝑝 =
𝑈𝐶𝐿+𝐿𝐶𝐿
2
, UCL: Upper Confidence Limit
Margin of error: 𝐸 =
𝑈𝐶𝐿−𝐿𝐶𝐿
2
, LCL: Lower Confidence Limit
Margin of Error & Confidence Interval for Estimating a Population
Proportion p & Determining Sample Size:
When data from a simple random sample are used to estimate a population
proportion p, the margin of error (maximum error of the estimate ),
denoted by E, is the maximum likely difference (with probability 1 – α, such
as 0.95) between the observed (sample) proportion 𝑝 and the true value of
the population proportion p.
4
ˆ
ˆ ˆ
p E
p E p p E

    2 2
ˆ ˆ ˆ ˆ
ˆ ˆ   
pq pq
p z p p z
n n
 
Recall: 7.1 Estimating a Population Proportion
2
ˆ ˆpq
E z
n

2
2
2
ˆ ˆ( )z pq
n
E


2
2
2
( ) 0.25z
n
E

When no estimate of 𝒑
is known: 𝒑 = 𝒒 =0.5
Determine the sample
size n required to
estimate the value of a
population proportion p
TI Calculator:
Confidence Interval:
proportion
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3. 1-prop ZINT
4. Enter: x, n & CL

5. 7.2 Estimating a Population Mean
Key Concept: A point estimate is a specific numerical value estimate of a parameter.
(A point estimate is a single value (or point) used to approximate a population
parameter. )
The sample mean 𝑥 is the best point estimate of the population mean 𝜇.
• Confidence Interval: Use sample data to construct and interpret a confidence
interval estimate of the true value of a population mean µ.
• Sample Size: Find the sample size necessary to estimate a population mean.
5
1. Verify that the two requirements are satisfied: The sample is a simple
random sample and the population is normally distributed or n > 30.
2. σ is unknown: Use t-distribution
3. σ is known: Use z-distribution

6. Confidence Interval for Estimating a Population Mean with σ Not Known: Requirements
1. The sample is a simple random sample.
2. Either or both of these conditions are satisfied: The population is normally distributed or n > 30.
µ = population mean
n = number of sample values
𝑥 = sample mean
E = margin of error
s = sample standard deviation
7.2 Estimating a Population Mean
Margin of Error
Confidence
Interval
2 2
   
      
   
s s
X t X t
n n
 
Degrees of Freedom: df = n − 1 is the number of degrees of freedom used when finding the critical value.
Critical value: tα/2 is the critical value separating an area of α/2 in the right tail of the student t
distribution.
2
s
E t
n

 
  
  6
𝐶𝐼: 𝑥 ± 𝐸 →

7. If a population has a normal distribution, then the distribution of
𝑡 =
𝑥−𝜇
𝑠/ 𝑛
is a Student t distribution for all samples of size n. A
Student t distribution is commonly referred to as a t distribution.
In general, the number of degrees of freedom for a collection of
sample data is the number of sample values that can vary after
certain restrictions have been imposed on all data values. Degrees
of freedom: df = n − 1 (if need be use he closes value in the
table or to be conservative use the next lower number of df.)
7.2 Estimating a Population Mean, Student t Distribution
The t distribution is similar to SND:
1. It is bell-shaped.
2. It is symmetric about the mean.
3. The mean, median, and mode are
equal to 0 and are located at the
center of the distribution.
4. The curve never touches the x axis. 7
The t distribution differs from SND:
1. The variance is greater than 1.
2. The t distribution is actually a family
of curves based on the concept of
degrees of freedom, which is related
to sample size.
3. As the sample size increases, the t
distribution approaches the standard
normal distribution.

8. Find the Critical Value
tα/2 corresponding to
95% CI, n =15
8
Example 1
Solution
n = 15,df = n − 1 = 14.
95% CL: 𝛼 = 0.05 → Area of
0.025 in each of the two tails of
the t distribution
TI Calculator:
T- Distribution: find the t-score
1. 2nd + VARS
2. invT(
3. 2 entries (Left Area,df)
4. Enter

9. There is a claim that garlic lowers cholesterol levels. To test the effectiveness of garlic, 49 subjects were
treated with doses of raw garlic, and their cholesterol levels were measured before and after the
treatment. The changes in their levels of LDL cholesterol (in mg/dL) have a mean of 0.4 and a standard
deviation of 21.0. Use the sample statistics of n = 49, 𝑥 = 0.4, and s = 21.0 to construct a 95%
confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What
does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol?
Example 2
Solution: Given: simple random sample and n = 49 (i.e., n > 30).
n = 49, 𝑥 = 0.4, and s = 21.0 , 95% CI = ?
9
,x E x E x E    
We are 95% confident that the limits of –5.6 and 6.4 actually do contain the
value of μ, the mean of the changes in LDL cholesterol for the population.
21
2.009
49
E
 
  
 
2
s
E t
n

 
  
 
n = 49 ⇾ df = 49 – 1 = 48,
Closest df = 50, two tails: 𝑡 𝛼/2 = 2.009
6.027
0 4 6 027 0 4 6 027   . . . .
TI Calculator:
T- interval
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3. T - Interval
4. Enter Data or Stats
5. Enter
5 627 6 427  . .
The confidence interval limits contain the value of 0, so it is
possible that the mean of the changes in LDL cholesterol is
equal to 0. This suggests that the garlic treatment did not
affect the LDL cholesterol levels. It does not appear that the
garlic treatment is effective in lowering LDL cholesterol.

10. Ten randomly selected people were asked how long they slept at night.
The mean time was 7.1 hours, and the standard deviation was 0.78 hour.
Find the 95% confidence interval of the mean time. Assume the variable is
normally distributed.
Example 3
Solution: Given: ND, n = 10
𝑥 = 7.1 hrs, and s = 0.78 hrs , 95% CI = ?
10
n = 10 ⇾ df = 10 – 1 = 9 ⇾ 𝑡 𝛼/2 = 2.262
2 2
   
      
   
s s
X t X t
n n
 
0.78 0.78
7.1 2.262 7.1 2.262
10 10
   
      
   

7.1 0.56 7.1 0.56    6.54 7.66 
,x E x E x E     2
s
E t
n

 
  
 
TI Calculator:
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11. Finding the Point Estimate and E from a Confidence Interval
Point estimate of µ: 𝑥 =
𝑈𝐶𝐿+𝐿𝐶𝐿
2
, UCL: Upper Confidence Limit
Margin of error: 𝐸 =
𝑈𝐶𝐿−𝐿𝐶𝐿
2
,LCL: Lower Confidence Limit
11
7.2 Estimating a Population Mean
Determine the sample size n required to estimate the value of a population
mean µ.
Confidence Interval for Estimating a Population Mean with σ Known
2
2  
  
 
z
n
E
 
2 2
   
      
   
X z X z
n n
 
 
𝐶𝐼: 𝑥 ± 𝐸 →2E z
n

 
  
 

12. People have died in boat and aircraft accidents because an obsolete estimate of the mean weight
of men was used. The mean weight of men has increased considerably, so we need to update our
estimate of that mean so that boats, aircraft, elevators, and other such devices do not become
dangerously overloaded. Using the weights of men from a random sample, we obtain these sample
statistics for the simple random sample: n = 40 and 𝑥 = 172.55 lb. Research from several other sources
suggests that the population of weights of men has a standard deviation given by σ = 26 lb.
Example 4
Solution: Given: Random sample and n = 40 ( n > 30), n = 40, 𝑥 = 172.55lb, and σ = 26 lb, 95% CI = ?
12
26
1.96
40
E
 
  
 
α = 0.05, so zα/2= ± 1.96
8.0575
172 55 8 0575 172 55 8 0575   . . . .
2E z
n

 
  
 
PE of 𝝁 = 𝒙 = 172.55lb
,x E x E x E    
164 49 180 61 . .
a. Find the point estimate (PE) of the mean weight of the population of all men.
c. What do the results suggest about the mean weight
of 166.3 lb that was used to determine the safe
passenger capacity of water vessels in 1960 ?
b. Construct a 95% confidence interval estimate of the mean weight of all men.
Based on the confidence interval, it is possible that the mean weight of 166.3 lb used in 1960 could
be the mean weight of men today. However, the best point estimate of 172.55 lb suggests that the
mean weight of men is now considerably greater than 166.3 lb. An underestimate of the mean
weight of men could result in lives lost through overloaded boats and aircraft, these results strongly
suggest that additional data should be collected.
TI Calculator:
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13. A researcher wishes to estimate the number of days it takes a car dealer to sell a particular brand of car. A
sample of 50 such cars had a mean time on the dealer’s lot of 54 days. Assume the population standard
deviation to be 6.0 days. Find the best point estimate of the population mean and the 95% confidence
interval of the population mean.
Example 5
Solution: Given: Random sample and n = 50 (n > 30)
n = 50, 𝑥 = 54, and σ = 6, 95% CI = ?
13
54, 6.0, 50,95% 1.96X n z    
PE of 𝜇 = 𝑥 = 54
6.0 6.0
54 1.96 54 1.96
50 50
   
      
   

54 1.7 54 1.7   
52.3 55.7 
52 56 
2E z
n

 
  
 
,x E x E x E    

14. A survey of 30 emergency room patients found that the average waiting time for
treatment was 174.3 minutes. Assuming that the population standard deviation is 46.5
minutes, find the best point estimate of the population mean and the 99% confidence of
the population mean.
Example 6
Solution: Given: Random sample and n = 30 ,
𝑥 = 174.3 min, and σ = 46.5 min, 99% CI = ?
14
99% 2.575z 
PE of 𝜇 = 𝑥 = 174.3min
46.5 46.5
174.3 2.575 174.3 2.575
30 30

   
      
   
174.3 21.861 174.3 21.861    152.439 196.161 
0.0050.005
2E z
n

 
  
 
,x E x E x E    

15. How many statistics students must be randomly selected for IQ
tests if we want 95% confidence that the sample mean is within 3 IQ points of
the population mean? Assume normal distribution with σ =15
15
Example 7
2
2  
  
 
z
n
E
  2
1.96 15
3
 
  
 
96.04 97n 
A scientist wishes to estimate the average depth of a river. He wants to be 99%
confident that the estimate is accurate within 2 feet. From a previous study, the
standard deviation of the depths measured was 4.33 feet. How large a sample is
required?
Example 8
99% 2.58, 2, 4.33z E    
2
2  
  
 
z
n
E
  2
2.575 4.33
2
 
  
 
31.07 32n 
95% 1.96
3, 15
z
E 
 
 
2
2z
n
E
  
  
 

16. 16
Example 9 (Time)
Solution: Requirement Check:
(1) The sample is a simple random sample.
(2) n = 15 < 30, so we need to investigate normality.
Sample data appear to
be from a normally
distributed population.
𝐶𝐼: 𝑥 ± 𝐸 →
2E z
n

 
  
 
,x E x E x E    
30.9 1.4760
29.4 32.4hg hg

 
2.9
1.96 1.4760
15
E
 
  
 
Given the following Random sample data for n =
15 , 𝑥 = 30.9 hg, and σ = 2.9 hg for birth
weights of, construct a 959% CI of the mean
birth weight of all girls.
33 28 33 37 31 32 31
28 34 28 33 26 30
31 28

17. Choosing an Appropriate Distribution: t or Z
Conditions Method
σ not known and normally distributed population or σ not known
and n > 30
Use student t distribution
σ known and normally distributed population or σ known and n >
30 (In reality, σ is rarely known.)
Use normal (z) distribution.
Population is not normally distributed and n ≤ 30. Use the bootstrapping method or
a nonparametric method.
17

18. Dealing with Unknown σ When Finding Sample Size
1. Use the range rule of thumb to estimate the standard deviation as follows: σ
 range/4, where the range is determined from sample data.
2. Start the sample collection process without knowing σ and, using the first
several values, calculate the sample standard deviation s and use it in place
of σ. The estimated value of σ can then be improved as more sample data
are obtained, and the sample size can be refined accordingly.
3. Estimate the value of σ by using the results of some other earlier study.
18

19. Dealing with Unknown σ When Finding Critical values
a. Find the critical t value for α = 0.05 with df = 16 for a right-tailed t test.
Find the 0.05 column in the top row and 16 in the left-hand column.
The critical t value is +1.746.
19
b. Find the critical t value for α = 0.01 with d.f. = 22 for a left-tailed test.
Find the 0.01 column in the One tail
row, and 22 in the d.f. column.
The critical value is t = –2.508 since
the test is a one-tailed left test.
c. Find the critical value for α = 0.10 with d.f. = 18 for a
two-tailed t test.
Find the 0.10 column in the Two tails row, and 18 in the d.f.
column. The critical values are 1.734 and –1.734.