1. Estimating 𝜇 When 𝜎 Is Known
Because of time and money constraints, difficulty in finding population members, an so forth, we usually do not have access to all measurements in an entire population.

2. Estimating 𝜇 When 𝜎 Is Known
Because of time and money constraints, difficulty in finding population members, an so forth, we usually do not have access to all measurements in an entire population.
Instead, we rely on information from a sample.

3. Estimating 𝜇 When 𝜎 Is Known
Because of time and money constraints, difficulty in finding population members, an so forth, we usually do not have access to all measurements in an entire population.
Instead, we rely on information from a sample.
In this section we will develop techniques for estimating the population mean 𝜇 using sample data. We assume the standard deviation 𝜎 is known.

4. Estimating 𝜇 When 𝜎 Is Known
Because of time and money constraints, difficulty in finding population members, an so forth, we usually do not have access to all measurements in an entire population.
Instead, we rely on information from a sample.
In this section we will develop techniques for estimating the population mean 𝜇 using sample data. We assume the standard deviation 𝜎 is known.
Some basic assumptions about the random variable 𝑥

5. Estimating 𝜇 When 𝜎 Is Known
Because of time and money constraints, difficulty in finding population members, an so forth, we usually do not have access to all measurements in an entire population.
Instead, we rely on information from a sample.
In this section we will develop techniques for estimating the population mean 𝜇 using sample data. We assume the standard deviation 𝜎 is known.
Some basic assumptions about the random variable 𝑥
1. We have a simple random sample of size 𝑛 drawn from a population of 𝑥 values

6. Estimating 𝜇 When 𝜎 Is Known
Because of time and money constraints, difficulty in finding population members, an so forth, we usually do not have access to all measurements in an entire population.
Instead, we rely on information from a sample.
In this section we will develop techniques for estimating the population mean 𝜇 using sample data. We assume the standard deviation 𝜎 is known.
Some basic assumptions about the random variable 𝑥
1. We have a simple random sample of size 𝑛 drawn from a population of 𝑥 values
2. The value of 𝜎, the population standard deviation of 𝑥, is known.

7. Estimating 𝜇 When 𝜎 Is Known
Because of time and money constraints, difficulty in finding population members, an so forth, we usually do not have access to all measurements in an entire population.
Instead, we rely on information from a sample.
In this section we will develop techniques for estimating the population mean 𝜇 using sample data. We assume the standard deviation 𝜎 is known.
Some basic assumptions about the random variable 𝑥
1. We have a simple random sample of size 𝑛 drawn from a population of 𝑥 values
2. The value of 𝜎, the population standard deviation of 𝑥, is known.
3. If the 𝑥 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑖𝑠 𝑛𝑜𝑟𝑚𝑎𝑙, then our methods work for any sample size 𝑛

8. Estimating 𝜇 When 𝜎 Is Known
Because of time and money constraints, difficulty in finding population members, an so forth, we usually do not have access to all measurements in an entire population.
Instead, we rely on information from a sample.
In this section we will develop techniques for estimating the population mean 𝜇 using sample data. We assume the standard deviation 𝜎 is known.
Some basic assumptions about the random variable 𝑥
1. We have a simple random sample of size 𝑛 drawn from a population of 𝑥 values
2. The value of 𝜎, the population standard deviation of 𝑥, is known.
3. If the 𝑥 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑖𝑠 𝑛𝑜𝑟𝑚𝑎𝑙, then our methods work for any sample size 𝑛
4. If 𝑥 has an unknown distribution, then we require a sample size 𝑛≥30.
However, if the 𝑥 distribution is distinctly skewed and definitely not mound – shaped, a
sample size of 50 or even 100 or higher may be necessary.

9. Estimating 𝜇 When 𝜎 Is Known
point estimate – an estimate of a population parameter given by a single number
𝑥 is the point estimate for 𝜇

10. Estimating 𝜇 When 𝜎 Is Known
point estimate – an estimate of a population parameter given by a single number
𝑥 is the point estimate for 𝜇
margin of error – when using 𝑥 as a point estimate for 𝜇, it is the magnitude of
𝑥 −𝜇 OR 𝑥 −𝜇

11. Estimating 𝜇 When 𝜎 Is Known
point estimate – an estimate of a population parameter given by a single number
𝑥 is the point estimate for 𝜇
margin of error – when using 𝑥 as a point estimate for 𝜇, it is the magnitude of
𝑥 −𝜇 OR 𝑥 −𝜇
confidence level (𝑐) – the reliability of an estimate
- usually set at 0.90 , 0.95 , 0.99

12. Estimating 𝜇 When 𝜎 Is Known
point estimate – an estimate of a population parameter given by a single number
𝑥 is the point estimate for 𝜇
margin of error – when using 𝑥 as a point estimate for 𝜇, it is the magnitude of
𝑥 −𝜇 OR 𝑥 −𝜇
confidence level (𝑐) – the reliability of an estimate
- usually set at 0.90 , 0.95 , 0.99
critical value 𝑧 𝑐 - the number such that the area under the standard normal curve
between − 𝑧 𝑐 and 𝑧 𝑐 equals 𝑐

13. Estimating 𝜇 When 𝜎 Is Known
point estimate – an estimate of a population parameter given by a single number
𝑥 is the point estimate for 𝜇
margin of error – when using 𝑥 as a point estimate for 𝜇, it is the magnitude of
𝑥 −𝜇 OR 𝑥 −𝜇
confidence level (𝑐) – the reliability of an estimate
- usually set at 0.90 , 0.95 , 0.99
critical value 𝑧 𝑐 - the number such that the area under the standard normal curve
between − 𝑧 𝑐 and 𝑧 𝑐 equals 𝑐
𝑃 − 𝑧 𝑐 <𝑧< 𝑧 𝑐 =c
Area = 𝒄
−𝑧 𝑐
𝑧 𝑐

14. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : Find 𝑧 such that 99% of the area under the normal curve lies between
− 𝑧 and 𝑧 0.99

15. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : Find 𝑧 such that 99% of the area under the normal curve lies between
− 𝑧 and 𝑧 0.99
𝑃 − 𝑧 0.99 <𝑧< 𝑧 =0.99

16. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : Find 𝑧 such that 99% of the area under the normal curve lies between
− 𝑧 and 𝑧 0.99
𝑃 − 𝑧 0.99 <𝑧< 𝑧 =0.99
SOLUTION : If A is the area between −𝑧 and 𝑧, then we can use 1−𝐴 2 to find the
area in the left tail
Area = 𝟎.𝟗𝟗
−𝑧 0.99
𝑧 0.99

17. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : Find 𝑧 such that 99% of the area under the normal curve lies between
− 𝑧 and 𝑧 0.99
𝑃 − 𝑧 0.99 <𝑧< 𝑧 =0.99
SOLUTION : If A is the area between −𝑧 and 𝑧, then we can use 1−𝐴 2 to find the
area in the left tail
1− =0.005
Area = 𝟎.𝟗𝟗
−𝑧 0.99
𝑧 0.99

18. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : Find 𝑧 such that 99% of the area under the normal curve lies between
− 𝑧 and 𝑧 0.99
𝑃 − 𝑧 0.99 <𝑧< 𝑧 =0.99
SOLUTION : If A is the area between −𝑧 and 𝑧, then we can use 1−𝐴 2 to find the
area in the left tail
1− =0.005
What z – score corresponds with ?
Area = 𝟎.𝟗𝟗
−𝑧 0.99
𝑧 0.99

19. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : Find 𝑧 such that 99% of the area under the normal curve lies between
− 𝑧 and 𝑧 0.99
𝑃 − 𝑧 0.99 <𝑧< 𝑧 =0.99
SOLUTION : If A is the area between −𝑧 and 𝑧, then we can use 1−𝐴 2 to find the
area in the left tail
1− =0.005
What z – score corresponds with ?
From the table 𝑧=−2.58
Area = 𝟎.𝟗𝟗
−𝑧 0.99
𝑧 0.99
−2.58

20. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : Find 𝑧 such that 99% of the area under the normal curve lies between
− 𝑧 and 𝑧 0.99
𝑃 − 𝑧 0.99 <𝑧< 𝑧 =0.99
SOLUTION : If A is the area between −𝑧 and 𝑧, then we can use 1−𝐴 2 to find the
area in the left tail
1− =0.005
What z – score corresponds with ?
From the table 𝑧=−2.58
For the right tail we need too find the
z – score that corresponds to …
Area = 𝟎.𝟗𝟗
−𝑧 0.99
𝑧 0.99
−2.58

21. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : Find 𝑧 such that 99% of the area under the normal curve lies between
− 𝑧 and 𝑧 0.99
𝑃 − 𝑧 0.99 <𝑧< 𝑧 =0.99
SOLUTION : If A is the area between −𝑧 and 𝑧, then we can use 1−𝐴 2 to find the
area in the left tail
1− =0.005
What z – score corresponds with ?
From the table 𝑧=−2.58
For the right tail we need too find the
z – score that corresponds to …
From the table 𝑧=2.58
Area = 𝟎.𝟗𝟗
−𝑧 0.99
𝑧 0.99
−2.58
2.58

22. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : Find 𝑧 such that 99% of the area under the normal curve lies between
− 𝑧 and 𝑧 0.99
𝑃 − 𝑧 0.99 <𝑧< 𝑧 =0.99
SOLUTION : If A is the area between −𝑧 and 𝑧, then we can use 1−𝐴 2 to find the
area in the left tail
1− =0.005
What z – score corresponds with ?
From the table 𝑧=−2.58
For the right tail we need too find the
z – score that corresponds to …
From the table 𝑧=2.58
** in general, these values will be quite close if not the same
Area = 𝟎.𝟗𝟗
−𝑧 0.99
𝑧 0.99
−2.58
2.58

23. Estimating 𝜇 When 𝜎 Is Known
Here is a table for some confidence levels and their corresponding z values.
Level of Confidence (𝒄)
Critical Value 𝒛 𝒄
0.70 or 70%
1.04
0.75 or 75%
1.15
0.80 or 80%
1.28
0.85 or 85%
1.44
0.90 or 90%
1.645
0.95 or 95%
1.96
0.98 or 98%
2.33
0.99 or 99%
2.58

24. Estimating 𝜇 When 𝜎 Is Known
An estimate is not very valuable unless we have some kind of measure to tell us how good that estimate actually is.

25. Estimating 𝜇 When 𝜎 Is Known
An estimate is not very valuable unless we have some kind of measure to tell us how good that estimate actually is.
𝑃 − 𝑧 𝑐 ∙ 𝜎 𝑛 < 𝑥 −𝜇< 𝑧 𝑐 ∙ 𝜎 𝑛 =𝑐
This equation allows us to calculate an error of tolerance E that serves as a bound
on the margin of error.

26. Estimating 𝜇 When 𝜎 Is Known
An estimate is not very valuable unless we have some kind of measure to tell us how good that estimate actually is.
𝑃 − 𝑧 𝑐 ∙ 𝜎 𝑛 < 𝑥 −𝜇< 𝑧 𝑐 ∙ 𝜎 𝑛 =𝑐
This equation allows us to calculate an error of tolerance E that serves as a bound
on the margin of error.
We can say that the point estimate 𝑥 differs from the population mean 𝜇 by a
maximal margin of error
𝐸= 𝑧 𝑐 ∙ 𝜎 𝑛

27. Estimating 𝜇 When 𝜎 Is Known
An estimate is not very valuable unless we have some kind of measure to tell us how good that estimate actually is.
𝑃 − 𝑧 𝑐 ∙ 𝜎 𝑛 < 𝑥 −𝜇< 𝑧 𝑐 ∙ 𝜎 𝑛 =𝑐
𝑃 −𝐸< 𝑥 −𝜇<𝐸 =𝑐
Substituting this into our equation above yields :
𝐸= 𝑧 𝑐 ∙ 𝜎 𝑛

28. Estimating 𝜇 When 𝜎 Is Known
An estimate is not very valuable unless we have some kind of measure to tell us how good that estimate actually is.
𝑃 − 𝑧 𝑐 ∙ 𝜎 𝑛 < 𝑥 −𝜇< 𝑧 𝑐 ∙ 𝜎 𝑛 =𝑐
𝑃 −𝐸< 𝑥 −𝜇<𝐸 =𝑐
With a little bit of Algebra we get :
𝑥 −𝐸=𝜇= 𝑥 +𝐸
This is the confidence interval for 𝜇 when 𝜎 is known, where
𝐸= 𝑧 𝑐 ∙ 𝜎 𝑛

29. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : Julia enjoys jogging. Usually she jogs 2 miles per day. The standard
deviations of her times is 1.80 minutes. During the past year, Julia has
recorded her times to run 2 miles. She has a random sample of 90 of
these times with a mean of 𝑥 =15.60 minutes. Let 𝜇 be the jogging time
for the entire distribution of Julia’s 2 – mile runs. Find a 95% confidence
interval for 𝜇.

30. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : Julia enjoys jogging. Usually she jogs 2 miles per day. The standard
deviations of her times is 1.80 minutes. During the past year, Julia has
recorded her times to run 2 miles. She has a random sample of 90 of
these times with a mean of 𝑥 =15.60 minutes. Let 𝜇 be the jogging time
for the entire distribution of Julia’s 2 – mile runs. Find a 95% confidence
interval for 𝜇.
SOLUTION : 95% = 0.95 = 𝑐
𝑧 𝑐 =1.96 ( from the table on the previous page )
𝑥 =15.60 , 𝑛=90 and 𝜎=1.8

31. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : Julia enjoys jogging. Usually she jogs 2 miles per day. The standard
deviations of her times is 1.80 minutes. During the past year, Julia has
recorded her times to run 2 miles. She has a random sample of 90 of
these times with a mean of 𝑥 =15.60 minutes. Let 𝜇 be the jogging time
for the entire distribution of Julia’s 2 – mile runs. Find a 95% confidence
interval for 𝜇.
SOLUTION : 95% = 0.95 = 𝑐
𝑧 𝑐 =1.96 ( from the table on the previous page )
𝑥 =15.60 , 𝑛=90 and 𝜎=1.8
1. Calculate 𝐸 𝐸=1.96∙ = =0.37

32. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : Julia enjoys jogging. Usually she jogs 2 miles per day. The standard
deviations of her times is 1.80 minutes. During the past year, Julia has
recorded her times to run 2 miles. She has a random sample of 90 of
these times with a mean of 𝑥 =15.60 minutes. Let 𝜇 be the jogging time
for the entire distribution of Julia’s 2 – mile runs. Find a 95% confidence
interval for 𝜇.
SOLUTION : 95% = 0.95 = 𝑐
𝑧 𝑐 =1.96 ( from the table on the previous page )
𝑥 =15.60 , 𝑛=90 and 𝜎=1.8
1. Calculate 𝐸 𝐸=1.96∙ = =0.37
2. Compute 𝑥 −𝐸<𝜇< 𝑥 +𝐸

33. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : Julia enjoys jogging. Usually she jogs 2 miles per day. The standard
deviations of her times is 1.80 minutes. During the past year, Julia has
recorded her times to run 2 miles. She has a random sample of 90 of
these times with a mean of 𝑥 =15.60 minutes. Let 𝜇 be the jogging time
for the entire distribution of Julia’s 2 – mile runs. Find a 95% confidence
interval for 𝜇.
SOLUTION : 95% = 0.95 = 𝑐
𝑧 𝑐 =1.96 ( from the table on the previous page )
𝑥 =15.60 , 𝑛=90 and 𝜎=1.8
1. Calculate 𝐸 𝐸=1.96∙ = =0.37
2. Compute 𝑥 −𝐸<𝜇< 𝑥 +𝐸
15.60−0.37<𝜇<
15.23<𝜇<15.97

34. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : Julia enjoys jogging. Usually she jogs 2 miles per day. The standard
deviations of her times is 1.80 minutes. During the past year, Julia has
recorded her times to run 2 miles. She has a random sample of 90 of
these times with a mean of 𝑥 =15.60 minutes. Let 𝜇 be the jogging time
for the entire distribution of Julia’s 2 – mile runs. Find a 95% confidence
interval for 𝜇.
SOLUTION : 95% = 0.95 = 𝑐
𝑧 𝑐 =1.96 ( from the table on the previous page )
𝑥 =15.60 , 𝑛=90 and 𝜎=1.8
1. Calculate 𝐸 𝐸=1.96∙ = =0.37
2. Compute 𝑥 −𝐸<𝜇< 𝑥 +𝐸
15.60−0.37<𝜇<
15.23<𝜇<15.97
** we conclude with 95% confidence that the interval from minutes to 15.97
minutes is one that contains the population mean 𝜇 of jogging times for Julia

35. Estimating 𝜇 When 𝜎 Is Known
The next thing we need to consider is the appropriate sample size. In the design stages of a statistical experiment, you need to decide on the confidence level you wish to use, the maximal margin of error (𝐸) and also the minimal sample size.

36. Estimating 𝜇 When 𝜎 Is Known
The next thing we need to consider is the appropriate sample size. In the design stages of a statistical experiment, you need to decide on the confidence level you wish to use, the maximal margin of error (𝐸) and also the minimal sample size.
How to find the sample size 𝑛 for estimating 𝜇 when 𝜎 is known

37. Estimating 𝜇 When 𝜎 Is Known
The next thing we need to consider is the appropriate sample size. In the design stages of a statistical experiment, you need to decide on the confidence level you wish to use, the maximal margin of error (𝐸) and also the minimal sample size.
How to find the sample size 𝑛 for estimating 𝜇 when 𝜎 is known
- the distribution of sample means 𝑥 is approximately normal

38. Estimating 𝜇 When 𝜎 Is Known
The next thing we need to consider is the appropriate sample size. In the design stages of a statistical experiment, you need to decide on the confidence level you wish to use, the maximal margin of error (𝐸) and also the minimal sample size.
How to find the sample size 𝑛 for estimating 𝜇 when 𝜎 is known
- the distribution of sample means 𝑥 is approximately normal
- formula for sample size
𝑛= 𝑧 𝑐 ∙ 𝜎 𝐸 2
where 𝐸= specified maximal margin of error
𝜎= population standard deviation
𝑧 𝑐 = critical value from the normal distribution for the desired
confidence level 𝑐

39. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : A preliminary study of a random sample of 50 salmon showed 𝑠≈2.15
pounds. How large of a sample should be taken to be 99% confident that
the sample mean 𝑥 is within 0.20 pound of the true mean weight 𝜇 ?

40. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : A preliminary study of a random sample of 50 salmon showed 𝑠≈2.15
pounds. How large of a sample should be taken to be 99% confident that
the sample mean 𝑥 is within 0.20 pound of the true mean weight 𝜇 ?
SOLUTION : For a 99% confidence interval 𝑧 𝑐 = 𝑧 0.99 =2.58 ( from the table )

41. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : A preliminary study of a random sample of 50 salmon showed 𝑠≈2.15
pounds. How large of a sample should be taken to be 99% confident that
the sample mean 𝑥 is within 0.20 pound of the true mean weight 𝜇 ?
SOLUTION : For a 99% confidence interval 𝑧 𝑐 = 𝑧 0.99 =2.58 ( from the table )
𝐸=0.20

42. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : A preliminary study of a random sample of 50 salmon showed 𝑠≈2.15
pounds. How large of a sample should be taken to be 99% confident that
the sample mean 𝑥 is within 0.20 pound of the true mean weight 𝜇 ?
SOLUTION : For a 99% confidence interval 𝑧 𝑐 = 𝑧 0.99 =2.58 ( from the table )
𝐸=0.20
𝜎=2.15 ( 50 fish is large enough to permit a good approximation )

43. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : A preliminary study of a random sample of 50 salmon showed 𝑠≈2.15
pounds. How large of a sample should be taken to be 99% confident that
the sample mean 𝑥 is within 0.20 pound of the true mean weight 𝜇 ?
SOLUTION : For a 99% confidence interval 𝑧 𝑐 = 𝑧 0.99 =2.58 ( from the table )
𝐸=0.20
𝜎=2.15 ( 50 fish is large enough to permit a good approximation )
𝑛= 𝑧 𝑐 ∙ 𝜎 𝐸 2 = (2.58 )(2.15)

44. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : A preliminary study of a random sample of 50 salmon showed 𝑠≈2.15
pounds. How large of a sample should be taken to be 99% confident that
the sample mean 𝑥 is within 0.20 pound of the true mean weight 𝜇 ?
SOLUTION : For a 99% confidence interval 𝑧 𝑐 = 𝑧 0.99 =2.58 ( from the table )
𝐸=0.20
𝜎=2.15 ( 50 fish is large enough to permit a good approximation )
𝑛= 𝑧 𝑐 ∙ 𝜎 𝐸 2 = (2.58 )(2.15) = =769.2

45. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : A preliminary study of a random sample of 50 salmon showed 𝑠≈2.15
pounds. How large of a sample should be taken to be 99% confident that
the sample mean 𝑥 is within 0.20 pound of the true mean weight 𝜇 ?
SOLUTION : For a 99% confidence interval 𝑧 𝑐 = 𝑧 0.99 =2.58 ( from the table )
𝐸=0.20
𝜎=2.15 ( 50 fish is large enough to permit a good approximation )
𝑛= 𝑧 𝑐 ∙ 𝜎 𝐸 2 = (2.58 )(2.15) = =769.2
** Any decimal fraction will be rounded to the next whole number

46. Estimating 𝜇 When 𝜎 Is Known
EXAMPLE : A preliminary study of a random sample of 50 salmon showed 𝑠≈2.15
pounds. How large of a sample should be taken to be 99% confident that
the sample mean 𝑥 is within 0.20 pound of the true mean weight 𝜇 ?
SOLUTION : For a 99% confidence interval 𝑧 𝑐 = 𝑧 0.99 =2.58 ( from the table )
𝐸=0.20
𝜎=2.15 ( 50 fish is large enough to permit a good approximation )
𝑛= 𝑧 𝑐 ∙ 𝜎 𝐸 2 = (2.58 )(2.15) = =769.2
** Any decimal fraction will be rounded to the next whole number
So we conclude that a sample size of 770 will be large enough