1. Confidence Interval with z
“Based on the sample,
we are ____% confident that
the population mean, 𝜇,
is between _____ and ____.”
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2. Confidence Interval with z
Inputs
Outputs
A sample of 𝑛 items
A list of 𝑛 data values measured in the sample
The mean of the sample data, 𝑥
The population standard deviation, 𝜎, is known
A chosen “Confidence Level”, like 90%, 95%, 99%
“Margin of Error”, 𝐸=𝑧 𝛼/2 ∙ 𝜎 𝑛
A low-to-high confidence interval, centered at your sample mean: 𝑥 −𝐸 to 𝑥 +𝐸
“I’m ___% sure that the population mean, is in this interval.”
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3. When can you do this legally?
Anytime that the population standard deviation, 𝜎 is known,
“Known” may mean “previous studies indicate that 𝜎 is…”
~ ~ ~ AND ~ ~ ~
At least one of these conditions is true:
It’s a “large” sample, sample size 𝑛≥30
Or you know that the population is normally distributed
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4. 95% Confidence Interval of the Mean from Bluman’s slides © McGraw Hill

5. Example – Hours of studying
Problem
By-hand solution
Sample of 𝑛=78 students surveyed
Sample mean 𝑥 =15.0 hours of studying per week
Suppose 𝜎=2.3 hours is known.
Find the 95% confidence interval for hours studied.
From page 361 of Beginning Statistics, by Warren, Denley, and Atchley, © 2008 Hawkes Learning Systems.
Find 𝑧 𝛼/2 corresponding to 95% confidence interval.
Find 𝐸= 𝑧 𝛼/2 ∙ 𝜎 𝑛
Form the confidence interval: 𝑥 −𝐸<𝜇< 𝑥 +𝐸
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6. Example – Hours of studying
Details
By-hand solution
95% in the middle
area in the middle
– = in two tails
/ 2 = in each tail
What z has area to its left? z = -1.96
So use 𝑧=1.96, positive
Find 𝑧 𝛼/2 corresponding to 95% confidence interval.
Find 𝐸= 𝑧 𝛼/2 ∙ 𝜎 𝑛
Form the confidence interval: 𝑥 −𝐸<𝜇< 𝑥 +𝐸
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7. Example – Hours of studying
Details
By-hand solution
𝐸=1.96∙
𝐸=0.51
Confidence interval is 15−0.51<𝜇<
14.49<𝜇<15.51 hours of studying per week
Find 𝑧 𝛼/2 corresponding to 95% confidence interval.
Find 𝐸= 𝑧 𝛼/2 ∙ 𝜎 𝑛
Form the confidence interval: 𝑥 −𝐸<𝜇< 𝑥 +𝐸
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8. What does it mean? Details Interpretation 𝐸=1.96∙ 2.3 78 𝐸=0.51
𝐸=1.96∙
𝐸=0.51
Confidence interval is 15−0.51<𝜇<
14.49<𝜇<15.51 hours of studying per week
The true mean is within 0.51 hours, high or low, of our sample mean
We’re 95% confident of that.
We’re 95% confident that the true mean number of hours studied is between and hours/wk.
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9. Example – Hours of studying
Problem
TI-84 Solution
Sample of 𝑛=78 students surveyed
Sample mean 𝑥 =15.0 hours of studying per week
Suppose 𝜎=2.3 hours is known.
Find the 95% confidence interval for hours studied.
From page 361 of Beginning Statistics, by Warren, Denley, and Atchley, © 2008 Hawkes Learning Systems.
STAT, TESTS, 7:Zinterval
Note Inpt: Stats
Highlight Calculate
Press ENTER
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10. Example – Hours of studying
Problem
TI-84 Solution
Sample of 𝑛=78 students surveyed
Sample mean 𝑥 =15.0 hours of studying per week
Suppose 𝜎=2.3 hours is known.
Find the 95% confidence interval for hours studied.
From page 361 of Beginning Statistics, by Warren, Denley, and Atchley, © 2008 Hawkes Learning Systems.
11/20/2018

11. 90% vs. 95% vs. 99% Confidence
Recall: Margin of Error is 𝐸= 𝑧 𝛼/2 ∙ 𝜎 𝑛
The Level of Confidence determines 𝑧 𝛼/2
If Level of Confidence is 90%, 𝑧 𝛼/2 =1.645
If Level of Confidence is 95%, 𝑧 𝛼/2 =1.96
If Level of Confidence is 98%, 𝑧 𝛼/2 =2.33
If Level of Confidence is 99%, 𝑧 𝛼/2 =2.575
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12. 90% vs. 95% vs. 99% Confidence
Recall: Margin of Error is 𝐸= 𝑧 𝛼/2 ∙ 𝜎 𝑛
The Level of Confidence determines 𝑧 𝛼/2
If you choose a higher Level of Confidence %,
The 𝑧 𝛼/2 value is higher.
Which causes a higher Margin of Error.
Which makes the confidence interval wider.
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13. 90% vs. 95% vs. 99% Confidence Again with 𝑛=78, 𝑥 =15.0, 𝜎=2.3
90% confidence interval
narrow confidence interval, (14.572,15.428)
95% confidence interval
medium confidence interval, (14.49, 15.51)
99% confidence interval
Wider confidence interval, (14.329, )
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14. How big of a sample do I need?
Inputs
Calculations
“I want a ____% confidence level.” (which determines the 𝑧 𝛼/2 value)
The population standard deviation is 𝜎.
“I want the margin of error to be no bigger than 𝐸.”
𝑛= 𝑧∙𝜎 𝐸 2
Always bump up 𝑛 to the next highest integer. Bump, don’t round. (Unless 𝑛 just happened to come out to an exact integer, very rare.)
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15. Sample Size Example 𝑛= 𝑧∙𝜎 𝐸 2 𝑛= (1.96)∙(3.25) (0.5) 2
Inputs
Calculations
“How many credit cards do you have?”
Suppose you know that the standard deviation 𝜎=3.25 cards.
And you can tolerate an error of 0.50 (half a card).
And you want a 95% confidence interval.
(Taken from Hawkes, page 362)
𝑛= 𝑧∙𝜎 𝐸 2
𝑛= (1.96)∙(3.25) (0.5) 2
𝑛= Bump it up!
We need a sample of 163 people.
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16. Vocabulary: “Point Estimate”
Our sample mean, 𝑥 , is a point estimate of the population mean, 𝜇.
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17. When you have only the raw data
Many book problems are nice
Raw data only more real-life
Textbook problems are nice to you, usually
They usually just tell you the 𝑥 , the 𝑛, the 𝜎, and the desired confidence interval %.
They do this to save time
They do this so you can focus on the big picture, finding the confidence interval
You’re doing your own real-life statistical research
All you have is the raw data, a bunch of measurements.
But if you have only the raw data, you have to calculate the 𝑥 and the 𝑛.
Book tells you only 𝜎 and which confidence level %
And then apply the formula.
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18. When you have raw data and TI-84
Put the data into a TI-84 list, such as L1.
If there are frequencies, put them into list L2.
Choose Inpt: Data, instead of Stats
It still asks for population 𝜎.
Tell it which List (like 2ND 1 for L1)
If no frequencies, keep Freq:1
C-Level decimal as usual.
Highlight Calculate
Press ENTER.
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19. Example 7-3: Credit Union Assets (from Bluman © McGraw Hill)
The following data represent a sample of the assets (in millions of dollars) of 30 credit unions in southwestern Pennsylvania.
Find the 90% confidence interval of the mean.
(Assume that the population is normally distributed.)
The data:
Bluman, Chapter 7

20. Example 7-3: Credit Union Assets
Step 4: Substitute in the formula. (BUT TRY TI-84 LIST INSTEAD)
One can be 90% confident that the population mean of the assets of all credit unions is between $6.752 million and $ million, based on a sample of 30 credit unions.
Bluman, Chapter 7