1. Confidence Interval with t
“Based on the sample,
we are ____% confident that
the population mean, 𝜇,
is between _____ and ____.”
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2. Confidence Interval with t, instead of 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 unknown
A chosen “Confidence Level”, like 90%, 95%, 99%
“Margin of Error”, 𝐸=𝑡 𝛼/2 ∙ 𝒔 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 should you use 𝑡 instead of 𝑧?
Anytime that the population standard deviation, 𝜎 is unknown, but we can use our sample standard deviation, 𝑠, in its place.
~ ~ ~ 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. What is this new 𝑡, anyway?
When the population standard deviation 𝜎 is unknown, we have more uncertainty.
𝑡 is kind of like 𝑧 but it takes this extra uncertainty into account.
So 𝑡 𝛼/2 will be a little bigger than the 𝑧 𝛼/2 .
Other than that, it’s all going to work the same as the 𝑧 worked.
Your challenge is to know when to use which.
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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= is unknown.
But supppose our sample standard deviation 𝒔=𝟐.𝟑
Find the 95% confidence interval for hours studied.
From page 361 of Beginning Statistics, by Warren, Denley, and Atchley, © 2008 Hawkes Learning Systems.
Find 𝑡 𝛼/2 corresponding to 95% confidence interval and the degrees of freedom, 𝑑.𝑓.=𝑛−1
Find 𝐸= 𝑡 𝛼/2 ∙ 𝒔 𝑛
Form the confidence interval: 𝑥 −𝐸<𝜇< 𝑥 +𝐸
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6. Example – Hours of studying
Details
By-hand solution
Table F: The t Distributions
95% in the middle
5%, or 0.05 in two tails
“Degrees of Freedom”, 𝑑𝑓 = 𝑛 – 1 = 78−1 = 77
Column “Two tails 0.05”
Row 𝑑𝑓=75 is closest
𝑡 𝛼/2 =1.992
Find 𝑡 𝛼/2 corresponding to 95% confidence interval.
Find 𝐸= 𝑡 𝛼/2 ∙ 𝑠 𝑛
Form the confidence interval: 𝑥 −𝐸<𝜇< 𝑥 +𝐸
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7. Example – Hours of studying
Details
By-hand solution
𝐸=1.992∙
𝐸=0.52
Confidence interval is 15−0.52<𝜇<
14.48<𝜇<15.52 hours of studying per week
Find 𝑡 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=1.992∙ 2.3 78 𝐸=0.52
𝐸=1.992∙
𝐸=0.52
Confidence interval is 15−0.52<𝜇<
14.48<𝜇<15.52 hours of studying per week
The true mean is within 0.52 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. Exact 𝑡 𝛼/2 values The printed table is limited.
We had to take the closest row.
Exact value using TI-84 invT(area to left, df)
Compare to our closest row 1.992
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10. Example 7-5: Using Table F
(From Bluman slides, © McGraw Hill)
Find the tα/2 value for a 95% confidence interval when the sample size is Degrees of freedom are d.f. = 21.
Bluman, Chapter 7

11. Compare that to invT 95% confidence interval, df = 21
The table says 𝑡 𝛼/2 =2.080
TI-84 invT gives exact value
Compare to 𝑧 𝛼/2 value
You can see how 𝑡 is “wider”, building in the added uncertainty because 𝛼 is unknown.
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12. 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 is unknown, but we have 𝑠=2.3.
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, 8:TIinterval
Note Inpt: Stats
Highlight Calculate
Press ENTER
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13. 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 is unknown, but we have 𝑠=2.3.
Find the 95% confidence interval for hours studied.
From page 361 of Beginning Statistics, by Warren, Denley, and Atchley, © 2008 Hawkes Learning Systems.
11/17/2018

14. How big of a sample do I need?
Calculations with z
What about with t?
“I want a ____% confidence level.” (which determines the 𝑧 𝛼/2 value) with a margin of error that’s no larger than 𝐸.”
𝑛= 𝑧∙𝜎 𝐸 2 and bump up
All of the formula’s inputs are conveniently known in advance!
We don’t do this with t.
It’s too complicated for us right now.
Trouble spots:
𝑛 and 𝑡 𝛼/2 are interdependent (via d.f.)
We don’t know 𝑠 until we have the sample.
But we were depending on this to give us the sample size to take!
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15. 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 𝑠, 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 and the 𝑠.
Book tells you only 𝜎 and which confidence level %
And then apply the formula.
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16. 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
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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17. 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 not normally distributed and that 𝜎 is not known, so use t.)
The data: (see, 𝑛 = 30)
Bluman, Chapter 7

18. Example 7-3: Credit Union Assets
Step 4: Substitute in the formula. (BUT TRY TI-84 LIST INSTEAD)
Recall Bluman’s 𝒛 version got this:
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.
Our TI-84 𝑡 version – compare to the Zinterval result:
Bluman, Chapter 7