Math 160 - Cooley
TI Calculator Handout #6
OCC
Constructing Confidence Intervals on the TI–84+
A confidence interval estimate for a parameter provides a range of numbers along with a percentage confidence that the
parameter lies in that range. If independent samples are taken repeatedly from the same population, and a confidence
interval calculated for each sample, then a certain percentage (confidence level) of the intervals will include the unknown
population parameter. As the width of the confidence interval increases so does the probability (percentage confidence)
that the unknown population parameter lies in that range.
The TI–84+ can construct seven different types of confidence intervals. We will only focus on the following five
confidence interval functions below:
Z Interval ‒ This function constructs a “One-Mean z-Confidence Interval”.
Assumptions:
1. Simple random sample.
2. Normal population or large sample.
3.  is known.
Input:  , x , n , C-Level
 = population standard deviation
x = sample mean
n = sample size
C-Level = confidence level
T Interval ‒ This function constructs a “One-Mean t-Confidence Interval”.
Assumptions:
1. Simple random sample.
2. Normal population or large sample.
3.  is unknown.
Input: x , Sx, n , C-Level
x = sample mean
Sx = sample standard deviation
n = sample size
C-Level = confidence level
2-SampT Int ‒ This function constructs a “Two-Mean t-Confidence Interval”. Pooled or non-pooled.
Assumptions (pooled):
1. Simple random sample.
2. Independent samples.
3. Normal population or large sample.
4. Equal population standard deviations.
Assumptions (non-pooled):
1. Simple random sample.
2. Independent samples.
3. Normal population or large sample.
Input: x1 , Sx1, n1 , x2 , Sx 2, n2 ,C-Level, Pooled/Unpooled
x1 = # of members in the 1st sample that have the specified attribute
Sx1 = standard deviation of first sample
n1 = 1st sample size
x2 = # of members in the 2nd sample that have the specified attribute
Sx1 = standard deviation of second sample
n2 = 2nd sample size
C-Level = confidence level
Pooled = choice of pooled or non-pooled standard deviations
1-PropZ Int ‒ This function constructs a “One-Proportion z-Confidence Interval”.
Assumptions:
1. Simple random sample.
2. x  5 and n  x  5 .
Input: x, n , C-Level
x = # of members in the sample that have the specified attribute
n = sample size
C-Level = confidence level
2-PropZ Int ‒ This function constructs a “Two-Proportion z-Confidence Interval”.
Assumptions:
1. Simple random sample.
2. x1  5 , n1  x1  5 , x2  5 , and n2  x2  5 .
Input: x1 , n1 , x2 , n2 ,C-Level
x1 = # of members in the 1st sample that have the specified attribute
n1 = 1st sample size
x2 = # of members in the 2nd sample that have the specified attribute
n2 = 2nd sample size
C-Level = confidence level
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Math 160 - Cooley
TI Calculator Handout #6
OCC
Example #1:
A total of 30 people are selected randomly from a certain town. If their mean age is 60.5 and   4.6 ,
find a 95% confidence interval for the true mean age,  , of everyone in the town.
Example #2:
Forty randomly selected students took the calculus final. From the sample, if the mean was 93 and the
standard deviation was 13.3, construct a 99% confidence interval for the mean score of all students.
Example #3:
A paint manufacturer wanted to compare the drying times of two different types of paint. Independent
simple random samples of 11 cans of type A and 9 cans of type B were selected and applied to similar
surfaces. The drying times, in hours, were recorded. The summary statistics are as follows.
Type A
Type B
x1  71.6
s1  3.3
n1  11
x2  68.5
s2  3.0
n1  9
Determine a 99% confidence interval for the difference between the mean drying time of Type A and the
mean drying time of Type B. Assume the mean drying times come from a normal distribution.
Example #4:
A researcher was interested in comparing the amount of time spent watching television by women and by
men. Independent simple random samples of 34 women and 37 men were selected, and each person was
asked how many hours he or she had watched television during the previous week. The summary
statistics are as follows.
Women
Men
x1  12.9
s1  3.9
n1  34
x2  14.2
s2  6.2
n1  37
Determine a 98% confidence interval for the difference, 1  2 , between the mean television watching
times for women and men.
Example #5:
When 294 college students are randomly selected and surveyed, it is found that 130 own a car. Find a
99% confidence interval for the proportion of all college students who own a car.
Example #6:
In a random sample of 49 Republicans from the city, 24 approved of the mayor’s performance. In a
random sample of 45 Democrats from one city, 10 approved of the mayor’s performance. Find a 90%
confidence interval for the difference between the proportions of Republicans and Democrats who
approve of the mayor’s performance.
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Math 160 - Cooley
Solution
#1
TI Calculator Handout #6
OCC
From the context of the question, we are working with one mean. So, we must check our three
assumptions in order to use the correct procedure. First, we have a simple random sample. Second,
we have a large sample, ( n  30 ). Third, the population standard deviation or  is KNOWN. Thus,
we will use “The One-Mean z-Interval Procedure”.
All versions of TI-83+ & TI-84+
▒ Key in: STAT. Scroll to TESTS. Select Z Interval ENTER.
Since the original data was not given and the mean and standard deviation was calculated, then we need to switch
the input from data to stats. So,
▒ Scroll to STATS ENTER.
Now, this should be the screen you see.
Now, we simply enter the required information. So, scroll down to  :
:
▒ Key in: 4.6 ENTER
x:
▒ Key in: 60.5 ENTER
n:
▒ Key in: 30 ENTER
C-Level : ▒ Key in: .95 ENTER
Your cursor is on Calculate, so, press ENTER again.
Thus, the 95% confidence interval constructed is (58.854 , 62.146).
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Math 160 - Cooley
Solution
#2
TI Calculator Handout #6
OCC
From the context of the question, we are working with one mean. So, we must check our three
assumptions in order to use the correct procedure. First, we have a simple random sample. Second,
we have a large sample, ( n  30 ). Third, the population standard deviation or  is UNKNOWN.
Thus, we will use “The One-Mean t-Interval Procedure”.
(Note:  is unknown, since the standard deviation given was from the sample. There was no
mention of  or a population standard deviation anywhere.)
All versions of TI-83+ & TI-84+
▒ Key in: STAT. Scroll to TESTS. Select T Interval ENTER.
Since the original data was not given and the mean and standard deviation was calculated, then we need to switch
the input from data to stats. So,
▒ Scroll to STATS ENTER.
Now, this should be the screen you see.
Now, we simply enter the required information. So, scroll down to x :
x:
▒ Key in: 93 ENTER
Sx :
▒ Key in: 13.3 ENTER
n:
▒ Key in: 40 ENTER
C-Level : ▒ Key in: .99 ENTER
Your cursor is on Calculate, so, press ENTER again.
Thus, the 99% confidence interval constructed is (87.305 , 98.695).
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Math 160 - Cooley
Solution
#3
TI Calculator Handout #6
OCC
From the context of the question, we are working with two means. So, we must check the
assumptions needed in order to use the correct procedure. First, we have a simple random sample.
Second, we are working with independent samples. Third, we know the variable comes from a
normal distribution. Now, we need to check if we can assume equal population standard deviations
or not. Since the two sample standard deviations appear to be relatively close to each other, then we
can assume that the two population standard deviations are equal. Thus, we will use “The TwoMean t-Interval Pooled Procedure”.
All versions of TI-83+ & TI-84+
▒ Key in: STAT. Scroll to TESTS. Select 2‒SampT Int ENTER.
Since the original data was not given and the mean and standard deviation was calculated, then we need to switch
the input from data to stats. So,
▒ Scroll to STATS ENTER.
Now, we simply enter the required information. So, scroll down to x1:
x1:
▒ Key in: 71.6 ENTER
Sx1:
▒ Key in: 3.3 ENTER
n1:
▒ Key in: 11 ENTER
x2:
▒ Key in: 68.5 ENTER
Sx 2 :
▒ Key in: 3 ENTER
n2 :
▒ Key in: 9 ENTER
C-Level :
▒ Key in: .99 ENTER
Pooled :
▒ Scroll to YES. Press ENTER.
Scroll down to Calculate. Press ENTER.
Thus, the 99% confidence interval constructed is (‒1.001, 7.202).
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Math 160 - Cooley
Solution
#4
TI Calculator Handout #6
OCC
From the context of the question, we are working with two means. So, we must check the
assumptions needed in order to use the correct procedure. First, we have a simple random sample.
Second, we are working with independent samples. Third, both samples are large,
( n1  30, n2  30 ). Now, we need to check if we can assume equal population standard deviations
or not. Since the two sample standard deviations appear to be significantly different from each
other, then we can assume that the two population standard deviations are not equal. Thus, we will
use “The Two-Mean t-Interval Non-Pooled Procedure”.
All versions of TI-83+ & TI-84+
▒ Key in: STAT. Scroll to TESTS. Select 2‒SampT Int ENTER.
Since the original data was not given and the mean and standard deviation was calculated, then we need to switch
the input from data to stats. So,
▒ Scroll to STATS ENTER.
Now, we simply enter the required information. So, scroll down to x1:
x1:
▒ Key in: 12.9 ENTER
Sx1:
▒ Key in: 3.9 ENTER
n1:
▒ Key in: 34 ENTER
x2:
▒ Key in: 14.2 ENTER
Sx 2 :
▒ Key in: 6.2 ENTER
n2 :
▒ Key in: 37 ENTER
C-Level :
▒ Key in: .98 ENTER
Pooled :
▒ Scroll to NO. Press ENTER.
Scroll down to Calculate. Press ENTER.
Thus, the 98% confidence interval constructed is (‒4.212, 1.612).
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Math 160 - Cooley
Solution
#5
TI Calculator Handout #6
OCC
From the context of the question, we are working with one proportion. So, we must check the
assumptions needed in order to use the correct procedure. First, we have a simple random sample.
Second, we need to verify that x  5 and n  x  5 , which it clearly is. Thus, we will use “The OneProportion z-Interval Procedure”.
All versions of TI-83+ & TI-84+
▒ Key in: STAT. Scroll to TESTS. Select 1‒PropZ Int ENTER.
Now, we simply enter the required information:
x:
▒ Key in: 130 ENTER
n:
▒ Key in: 294 ENTER
C-Level :
▒ Key in: .99 ENTER
Your cursor is on Calculate, so, press ENTER again.
Thus, the 99% confidence interval constructed is (0.368, 0.517).
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Math 160 - Cooley
Solution
#6
TI Calculator Handout #6
OCC
From the context of the question, we are working with two proportions. So, we must check the
assumptions needed in order to use the correct procedure. First, we have a simple random sample.
Second, we need to verify that x1  5 , n1  x1  5 , x2  5 , and n2  x2  5 , which it clearly is. Thus,
we will use “The Two-Proportion z-Interval Procedure”.
All versions of TI-83+ & TI-84+
▒ Key in: STAT. Scroll to TESTS. Select 2‒PropZ Int ENTER.
Now, we simply enter the required information.
(Let n1 represent the sample of Republicans and n2 represent the sample of Democrats)
x1:
▒ Key in: 24 ENTER
n1:
▒ Key in: 49 ENTER
x2 :
▒ Key in: 10 ENTER
n2 :
▒ Key in: 45 ENTER
C-Level :
▒ Key in: .90 ENTER
Your cursor is on Calculate, so, press ENTER again.
Thus, the 90% confidence interval constructed is (0.112, 0.423).
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Math 160 - Cooley
TI Calculator Handout #6
OCC
Sample TI Calculator Quiz on Constructing Confidence Intervals
Exercises:
Find each of the following using a TI-calculator.
1)
A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 188 milligrams with
s  12.7 . Construct a 95% confidence interval for the true mean cholesterol content of all such eggs. Assume that
the mean amount of cholesterol comes from a normal distribution.
1) ________________________
2)
A random sample of 149 full-grown lobsters had a mean weight of 15 ounces. Assume that   3.6 ounces.
Construct a 95% confidence interval for the population mean μ.
2) ________________________
3)
A researcher was interested in comparing the salaries (in dollars) of female and male employees at a particular
company. Independent simple random samples of 8 female employees and 15 male employees yielded the
following results:
Female
Male
x1  705.375
s1  183.855
x2  817.067
s2  330.146
Determine a 98% confidence interval for the difference between the mean weekly salary of female employees and
the mean weekly salary of male employees at this particular company. Assume that the salaries come from a
normal distribution.
3) ________________________
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Math 160 - Cooley
TI Calculator Handout #6
OCC
Sample TI Calculator Quiz on Constructing Confidence Intervals
Exercises:
Find each of the following using a TI-calculator. Round all answers to three decimal places.
4)
A researcher was interested in comparing the resting pulse rates of people who exercise regularly and people who
do not exercise regularly. Independent simple random samples of 16 people ages 30-40 who do not exercise
regularly and 12 people ages 30-40 who do exercise regularly were selected, and the resting pulse rate (in beats
per minute) of each person was measured. The summary statistics are as follows:
Do Not Exercise
Do Exercise
x1  72.7
s1  10.1
n1  16
x2  69.4
s2  8.9
n1  12
Determine a 90% confidence interval for the difference, 1  2 , between the mean pulse rate of people who do
not exercise and the mean pulse rate of people who do exercise. Assume the resting rates come from a normal
distribution.
4) ________________________
5)
In a random sample of 300 women, 66% favored stricter gun control legislation. In a random sample of 200 men,
58% favored stricter gun control legislation. Construct a 98% confidence interval for the difference between the
proportions of women and men who favor stricter gun control legislation. (Note: You will have to calculate x1
and x2 ).
5) ________________________
6)
Of 102 patients, selected at random from a clinic, 34 were found to have high blood pressure. Construct a 95%
confidence interval for the percentage of all patients at this clinic that have high blood pressure.
6) ________________________
Solutions:
1) (179.93, 196.07)
5) (‒.0231, 0.1832)
2) (14.422, 15.578)
6) (0.242, 0.425)
3) (‒381.7, 158.3)
4) (‒2.96, 9.56)
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