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Z
1-Sample
Interval .... [Interval] - [One-Sample Z Int]
Calculates the confidence interval for the population mean based on a sample mean and known population
standard deviation.
0708
To specify the data below and perform a 1-Sample
list1: {299.4, 297.7, 301, 298.9, 300.2, 297}
Population standard deviation: 3
Significance level: 5% ( = confidence level: 95%)
Z
2-Sample
Interval .... [Interval] - [Two-Sample Z Int]
Calculates the confidence interval for the difference between population means based on the difference
between sample means when the population standard deviations are known.
Z
1-Proportion
Interval .... [Interval] - [One-Prop Z Int]
Calculates the confidence interval for the population proportion based on a single sample proportion.
Z
2-Proportion
Interval .... [Interval] - [Two-Prop Z Int]
Calculates the confidence interval for the difference between population
proportions based on the difference between two sample proportions.
t
1-Sample
Interval .... [Interval] - [One-Sample
Calculates the confidence interval for the population mean based on a sample mean and a sample standard
deviation when the population standard deviation is not known.
t
2-Sample
Interval .... [Interval] - [Two-Sample
Calculates the confidence interval for the difference between population means based on the difference
between sample means and sample standard deviations when the population standard deviations are not
known.
When the two population standard deviations
are equal (pooled)
When the two population standard deviations
are not equal (not pooled)
General Confidence Interval Precautions
If you input a C-Level (confidence level) value in the range of 0 s C-Level < 1, the value you input is used. To
specify a C-Level of 95%, for example, input "0.95".
Z
Interval calculation
t
Int]
t
Int]
2
df
n
= 1/(C
/(
– 1) + (1 – C)
1
Lower, Upper
Lower, Upper
= (o
Lower, Upper
Lower, Upper
x
x
1
2
–
n
n
1
2
Lower, Upper
Lower, Upper
= o
² o
2
Q
s
= ((
– 1)s
S
[
1
1
– o
= (o
Lower, Upper
1
2
n
/(
– 1))
C = (s
2
Chapter 7: Statistics Application 150
Z α
= o
2
σ
Z α
2
– o
1
)
+
1
2
n
2
1
α
x
1
x
=
Z
1–
n
n
n
2
=
x
x
x
1
1
2
1– n
1– n
n
n
Z α
1
1
2
+
n
n
2
1
2
α
= o
t
n –1
2
_
W
s
S
Q
Q
²
Q
2
Q
Q
Q
+ (
– 1)s
)/(
+
– 2)
[
2
2
1
2
α
s
s
2
x
1
)
t
+
n
n
2
df
2
1
2
2
2
n
n
/
)/(s
/
+ s
x
x
x
1
1
1
1
2
σ
n
σ
2
2
n
2
x
n
x
2
2
s
x
n
Q
2
x
2
2
n
/
)
2
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