# Inferences Concerning Two Means, - HP 49g+ User Manual

Graphing calculator.

If using z,
If using t,
Example 2 --
Test the null hypothesis H
alternative hypothesis, H
0.05, using a sample of size n = 25 with a mean x = 22.0 and a standard
deviation s = 3.5. Again, we assume that we don't know the value of the
population standard deviation, therefore, the value of the t statistic is the same
as in the two-sided test case shown above, i.e., t
ν = 25 - 1 = 24 degrees of freedom is
P-value = UTPT(24, |-0.7142|) = UTPT(24,0.7124) = 0.2409,
since 0.2409 > 0.05, i.e., P-value > α, we cannot reject the null hypothesis
: µ = 22.0.
H
o
Inferences concerning two means
The null hypothesis to be tested is H
α)100%, or significance level α, using two samples of sizes, n
values x
and x
, and standard deviations s
1
2
standard deviations corresponding to the samples, σ
if n
> 30 and n
> 30 (large samples), the test statistic to be used is
1
2
If n
< 30 or n
< 30 (at least one small sample), use the following test statistic:
1
2
t
=
(
n
1
P-value = UTPN(0,1,z
o
P-value = UTPT(ν,t
)
o
: µ = 22.0 ( = µ
o
: µ >22.5 at a level of confidence of 95% i.e., α =
1
: µ
= δ, at a level of confidence (1-
o
1
2
(
x
x
)
δ
z
=
1
2
o
2
2
σ
σ
+
1
2
n
n
1
2
(
x
x
)
δ
n
1
2
2
2
) 1
s
+
(
n
) 1
s
1
2
2
)
), against the
o
= -0.7142, and P-value, for
o
and n
1
2
and s
. If the populations
1
2
and σ
, are known, or
1
2
n
(
n
+
n
) 2
1
2
1
2
n
+
n
1
2
Page 18-39
, mean