Hypothesis Test of Mean


Example: A sample of size 20 has a mean
of 110 and a standard deviation of 16.
Use the TI83 calculator to test the hypothesis that the population mean
is greater than 100 with a level of significance of a =
5%.
Solution: “The population mean is greater than 100” means
the alternate hypothesis is H_{1}: m > 100, and the null hypothesis is H_{0}:
m £ 100. Follow
the steps below to solve the problem using the TI83. [NOTE: If the pvalue <
a, reject the null
hypothesis; otherwise, do not reject the null hypothesis. 

Press STAT and the right arrow twice to select TESTS. To select the highlighted

Use right arrow to select Stats (summary values rather
than raw data) and Press ENTER. Press down arrow to select Calculate and press ENTER. 
Results: Since the pvalue is 0.0058, reject the null hypothesis
with an alpha value larger than 0.0058 (0.58% level of significance or
larger). 
Hypothesis Test of Mean


Example: Two samples were taken, one from
each of two populations. Use the TI83
calculator to test the hypothesis that the two population means are equal
with a level of significance of a = 2%.
Solution: For the two samples, we have the
following summary data:

_{}

_{}

s_{x1 }= 5

s_{x2 }= 7

n_{1 }= 8

n_{2 }= 5

H_{0}: m_{1 }= m_{2}


H_{1}: m_{1 }¹ m_{2}


Use a =
2%

Press STAT and the right arrow twice to select TESTS.
Use the down arrow to select
4:2SampTTest…
Press ENTER.
Use right arrow to select Stats (summary values rather than raw
data). Enter sample mean, standard deviation, and sample size for
samples 1 and 2. Press down arrow to select Calculate and press ENTER.

Results: Since the pvalue is 0.384, do not reject the null
hypothesis with an alpha value of 0.02 (because 0.384 is not less than 0.02). Conclude that the two population means are not different. 
Hypothesis Test of Mean


Example: Effectiveness of Hypnotism in Reducing
Pain The values listed below are before and after hypnosis; the
measurements are in centimeters on a pain scale. Use a = 0.05
to test the claim that the sensory measurements are lower after hypnotism.
Solution: The data above should be tested
using the “matched pairs” approach. Let the difference d = before – after. If
“after” is lower than “before”, then the claim can be stated as d > 0. The
hypotheses are
H_{0}: µ_{d} = 0 and H_{1}: µ_{d} > 0
“… test the claim that the
sensory measurements are lower after hypnotism” Implies that d = before –
after is positive. Follow the steps
below to solve the problem using the TI83.
[NOTE: If the pvalue < a, reject
the null hypothesis; otherwise, do not reject the null hypothesis.


Enter the “Before” data into L1 and “After” data into L2. Highlight L3 and set L3 = L1
– L2.


Press ENTER to get the values of d = before – after in L3.


Using the alternative hypothesis (H_{1}: µ_{d} > 0), choose > µ_{0} . Highlight Calculate and press ENTER to get the results as shown in the screen below. 

Results: Since the pvalue is 0.0095 < a = 0.05, reject the null hypothesis with an alpha value of 0.05. Conclude that
the two population means are different. Support the claim that “the sensory measurements are lower after
hypnotism.” 