Hypothesis Test of Mean
for Student T-Test - One Small Sample

 

Example: A sample of size 20 has a mean of 110 and a standard deviation of 16.  Use the TI-83 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 H1: m > 100, and the null hypothesis is H0: m £ 100.  Follow the steps below to solve the problem using the TI-83.  [NOTE: If the p-value < 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
2:T-Test…
Press ENTER.

 

 

Use right arrow to select Stats (summary values rather than raw data) and Press ENTER.
Use the down arrow to Enter the hypothesized mean, sample mean, standard deviation, and sample size.
Select alternate hypothesis.

Press down arrow to select Calculate and press ENTER.

 

Results:

Since the p-value 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
for Student T-Test - Two Small Samples

 

Example: Two samples were taken, one from each of two populations.  Use the TI-83 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:

sx1 = 5

sx2 = 7

n1 = 8

n2 = 5

H0: m1 = m2

H1: m1 ¹ m2

Use a = 2%


“The two population means are equal” means the null and alternate hypotheses are H0: m1 = m2 and H1: m1 ¹ m2, respectively.  Follow the steps below to solve the problem using the TI-83.  [NOTE: If the p-value < a, reject the null hypothesis; otherwise, do not reject the null hypothesis.

Press STAT and the right arrow twice to select TESTS. 

 

Use the down arrow to select
4:2-SampTTest…
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.
Select alternate hypothesis.
Select pooled: Yes

Press down arrow to select Calculate and press ENTER.

 

 

Results:

Since the p-value 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
for Student T-Test - Two Matched Pairs Samples

 

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.

 

Subject

A

B

C

D

E

F

G

H

Before

6.6

6.5

9.0

10.3

11.3

8.1

6.3

11.6

After

6.8

2.4

7.4

  8.5

  8.1

6.1

3.4

  2.0

 

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

 

H0:  µd = 0 and H1:  µd > 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 TI-83.  [NOTE: If the p-value < 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 (H1:  µd > 0), choose   > µ0 .

Highlight Calculate and press ENTER to get the results as shown in the screen below.

 

 

Results:

Since the p-value 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.”