Hypothesis Testing for the Mean
Unknown
(T-test)

Given Data | Given Statistics

1. Given Data:

Test the claim that : < 80 for these data using = 0.10:

60
70
75
55
80
55
65
40
80
70
80
95
120
90
75
85
80
60

Note that is 80.

First enter the data in by pressing STAT and selecting 1:Edit by pressing ENTER. The result is shown below

Press STAT, then RIGHT ARROW twice to TESTS. Press 2 to select 2:T-Test. If Data is not already highlighted, LEFT ARROW to highlight it, and then press ENTER to select the data mode. ARROW DOWN once and enter 80 next to :. Make sure the defaults of for List: and 1 for Freq: are also listed.

 
 
 

ARROW DOWN to :. Use the arrow keys to highlight <, since the claim being tested is < 80, and press ENTER. ARROW DOWN to select Calculate, and then press ENTER. The results are shown below.

Conclusions based on t-value:
The test t value shown above is t = -1.381. Using a t distribution table for a 1-tailed test with= 0.10 and d.f.= 17, we find that the critical t value is t = -1.333.
This is a left-tailed test, so since -1.381 < -1.333, we are in the rejection region, so reject : 80, and accept the alternative hypothesis :< 80.

Conclusions based on P-value:
The test p value shown above is p = 0.0926361139. The decision is to reject : 80, since 0.0926361139 < 0.10; that is, p < .


2. Given Statistics.

Test the claim : = 8 when = 0.6, = 8.2, and n = 25 using = 0.05. We use : 8.

Press STAT, then RIGHT ARROW twice to TESTS. Press 2 to select 2:T-Test.

 
 
 

If STATS is not highlighted, ARROW RIGHT once and press ENTER to select STATS. ARROW DOWN once and enter the value of 8 for . ARROW DOWN once and enter 8.2 for . ARROW DOWN once and enter .6 for . ARROW DOWN once again, and enter 25 for n.

Now we must choose which test. Since this is a two-tailed test, ARROW DOWN once to : then use the LEFT or RIGHT ARROW to select , then press ENTER. ARROW DOWN one last time to Calculate and then press ENTER. The results are shown below.

 
 


Conclusions based on t-value:
The test t value shown above is t = 1.667. Using a t-distribution table for = 0.05, d.f.= 24 and the fact that this is a two-tailed test, we find that the critical value is t = 2.064. Thus, the rejection region is t < -2.064 or t > 2.064. Since -2.064 < 1.667 < 2.064, the decision is that we cannot reject : = 8.

Conclusions based on P-value:
The test p value shown about is p = 0.1085801058. Thus, we cannot reject : = 8 since 0.1085801058 > 0.05



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