In the constant quest to reduce variation and improve products,
companies need to evaluate different alternatives. A t test using
two paired samples compares two dependent sets of
test data. It helps determine if the means (i.e., averages) are
different from each other.
Weight Loss Paired t test
If a diet claims to cause more than a 10 lb weight loss over a
six month period, you could design a test using several individuals
before and after weights. The samples are "paired" by
each individual. You might want to know if the diet truly delivers
greater than a 10 lb weight loss. The null hypothesis is less than
or equal to 10. The alternate hypothesis is greater than 10.
H0 <= 10 lbs
Ha > 10 lbs
Since the null hypothesis stated as "less than or equal to",
this is a one-sided test.
Now, conduct a test with several individuals and enter the data
into Excel:
Then, select the data with the mouse and click on the QI Macros Menu to select the two
sample t test:
The QI Macros will prompt for a significance level (default = 0.05):
And a hypothesized mean difference (in this case 10 lbs):
The QI Macros paired t test two sample macro will perform the calculations
and interpret the results for you:
What's cool about QI Macros t-Test? When you run the t-Test, the QI Macros will compare the p-value (0.429) to the significance level (0.05) and tells you to "Accept the Null Hypothesis because p>0.05" and that the "Means are the same."
Interpreting the Paired t test results
If you want to evaluate the results manually:
If
Then
test statistic > critical value
(i.e. t> tcrit)
Reject the null hypothesis
test statistic < critical value
(i.e. t< tcrit)
Accept the null hypothesis
p value < a
Reject the null hypothesis
p value > a
Accept the null hypothesis
Since the null hypothesis is that weight loss is less than or equal
to 10, this is a one-sided test. Therefore, use the one-tail values
for your analysis.
Note: The two-sided values would apply if our null hypothesis
was that:
H0: mean difference = 10 lbs.)
Since the t statistic < t critical (.182< 1.753) and p
value > a ( 0.429> 0.05) ,
we can accept the null hypothesis that the weight loss is less than
or equal to 10.
Example of t test one-sample
We could have cast this as a t test one sample.
If we calculate the difference between the before and after weights,
we could test whether the difference is greater than 10 lbs.
Again, since the p value of .429172 is greater than 0.05,
we accept the null hypothesis that weight loss is less than or equal
to 10 lbs.
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