Using a slightly modified example problem from a prior exercise:
A scientist creates a chemical which he believes changes the temperature of water. He applies this chemical to water and takes the following measurements:
N = 8
SD = 2.167124
Mean = 72.875
He then measures temperature in samples which the chemical was not applied.
Summary Data Pertaining to the Second Set of Measurements
N = 8
SD = 1.669046
Mean = 75.25
Can the scientist conclude, with a 95% confidence interval, that his chemical is in some way altering the temperature of the water?
To perform this analysis, select “Compare Means” from the “Analyze” drop down, then select “Summary Independent-Samples T Test”.
You should be presented with the menu below. Simply enter the information that was provided in the example problem and click “OK”. You have the option to change the Confidence Level (%), but since we are assuming a 95% confidence level, doing so is un-necessary.
This presents the output:
From this output we can conclude:
With a p-value of 0.028 (.028 < .05), and a corresponding t-value of - 2.456, we can state that, at a 95% confidence interval, that the scientist's chemical is altering the temperature of the water.
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