Jennifer Brown Unit 3 IP    

Running Head: SAMPLES

Using the t Test and t Distribution

Jennifer Brown

American Intercontinental University

Using the t Test and t Distribution

        AIU claims that the average level of intrinsic job satisfaction of American workers is equal to 5. A sample of 25 AIU American workers was sampled. Is there enough evidence to reject AIU’s claim at α = 0.05? We need to determine the mean of the ratings that the 25 surveyed people gave. This is done by adding all the ratings together and dividing by 25 (3.2 + 3.5 + 4.8 + 4.9 + 5.5 + 4.3 + 6.1 + 6.5 + 6.6 + 3.8 + 4.6 + 5.6 + 5.7 + 5.7 + 3.4 + 3.6 + 3.7 + 4.0 + 4.2 + 4.5 + 5.6 + 5.9 + 6.0 + 6.2 + 6.4 = 124.3/25 = 4.972). Now, we can determine the standard deviation. This is done by subtracting 4.972 from each rating and squaring the number. Then you add up all of the calculations and divide by one less than the number of variable used. Then you take the square root of that number, and you have your standard deviation. For simplification, I used the standard deviation formulas built into Excel:

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To check these answers, I then used the descriptive statistics feature built into the data analysis part of Excel:

        Now, I can continue and work the problem:

Step 1         H0: The average satisfaction level of the American worker = 5 (claim) and H1: The average satisfaction level of the American worker  < or > 5

Step 2         The critical values are +2.060 and -2.060 for α = 0.05 and d.f. = 25. The α or level of significance was chosen reasonably at random. Normal levels of significance used for testing are either 0.05 or 0.01. For this ...

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