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The Internet versions of Quick Notes Statistics and Excel Statistics Lab Manual are Now Free! III One-Sample t-Test Using Excel
I.	Darin wants to determine at the .05 level of significance whether the average number of sick days taken by workers who did not graduate from high school increased from the 8 day average of last year. A simple random sample was drawn from a normally distributed population with an unknown standard deviation. Nothing happened during the year to cause Darin to feel the observations were not independent of each other.
	A.	Conduct a one-tail hypothesis test
		1.	Calculate p, the probability for the right tale of the data. If it is less than the level of significance, we will reject the null hypothesis. Why? If the sample mean of 10 is so far above the hypothesized mean of 8 that a number this large or larger happens less than 5% of the time, we will conclude Sick Days went up. Note we will not know Sick Days went up unless we take a census and that takes too much time and money.
			Done with Excel, the answer .000236193 is substantially less than .05 and the null hypothesis is rejected. Non-graduates took more days off! Reasoning, the difference between last year and our sample was so great that it happens less than 5% of the time.
		2.	Using the critical value approach to hypothesis will require determining the critical value for a one-tail alpha of .05.
			Done in Excel. the answer, called the critical value for t, of 1.795883691is the maximum allowable t for the .05 level of significance. Our data has a t value of 4.898979486 and the null hypothesis is again rejected. Non-graduates took more days off! Reasoning, the difference between last year and our sample was so great that t from the data was substantially larger than the t associated with a 5% level of significance. If our decision to reject is wrong, it will be so less than 5% of the time!
	B.	Conduct a two-tail hypothesis test

Proceed to 5. Pearson's Correlation Coefficient