# Conclusion edited as well

In order to compute the hypothesis test, a few determinations must be made first - Conclusion edited as well introduction. In doing this, we compiled our data of both months of March and November of the past 30 years into an Excel spreadsheet (appendix ). This compilation provided us with the x factor and our standard deviation. We then determined it was a lower one-tail test based on the use of the words “more than” as in; our alternate hypothesis of “it rains more in March than in November”. The formula used to calculate the test statistic was:

z = x1 – x2 /square root of the square of s1/ n1 + the square of s2/n2

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As part of a study for Washington State Climatology, a comparison was done of the average rainfall for the last 30 years to find out if Washington experienced more rainfall in March versus November. A sample of 31 years for the month of November revealed a mean of 6.26 inches of rain had fallen with a standard deviation sample of 3.21. In the month of March a sample of 31 years revealed a mean of 3.51 inches of rainfall with a standard deviation sample of 1.26. With a level of significance of .05, can we conclude that there has been more rainfall in the month of March versus the month of November over the last 31 years?

The following is the legend of values we used: x1: 6.26, x2: 3.51, s1: 3.21, s2: 1.46, n1: 31, n2: 31, giving us our hypothesis test with two samples:

z = 6.26 – 3.51 /square root of the square of 3.21/ 31 + the square of 1.46/31. The z critical value is –1.645 giving us a final answer of .145 allowing us to find our p-value.

Since we have previously determined that the level significance is .05, and according to table three, the Standard Normal Table (appendix ?), it was determined that the p value is .4443. Since the P value of .4443 is larger than the level of significance, we can then safely accept our null hypothesis.

In conclusion, and as you can see, we would accept our null hypothesis which in turn would prove our alternate hypothesis to be wrong.  The data shown in the above equations satisfy the parameters of our null hypothesis.  By carefully compiling the data over a 31 year period, we allowed for there to be enough variance to conclude the null hypothesis to be true, which proved the alternate to be untrue.  With the P value being larger than .05 there was no other outcome possible.

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