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Null & Alternate Hypothesis

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In order to compute the hypothesis test, a few determinations must be made first. 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:

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

            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 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 then the level of significance we can safely accept our null hypothesis.

As the tests and relevant data used in this experiment have revealed, the statistical significance of the results definitely confirm our null hypothesis which is that there is more rain in the month of November over the Washington Area.  Basically the date reveals that while the results for the tests conducted on the alternate hypothesis show a higher degree of rainfall level at 6.26 inches of rain, the standard deviation was also much higher at 3.21.  This high degree of uncertainty shows that the rainfall was sporadic during that period.  What it means is that while there is a higher measurement of rainfall in terms of inches, these samples indicate that it was never constant.  There are times when it would rain heavily and times when it would hardly rain at all which account for the higher rate of standard deviation.

In comparison, the null hypothesis test yielded results that were more consistent.  While the average rainfall only measured 3.51 inches, the lower standard deviation of 1.26 shows that the results are clustered closer together and more consistent.  The grouping of the samples are more or less along the same line and from this it can be concluded that the rainfall level for that period remained constant at around 3.52 inches.

Since we have previously determined that the level significance is .05, and according to table three, the Standard Normal Table it was determined that the p value is .4443. The P value of .4443 is larger then the level of significance we can safely accept our null hypothesis.  In conclusion as you can see we would accept our null hypothesis which in turn would prove our alternate hypothesis to be wrong.

Cite this Null & Alternate Hypothesis

Null & Alternate Hypothesis. (2017, Apr 18). Retrieved from https://graduateway.com/null-alternate-hypothesis/

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