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Biochemical Oxygen Demand Hypothesis Testing - Essay Example

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The present paper "Biochemical Oxygen Demand Hypothesis Testing" reveals a statistical study whereas the researcher would perform test on whether biochemical oxygen demand downstream is higher than the biochemical oxygen demand at the intake or at the upstream…
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Biochemical Oxygen Demand Hypothesis Testing
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PROBLEM STATEMENT: An aquaculture farm takes water from a stream and returns it after it has circulated through the fish tanks. The owner thinks that, since the water circulates rather quickly through the tanks, there is little organic matter in the effluent. To find out if this is true, he takes some samples of the water at the intake and other samples downstream the outlet and tests for Biochemical Oxygen Demand (BOD). If BOD increases, it can be said that the effluent contains more organic matter than the stream can handle. Test whether BOD downstream is higher than the BOD at the intake or at the upstream. Table below shows BOD in the stream Sample Upstream Downstream di 1 6.782 9.063 2.281 2 5.809 8.831 3.022 3 6.849 8.66 1.811 4 6.879 8.405 1.526 5 7.014 9.248 2.234 6 7.321 8.735 1.414 7 5.986 9.772 3.786 8 6.628 8.545 1.917 9 6.822 8.063 1.241 10 6.448 8.001 1.553 Mean 2.0785 Std dev 0.79533 Step 1: Identify the null hypothesis Ho and the alternative hypothesis H1 The null hypothesis refers to the hypothesis we wish to test and is denoted by Ho. It is a hypothesis which states that there is no difference between the procedures. Rejection of the null hypothesis leads to the acceptance of the alternative hypothesis. The alternative hypothesis states that there is a difference between the procedures. It is denoted by H1. For this problem we test the null and alternative hypotheses: Ho: µu = µd or µD = µd - µu = = d0 H1: µd > µu or µD = µd - µu > d0 Where µu = mean BOD upstream; µd= mean BOD downstream; d0 = 0 To check whether the effluent contains more organic matter than the stream could handle, we have to determine if there is significant increase in the downstream BOD compared to the upstream BOD. Step 2: Choose the level of significance, α Assuming that the upstream BOD and downstream BOD are normally distributed, we test using significance level of 0.05 whether BOD increases at the downstream. The significance level of a statistical hypothesis test is a fixed probability of rejecting the null hypothesis H0, when it is in fact true. It is called a type I error and is set by the investigator in relation to the consequences of such an error. We want to make the significance level as small as possible in order to protect the null hypothesis and to prevent, as far as possible, the investigator from inadvertently making false claims. The significance level is usually denoted by. Step 3: Select the test statistic and determine its value from the test data Test Statistic: It is the random variable X whose value is tested to arrive at a decision. The Central Limit Theorem states that for large sample sizes (n > 30) drawn randomly from a population, the distribution of the means of those samples will approximate normality, even when the data in the parent population are not distributed normally. A z statistic is usually used for large sample sizes (n > 30), but often large samples are not easy to obtain, in which case the t-distribution can be used. The population standard deviation s is estimated by the sample standard deviation, The observed t value is calculated from the sample used. Testing of means can be accomplished when the data are in the form of paired observations. We compute for the confidence interval of µd - µu in the situation with paired observations is based on the random variable Where and are random variables representing the sample mean and standard deviations of the differences of the observations in the experimental units. This two-sample problem is essentially reduced to a one-sample problem by using the computed difference d1, d2, d3… dn. Thus the hypothesis reduces to Ho: =do The computed test statistic is then given by: The Critical Region is the part of the sample space where the null hypothesis H0 is rejected. The size of this region is determined by the probability of the sample point falling in the critical region when H0 is true. It is also known as the level of significance, the probability of the value of the random variable falling in the critical region. Also it should be noted that the term "Statistical significance" refers only to the rejection of a null hypothesis at some level a. It implies only that the observed difference between the sample statistic and the mean of the sampling distribution did not occur by chance alone. Critical regions are constructed using the t-distribution with n-1 degrees of freedom. In this case we have 10 paired observations resulting to 9 (10-1) degrees of freedom. Critical Region: t > t0.05,9 t > 1.833 Step 4: Compare the observed value of the statistic to the critical value = 8.26423641 8.264 > 1.833, t falls within the critical region Step 5: Make the Decision Because the test statistic, t, falls within the critical region, we reject the null hypothesis and conclude that the BOD of water increases after circulating in the fish tanks. Therefore the effluent contains more organic matter than the stream can handle. Comparison of Results: 1st Assignment 2nd Assignment Null and alternative hypotheses Ho: µu = µd H1: µu < µd Ho: µu = µd or µD = µd - µu = 0 H1: µd > µu or µD = µd - µu > 0 Point of comparison Compared mean of upstream vs. mean of downstream Compared difference between upstream and downstream BOD vs. d0=0 mean µu = 6.6539, µd = 8.6874 µD = 2.0785 Standard deviation Variance = 0.2124 Pooled variance=0.2556 Standard deviation = 0.795329876 Degrees of freedom n1+n2-2 = 10 + 10 – 2 = 18 n-1 = 10-1 = 9 Test Statistic t=-8.9941 (within the critical region) t=8.264 (within the critical region) Critical region t1.833 Conclusion Reject the null hypothesis and conclude that the BOD upstream is less than the BOD downstream Reject the null hypothesis and conclude that the BOD of water increases after circulating in the fish tanks. In the first assignment the problem was solved as a two sample problem where the mean of the upstream is compared to the mean of the downstream BOD. In the second hypothesis test, the 10 observations are paired allowing us to reduce the problem to one sample where the difference between paired observations is compared to 0 or no increase/decrease. While different test statistic was used, results from both tests consistently show that the Biochemical Oxygen Demand (BOD) increases at the downstream. References: Walpole, R. E., & Myers, R. H. (1993). Probability and Statistics for Engineers and Scientists. NY: Macmillan Publishing University of Glaslow, Department of Statistics. Statistics Glossary, Retrieved September 5, 2006 from http://www.stats.gla.ac.uk/steps/glossary/index.html Read More
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