Need to solve problem 8.35 with same format as with our lecture. It has to have
ID: 3066076 • Letter: N
Question
Need to solve problem 8.35 with same format as with our lecture. It has to have the 1. Hypotheses 2. Test statistics 3. 95% acceptance region 4. Decision / conclusion š Wastewater treatment study. In Ecological Engineerin (Feb. 2004), the potential of floating aquatic plants to treat dairy manure wastewater was investigated. For one part of the study, 16 treated wastewater samples were randomly divided into two groups-a control algal was cultured in half the samples and water hyacinth was cultured in the other half. The rate of increase in the amount of total phos- phorus was measured in each water sample; a summary of the results is given in the accompanying table. Conduct a test to determine if there is a difference in mean rates of Increase of total phosphorus for the two aquatic plants. Use a 05. Water Hyacinth Control Algal Number of Water Samples Sample Mearn Standard Deviation Source: Sooknah 026 036 008 ), R., and Wilkie, A. "Nutrient removal by floating 006 .300 phytes cultured in anaerobically digested flushed dairy manure wastewater." Table 5) cological Engineering, Vol. 22, No. 1, Feb. 2004Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 = 2
Alternative hypothesis: 1 2
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 0.00354
DF = 14
t = [ (x1 - x2) - d ] / SE
t = 2.82
tcritical = + 2.145
We will reject the null hypothesis if - 2.145 < t < 2.145
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 14 degrees of freedom is more extreme than -2.82; that is, less than -2.82 or greater than 2.82.
Thus, the P-value = 0.0048.
Interpret results. Since the P-value (0.0048) is less than the significance level (0.05), we have to reject the null hypothesis.
From the above test we have sufficient evidence in the favor of the claim that there is significance difference in mean rates of total phophorus for two aquatic plants.
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