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In June 2008, chemical analysis of n1 =110 water samples from various parts of c

ID: 3312790 • Letter: I

Question

In June 2008, chemical analysis of n1 =110 water samples from various parts of catawba river were made. In June 2012, the experiment was repeated using n2 = 85 water samples. We have the following result:

1. An environmental agency claims that the mean chlorine content in river increased from 2008 to 2012. choose the appropriate hypotheses to test the claim

2. The standardized test statistic is about

4. Compute the observed significance of the test.

Chlorine Content (2008) Chlorine Content (2012) x1x1¯ = 17.8 x2x2¯ = 18.3 S1 = 1.8 S2 = 1.2

Explanation / Answer

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: 2008> 2012
Alternative hypothesis: 2008 < 2012

Note that these hypotheses constitute a one-tailed test.

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 = 0.2154

DF = 110 + 85 - 2

D.F = 193

t = [ (x1 - x2) - d ] / SE

t = - 2.32

tcritical = - 1.653

Rejection region is t < - 1.653

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 population means, and SE is the standard error.

The observed difference in sample means (10) produced a t statistic of -2.32. We use the t Distribution Calculator to find P(t < - 2.32).

Therefore, the P-value in this analysis is 0.011.

Interpret results. Since the P-value (0.011) is less than the significance level (0.05), we have to reject the null hypothesis.

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