As part of an analytical chemistry laboratory course, a student measured the Ca2
ID: 3205070 • Letter: A
Question
As part of an analytical chemistry laboratory course, a student measured the Ca2 content in two water samples, city-supplied drinking water and well-supplied drinking water, using two different analytical methods, flame atomic absorption spectrometry (FAAS) and EDTA complexometric titration. The results of this experiment are shown below as the mean Ca2 concentration (x) ± standard deviation (s) in units of parts per million (ppm). Each sample was measured five times (n = 5) by each method.
City-Supplied Drinking Water (Xt s) Well Supplied Drinking Water (xt s) Method FAAS 58.06 t 0.54 ppm 64.30 t 0.56 ppm 58.81 t 0.91 ppm 65.15 t 0.93 ppm EDTA Titration A) Method Comparison: For each drinking water sample (city and well), compare the Number Ca content measured with FAAS and O No EDTA titration. Calculate the t value (tcalc) City: calc for each sample. Do the methods produce statistically different results at the 95% confidence level when measuring the Number Ca content of the two drinking water samples (Yes or No)? A list of Student's t Well: values at several confidence levels can be calc found in the Student's t table. B) Sample Comparison: For each Number method (FAAS and EDTA titration), compare the Ca content measured FAAS: calc in the city-supplied and well-supplied drinking water samples. Calculate the t value (tcalc) for each method. Do the drinking water samples contain statistically different amounts of Ca Number at the 95% confidence level when measured by each of the methods EDTA calc L (Yes or No)?Explanation / Answer
Solution:-
1) State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: CityFAAS - City EDTA = 0
Alternative hypothesis: CityFAAS - City EDT 0
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 = 0.4732
DF = 6.506
D.F = 7
t = [ (x1 - x2) - d ] / SE
t = - 1.585
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.
This is two tailed test,
Thus, the P-value = 0.078493 + 0.078493 = 0.157
Interpret results. Since the P-value (0.157) is greater than the significance level (0.05), we have to accept the null hypothesis.
From this we do not have sufficient evidence in the favor of the claim that CityFAAS and 2City EDT are different.
2)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: Well FAAS - Well EDTA = 0
Alternative hypothesis: Well FAAS - Well EDT 0
Note that these hypothesis 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 = 0.4855
DF = 6.5636
D.F = 7
t = [ (x1 - x2) - d ] / SE
t = - 1.75
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.
This is two tailed test,
Thus, the P-value = 0.06179 + 0.06179 = 0.12358
Interpret results. Since the P-value (0.12358) is greater than the significance level (0.05), we have to accept the null hypothesis.
From this we have sufficient evidence in the favor of the claim that WellFAAS and Well EDT are different.
3)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: FAAS City - FAAS Well = 0
Alternative hypothesis: FAAS City - FAAS Well 0
Note that these hypothesis 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 = 0.348
DF = 7.98
D.F = 8
t = [ (x1 - x2) - d ] / SE
t = - 17.93
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.
This is two tailed test,
Thus, the P-value = 0.00001
Interpret results. Since the P-value (0.00001) is less than the significance level (0.05), we have to reject the null hypothesis.
From this we do not have sufficient evidence in the favor of the claim that City FAAS and Well FAAS are different.
4)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: EDTA City - EDTA Well = 0
Alternative hypothesis: EDTA City - EDTA Well 0
Note that these hypothesis 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 = 0.5819
DF = 7.962
D.F = 8
t = [ (x1 - x2) - d ] / SE
t = - 10.895
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.
This is two tailed test,
Thus, the P-value = 0.00001
Interpret results. Since the P-value (0.00001) is less than the significance level (0.05), we have to reject the null hypothesis.
From this we do not have sufficient evidence in the favor of the claim that City EDTA and Well EDTA are different.
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