In this example we are dealing with paired data.
\n\\(H_0:\\) \\(\\mu_d=0\\) i.e. The mean zinc concentration for surface water and bottom water is the same.
\n\\(H_1:\\) \\(\\mu_d<0\\) i.e The mean zinc concentration for surface water is less than the mean zinc concentration for bottom water.
\nWe must evaluate the differences: for each pair of values \\(d=x_1-x_2\\)
\n\\(\\var{diff}\\)
\nWe now have a sample of \\(n=\\var{sample_size}\\) differences.
\nRecall:
\nthe formula for the mean difference: \\(\\overline{d}=\\frac{\\sum {d}}{n}=\\var{diff_mean}\\)
\nthe formula for the standard deviation: \\(s=\\sqrt{\\frac{\\sum{(d-\\overline{d})^2}}{n-1}}=\\var{diff_stdev}\\)
\nthe formula for the t-statistic: \\(t=\\frac{\\overline{d}}{\\frac{s}{\\sqrt{n}}}=\\frac{\\var{diff_mean}}{\\frac{\\var{diff_stdev}}{\\sqrt{12}}}=\\var{test_statistic}\\)
\nThe t-table value will be for a one-tailed test and will have \\(n-1=11\\) degrees of freedom. Because of the alternative hypothesis the t-value chosen will be negative.
\n\\[\\begin{array}{r|rrrr}&0.10&0.05&0.01\\\\\\hline11&-\\var{t90}&-\\var{t95}&-\\var{t99}\\end{array}\\]
\nCompare the test statistic with the t-table values and choose your conclusion.
", "rulesets": {}, "extensions": ["polynomials", "stats"], "name": "Two sample t-test: paired data", "ungrouped_variables": ["r1", "diff_mean", "diff_stdev", "mu1", "sigm1", "test_statistic", "t95", "scenario", "decision_matrix", "t99", "t90", "r2", "mu2", "sigm2", "sample_size", "diff"], "functions": {}, "tags": [], "variablesTest": {"condition": "", "maxRuns": 100}, "variable_groups": [], "variables": {"t90": {"name": "t90", "group": "Ungrouped variables", "definition": "1.363", "description": "", "templateType": "number"}, "diff_mean": {"name": "diff_mean", "group": "Ungrouped variables", "definition": "precround(mean(diff),2)", "description": "", "templateType": "anything"}, "r2": {"name": "r2", "group": "Ungrouped variables", "definition": "repeat(round(normalsample(mu2,sigm2)),sample_size)", "description": "", "templateType": "anything"}, "scenario": {"name": "scenario", "group": "Ungrouped variables", "definition": "sum(map(abs(test_statistic)Trace metals in drinking water affect the flavor and an unusually high concentration can pose a health hazard. Twelve locations were selected and pairs of data were taken measuring zinc concentration in bottom water and surface water. (\\(\\mu g/l\\))
\nThe data is presented below:
\n\n
| \n | 1 | \n2 | \n3 | \n4 | \n5 | \n6 | \n7 | \n8 | \n9 | \n10 | \n11 | \n12 | \n
| Surface Water | \n{r1[0]} | \n{r1[1]} | \n{r1[2]} | \n{r1[3]} | \n{r1[4]} | \n{r1[5]} | \n{r1[6]} | \n{r1[7]} | \n{r1[8]} | \n{r1[9]} | \n{r1[10]} | \n{r1[11]} | \n
| Bottom Water | \n{r2[0]} | \n{r2[1]} | \n{r2[2]} | \n{r2[3]} | \n{r2[4]} | \n{r2[5]} | \n{r2[6]} | \n{r2[7]} | \n{r2[8]} | \n{r2[9]} | \n{r2[10]} | \n{r2[11]} | \n
\n
It is believed that the bottom water will contain more zinc per litre than the surface water.
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\nInput the appropriate sample standard deviation: [[1]]
\nEnter the value for the test statistic: t = [[2]]
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", "Reject the Null Hypothesis at the 5% significance level but accept the Null Hypothesis at the 1% significance level and conclude that the mean zinc concentration is the same for bottom water and surface water
", "Reject the Null Hypothesis at the 10% significance level but accept the Null Hypothesis at the 5% significance level and conclude that the mean zinc concentration is the same for bottom water and surface water
", "Accept the Null Hypothesis at the 10% significance level and conclude that the mean zinc concentration is the same for bottom water and surface water
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