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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\\))

The data is presented below:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Surface Water

{r1[0]}

{r1[1]}

{r1[2]}

{r1[3]}

{r1[4]}

{r1[5]}

{r1[6]}

{r1[7]}

{r1[8]}

{r1[9]}

{r1[10]}

{r1[11]}

Bottom Water

{r2[0]}

{r2[1]}

{r2[2]}

{r2[3]}

{r2[4]}

{r2[5]}

{r2[6]}

{r2[7]}

{r2[8]}

{r2[9]}

{r2[10]}

{r2[11]}

It is believed that the bottom water will contain more zinc per litre than the surface water.

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Input the appropriate sample mean: [[0]]

Input the appropriate sample standard deviation: [[1]]

Enter the value for the test statistic: t = [[2]]

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Reject the Null Hypothesis and conclude that the mean zinc concentration is greater for bottom water than surface water

", "

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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Having compared your test statistic with the table value, select one of the following conclusions that best describes your conclusion.

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In this example we are dealing with paired data.

\\(H_0:\\) \\(\\mu_d=0\\) i.e. The mean zinc concentration for surface water and bottom water is the same.

\\(H_1:\\) \\(\\mu_d<0\\) i.e The mean zinc concentration for surface water is less than the mean zinc concentration for bottom water.

We must evaluate the differences: for each pair of values \\(d=x_1-x_2\\)

\\(\\var{diff}\\)

We now have a sample of \\(n=\\var{sample_size}\\) differences.

Recall:

the formula for the mean difference: \\(\\overline{d}=\\frac{\\sum {d}}{n}=\\var{diff_mean}\\)

the formula for the standard deviation: \\(s=\\sqrt{\\frac{\\sum{(d-\\overline{d})^2}}{n-1}}=\\var{diff_stdev}\\)

the 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}\\)

The 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.

\\[\\begin{array}{r|rrrr}&0.10&0.05&0.01\\\\\\hline11&-\\var{t90}&-\\var{t95}&-\\var{t99}\\end{array}\\]

Compare the test statistic with the t-table values and choose your conclusion.

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