// Numbas version: exam_results_page_options {"name": "Two sample t-test: paired data", "extensions": ["polynomials", "stats"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"metadata": {"licence": "Creative Commons Attribution-NonCommercial 4.0 International", "description": ""}, "type": "question", "preamble": {"css": "", "js": ""}, "advice": "

In this example we are dealing with paired data.

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

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

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

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\$$\\var{diff}\$$

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

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Recall:

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

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

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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}\$$

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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.

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

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

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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\$$)

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The data is presented below:

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\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
 1 2 3 4 5 6 7 8 9 10 11 12 Surface Water {r1} {r1} {r1} {r1} {r1} {r1} {r1} {r1} {r1} {r1} {r1} {r1} Bottom Water {r2} {r2} {r2} {r2} {r2} {r2} {r2} {r2} {r2} {r2} {r2} {r2}
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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: []

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Input the appropriate sample standard deviation: []

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

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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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