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t

", "group": "Ungrouped variables", "templateType": "anything", "definition": "precround((sample_mean_2-mu1)/(sample_stdev_2/sqrt(sample_size)),2)", "name": "test_statistic"}, "t99": {"description": "", "group": "Ungrouped variables", "templateType": "number", "definition": "2.977", "name": "t99"}, "t90": {"description": "", "group": "Ungrouped variables", "templateType": "number", "definition": "1.761", "name": "t90"}, "sample_size": {"description": "", "group": "Ungrouped variables", "templateType": "number", "definition": "15", "name": "sample_size"}, "scenario": {"description": "", "group": "Ungrouped variables", "templateType": "anything", "definition": "sum(map(abs(test_statistic)The hypotheses that we are testing are:

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                                                                   $ H_0: \\mu=\\var{mu1} $ ,    $ H_1: \\mu \\neq \\var{mu1} $

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The test statistic for a one-sample t-test is given by:

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                                                         $ |t| = \\frac{\\bar{x}-\\mu_0}{s/\\sqrt{n}} $

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where $\\bar{x}$ is the sample mean, given by:

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                                                      $\\bar{x}= \\sum_{i=1}^{i=15} \\frac{x_i}{15} = \\var{sample_mean_2}$ $(2dp)$ 

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and $s$ is the sample standard deviation, given by:

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                                                         $ s = \\sqrt{\\frac{\\left (\\sum (x-\\bar{x})^2\\right)}{n-1}} = \\var{sample_stdev_2}$   $(2dp)$

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Therefore, the test statistic is:

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                                                         $ |{t}| = \\frac{\\var{sample_mean_2}-\\var{mu1}}{\\var{sample_stdev_2}/\\sqrt{15}}$

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                                                           $   = \\var{abs({test_statistic})} $ $(2dp)$

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The t-table values will be for a two-tailed test and will have $ n-1 = 14$ degrees of freedom:

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\\[\\begin{array}{r|rrrr}&0.10&0.05&0.01\\\\\\hline13&\\pm\\var{t90}&\\pm\\var{t95}&\\pm\\var{t99}\\end{array}\\]

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If $ |t| < t_{14}(0.1) $, then we accept $ H_0$ at the 10% level. 

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If $ |t| > t_{14}(0.1)$, then we reject $H_0$ at the 10% level.

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If $ |t| > t_{14}(0.05) $, we reject $H_0$ at the $5%$ level.

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If $ |t| > t_{14}(0.01)$, then we reject $H_0$ at the 1% level.

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The following is a set of BMI's (Body Mass Index) for 15 overweight children aged 5.

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The data is thought to follow a normal distribution.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
{r1[0]}{r1[1]}{r1[2]}{r1[3]}{r1[4]}{r1[5]}{r1[6]}{r1[7]}{r1[8]}{r1[9]}{r1[10]}{r1[11]}{r1[12]}{r1[13]}
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We want to test the hypothesis that the population mean equals \\(\\var{mu1}\\) .

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Calculate the sample mean.

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Calculate the sample standard deviation.

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Calculate the test statistic.

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For a one sample t-test, the test statistic is given by:    

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                                          $  |t| =  \\frac{\\bar{x}-\\mu}{s/\\sqrt{n}} $ 

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Reject the Null Hypothesis and conclude that mean value is not \\(\\var{mu1}\\) 

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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 mean value is \\(\\var{mu1}\\).

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Reject the Null Hypothesis at the 10% significance level but accept the Null Hypothesis at the 5% significance level and conclude that mean value is \\(\\var{mu1}\\).

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Accept the Null Hypothesis at the 10% significance level and conclude that mean value is \\(\\var{mu1}\\) .

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Using t-tables, compare your test statistic to the appropriate t-value, taking into account the correct degrees of freedom. From this comparison select one of the following statements that best describes your conclusion.

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