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Test to see if there has been a change in sales on average at the 5% level of significance.
\nWhat is the sample mean?: [[0]]
\nWhat is the population mean?: [[1]]
\nWhat is the sample standard deviation?: [[3]]
\nEnter the value for the test statistic: t = [[4]]
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", "type": "1_n_2", "scripts": {}, "variableReplacements": [{"variable": "r1", "part": "p0g2", "must_go_first": false}], "shuffleChoices": false, "showCorrectAnswer": true, "customMarkingAlgorithm": "", "useCustomName": false, "customName": "", "minMarks": "3", "displayType": "radiogroup", "choices": ["Reject the Null Hypothesis and conclude that mean hydrocarbon emission rate has decreased.
", "Reject the Null Hypothesis at the 5% significance level but accept the Null Hypothesis at the 1% significance level and conclude that mean hydrocarbon emission rate is the same for both models.
", "Reject the Null Hypothesis at the 10% significance level but accept the Null Hypothesis at the 5% significance level and conclude that mean hydrocarbon emission rate is the same for both models.
", "Accept the Null Hypothesis at the 10% significance level and conclude that mean hydrocarbon emission rate is the same for both models.
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t
", "definition": "precround((sample_mean_1-sample_mean_2)/(pooled_s*sqrt(2/15)),2)", "group": "Ungrouped variables", "name": "test_statistic", "templateType": "anything"}, "t99": {"description": "", "definition": "2.467", "group": "Ungrouped variables", "name": "t99", "templateType": "number"}, "mu1": {"description": "", "definition": "random(30000 .. 50000#50)", "group": "Ungrouped variables", "name": "mu1", "templateType": "randrange"}, "sample_mean_2": {"description": "", "definition": "precround(mean(r2),2)", "group": "Ungrouped variables", "name": "sample_mean_2", "templateType": "anything"}, "scenario": {"description": "", "definition": "sum(map(abs(test_statistic)\\(H_0:\\) \\(\\mu_1=\\mu_2\\) i.e. The mean emission rate is the same for both sets of cars.
\n\\(H_1:\\) \\(\\mu_1>\\mu_2\\) i.e. The mean emission rate of the older set of cars is greater than the mean emission rate of the newer set of cars.
\nFor each sample we must evaluate the sample mean and the sample standard deviation.
\nRecall when given a sample of size \\(n\\) :
\nthe formula for the sample mean: \\(\\overline{x}=\\frac{\\sum {x}}{n}\\)
\n\\(\\overline{x_1}=\\var{sample_mean_1}\\) and \\(\\overline{x_2}=\\var{sample_mean_2}\\)
\nthe formula for the sample standard deviation: \\(s=\\sqrt{\\frac{\\sum{(x-\\overline{x})^2}}{n-1}}\\)
\n\\(s_1=\\var{sample_stdev_1}\\) and \\(s_2=\\var{sample_stdev_2}\\)
\nWe must evaluate the pooled sample standard deviation using the two sample standard deviartions \\(s_1\\) and \\(s_2\\) and the formula
\n\\(s_p=\\sqrt{\\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}=\\sqrt{\\frac{14(\\var{sample_stdev_1})^2+14(\\var{sample_stdev_2})^2}{15+15-2}}=\\var{pooled_s}\\)
\nThe formula for the t-statistic:
\n\\(t=\\frac{\\overline{x_1}-\\overline{x_2}}{s_p\\sqrt{\\frac{1}{n_1}+\\frac{1}{n_2}}}=\\frac{\\var{sample_mean_1}-\\var{sample_mean_2}}{\\var{pooled_s}\\sqrt{\\frac{1}{14}+\\frac{1}{14}}}=\\var{test_statistic}\\)
\nThe t-table value will be for a one-tailed test and will have \\(n_1+n_2-2=28\\) degrees of freedom. Because of the alternative hypothesis the t-value chosen will be positive.
\n\\[\\begin{array}{r|rrrr}&0.10&0.05&0.01\\\\\\hline28&\\var{t90}&\\var{t95}&\\var{t99}\\end{array}\\]
\nCompare the test statistic with the t-table values and choose your conclusion.
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