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A recent poll of \\(\\var{n1}\\) people indicated that \\(\\var{prop1}\\) of them had delayed seeking healthcare treatment due to the associated cost.

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It has long been believed that \\(\\var{percentage}\\)% of people will delay seeking healthcare treatment due to the associated cost. 

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Does the data support this theory? 

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Enter the value of the sample proportion: \\(p=\\) [[1]]

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Enter the value for the appropriate test statistic: Z = [[0]]

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Reject the Null Hypothesis and conclude conclude that the proportion of people who will delay seeking healthcare treatment due to the associated cost is not given by \\(p=\\frac{\\var{percentage}}{100}\\).

", "

Reject the Null Hypothesis at the 5% significance level but accept the Null Hypothesis at the 1% significance level and conclude that the proportion of people who will delay seeking healthcare treatment due to the associated cost is given by \\(p=\\frac{\\var{percentage}}{100}\\).

", "

Reject the Null Hypothesis at the 10% significance level but accept the Null Hypothesis at the 5% significance level and conclude that the proportion of people who will delay seeking healthcare treatment due to the associated cost is given by \\(p=\\frac{\\var{percentage}}{100}\\).

", "

Accept the Null Hypothesis at the 10% significance level and conclude that the proportion of people who will delay seeking healthcare treatment due to the associated cost is given by \\(p=\\frac{\\var{percentage}}{100}\\). 

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Having compared your test statistic with the table values for a two-tailed Z-test, select one of the following conclusions that best describes your conclusion.

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\\(H_0:\\)  p =\\(\\simplify{{percentage}/100}\\).

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\\(H_1:\\) p \\(\\ne \\simplify{{percentage}/100}\\).

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Given a sample of size \\(n\\) recall:

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the formula for the sample proportion:      \\(\\overline{p}=\\frac{{x}}{n}\\) where \\(x\\) is the number of observations

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 \\(\\overline{p}=\\frac{\\var{prop1}}{\\var{n1}}=\\var{p}\\)

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the formula for the Z-statistic:     \\(Z=\\frac{\\overline{p}-p}{\\sqrt{\\frac{p(1-p)}{n}}}\\)

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\\(Z=\\frac{\\var{p}-\\var{pop_p}}{\\sqrt{\\frac{\\var{pop_p}(1-\\var{pop_p})}{\\var{n1}}}}\\)

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\\(Z=\\frac{\\simplify{{p}-{pop_p}}}{\\sqrt{\\simplify{{pop_p}*(1-{pop_p})/{n1}}}}=\\var{test_statistic}\\)

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The Z-table values will be for a two-tailed test are given below. 

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significance              10%                    5%                   1%

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      limits                \\(\\pm1.65\\)             \\(\\pm1.96\\)             \\(\\pm2.58\\)

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

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