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Calculate confidence interval for the proportion

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A company manufacturing a diagnostic kit sent {n1} samples of the kit to an independent laboratory to be analysed. The number of diagnostic kits that met industry standards was {prop1}.

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The estimate for the population proportion is $\\hat{p} =\\frac{{\\var{prop1}}}{\\var{n1}}=\\var{precround(p_hat,3)}$

The estimate for the standard error associated with $\\hat{p}$ is $=\\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}}=\\sqrt{\\frac{\\var{precround(p_hat,3)}\\times (1-\\var{precround(p_hat,3)})}{\\var{n1}}}=\\var{se}$

To construct a $\\var{percentage}$% confidence interval, $Z_\\frac{\\alpha}{2}=Z_\\frac{\\var{alpha2}}{2} = \\var{Z}$

Putting this information together, the lower bound of the $\\var{percentage}$% confidence interval is:

$\\hat{p} -Z_\\frac{\\alpha}{2}\\times\\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{\\var{n1}}}=\\var{precround(p_hat,3)} -\\var{Z}\\times \\var{se}=\\var{precround(lower,2)}$

The upper bound of the $\\var{percentage}$% confidence interval is:

$\\hat{p} +Z_\\frac{\\alpha}{2}\\times\\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{\\var{n1}}}=\\var{precround(p_hat,3)} +\\var{Z}\\times \\var{se}=\\var{precround(upper,2)}$

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The sample proportion $\\hat{p}$ is [[0]]

The standard error of the sample proportion is [[0]]

Calculate a {percentage}% confidence interval for the proportion of diagnostic kits that meet idustry standards.

Lower limit = [[0]]

Upper limit = [[1]]

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