// Numbas version: exam_results_page_options {"name": "Confidence Interval Doctor's Surgery", "extensions": ["stats"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"parts": [{"extendBaseMarkingAlgorithm": true, "showCorrectAnswer": true, "scripts": {}, "showFeedbackIcon": true, "sortAnswers": false, "prompt": "

Calculate the mean and standard deviation of the sample of patients visiting on Mondays.

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Enter your answers to 2 decimal places.

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Mean: [[0]]

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Standard Deviation: [[1]]

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Find the 95% confidence interval for the population mean $\\mu_m$ of the age of patients visiting on Monday, using the $t$-distribution with $\\var{num_monday-1}$ degrees of freedom.

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Enter your answers to 2 decimal places.

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Lower limit: [[0]]

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Upper limit: [[1]] 

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The ages of a random sample of patients coming into a doctor's surgery during the school holidays are noted down on each Monday and Friday thoughout August.

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{table([['Monday']+mon,['Friday']+fri],[])}

", "advice": "

a)

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The mean of the sample of patients visiting on Mondays is 

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\\[ \\bar{x} = \\frac{1}{n}(\\sum x_i) = \\var{precround(monday_mean,2)} \\text{ (to 2 d.p.)} \\]

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The standard deviation is

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\\[ s = \\sqrt{\\frac{1}{n-1}\\sum \\left(x_i-\\bar{x}\\right)^2} = \\var{precround(monday_sd,2)} \\text{ (to 2 d.p.)} \\]

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b)

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The sample mean $\\mu_m$ follows a $t$-distribution with $n-1 = \\var{num_monday-1}$ degrees of freedom centred on the population mean, with a standard error of $\\frac{s}{\\sqrt{n}} = \\simplify{{precround(monday_sd,2)}/sqrt({num_monday})} = \\var{precround(monday_se,2)}$.

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That means that the 95% confidence interval for the population mean is as follows:

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\\begin{array}{lrcl}
&\\left[ \\bar{x} - s \\times t_{\\var{num_monday-1},0.975} \\right. \\kern{-0.5em}&,&\\kern{-0.5em} \\left. \\bar{x} + s \\times t_{\\var{num_monday-1},0.975} \\right] \\\\
= & \\left[ \\simplify[]{{precround(monday_mean,2)} - {precround(monday_se,2)}*{precround(monday_t,2)}} \\right. \\kern{-0.5em}&,&\\kern{-0.5em} \\left. \\simplify[]{{precround(monday_mean,2)} + {precround(monday_se,2)}*{precround(monday_t,2)}} \\right] \\\\
= & \\left[ \\var{precround(monday_low,2)} \\right. \\kern{-0.5em}&,&\\kern{-0.5em} \\left. \\var{precround(monday_high,2)} \\right]
\\end{array}

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Lower end of the confidence interval for Monday's mean

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Data for Friday

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Data for Monday

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Compute a 95% confidence interval for the population mean given a small sample, and compare it with a confidence interval for a different population.

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