Mission10+11.ipynb

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    "## MISSION 10 &mdash; CONFIDENCE INTERVALS FOR POPULATION PROPORTIONS (p)\n",
    "\n",
    "### SUMMARY NOTES &mdash; MISSION 10\n",
    "\n",
    "#### FORMAT OF CONFIDENCE INTERVAL\n",
    "$ \\newcommand{\\olsi}[1]{\\,\\overline{\\!{#1}}} % overline short italic $ \n",
    "\n",
    "| Categorical Data (yes/no, success/failure): <br/> Determination of the percentage of the population with a particular characteristic | Quantitative Data: <br/> Determinination of the average value of a parameter of the population | \n",
    "|:-|:-|\n",
    "| __CI for Population Proportion:__ $p$ | __CI for Population Mean,__ $\\mu$ |  \n",
    "| __Sample Size:__ $n$ | __Sample Size:__ $n$ |  \n",
    "| __Sample Proportion:__ $\\hat{p}=\\frac{x}{n}$ <br/> (sample proportion of x successes in a sample size of $n$. | __Sample Mean:__ $\\olsi{X}$ <br/> __Sample Standard Deviation:__ $s$ |  \n",
    "| __Critical Value:__ $ (Z_{\\alpha\\,/\\,2})^* $ <br/> for $n$ sufficiently large ($n \\cdot p \\ge 5$) <br/> &nbsp; &nbsp; &nbsp; $(1-\\alpha) \\; \\text{CI:} \\; \\hat{p} \\pm (Z_{\\alpha\\,/\\,2})^* \\cdot \\left(\\sqrt{\\frac{\\hat{p} \\cdot (1-\\hat{p})}{n}}\\right) $| __Critical Value:__ $ (Z_{\\alpha\\,/\\,2})^* $ <br/> 1. If population standard deviation  is known: <br/> &nbsp; &nbsp; &nbsp; &nbsp; $(1-\\alpha) \\; \\text{CI:} \\; \\olsi{x} \\pm (Z_{\\alpha\\,/\\,2})^* \\cdot \\left( \\frac{\\sigma}{\\sqrt{n}} \\right)$ <br/> 2. If population standard deviation is unknown $\\land \\; n \\ge 30 $ <br/> &nbsp; &nbsp; &nbsp; &nbsp; $(1-\\alpha) \\; \\text{CI:} \\; \\olsi{x} \\pm (t_{\\alpha\\,/\\,2})^* \\cdot \\left( \\frac{s}{\\sqrt{n}} \\right)$ <br/> Or approximate by <br/> &nbsp; &nbsp; &nbsp; &nbsp; $(1-\\alpha) \\; \\text{CI:} \\; \\olsi{x} \\pm (Z_{\\alpha\\,/\\,2})^* \\cdot \\left( \\frac{s}{\\sqrt{n}} \\right)$ <br/> 3. If population standard deviation is unknown $\\land \\; n \\lt 30 $ <br> &nbsp; &nbsp; &nbsp; &nbsp; $(1-\\alpha) \\; \\text{CI:} \\; \\olsi{x} \\pm (t_{\\alpha\\,/\\,2})^* \\cdot \\left( \\frac{s}{\\sqrt{n}} \\right)$ |\n",
    "| __Approximate 95% Confidence Interval; Critical Value=2__ | | \n",
    "| $ \\hat{p} \\pm 2 \\cdot \\left( \\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}} \\right) $ | $ \\olsi{x} \\pm 2 \\cdot \\left( \\frac{s}{\\sqrt{n}} \\right); \\text{if} \\, \\sigma \\, \\text{is unknown,} \\, n \\ge30 $ |  \n",
    "| $ 95\\% \\, \\text{Confidence Interval} \\, (\\alpha = 0.05); $ <br/> $ \\text{Critical Value} = (Z_{0.05/2})^* = (Z_{0.025})^* = 1.96 $ ||  \n",
    "| $ \\hat{p} \\pm 1.96 \\cdot \\left( \\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}} \\right) $ | $ \\olsi{x} \\pm 1.96 \\cdot \\left( \\frac{s}{\\sqrt{n}} \\right); \\text{if} \\, \\sigma \\, \\text{is unknown,} \\, n \\ge30 $ |  \n",
    "| $ 90\\% \\, \\text{Confidence Interval} \\, (\\alpha = 0.10); $ <br/> $ \\text{Critical Value} = (Z_{0.10/2})^* = (Z_{0.05})^* = 1.645 $ ||  \n",
    "| $ \\hat{p} \\pm 1.645 \\cdot \\left( \\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}} \\right) $ | $ \\olsi{x} \\pm 1.645 \\cdot \\left( \\frac{s}{\\sqrt{n}} \\right); \\text{if} \\, \\sigma \\, \\text{is unknown,} \\, n \\ge30 $ |  \n",
    "| $ 99\\% \\, \\text{Confidence Interval} \\, (\\alpha = 0.01); $ <br/> $ \\text{Critical Value} = (Z_{0.01/2})^* = (Z_{0.005})^* = 2.575 $ ||  \n",
    "| $ \\hat{p} \\pm 2.575 \\cdot \\left( \\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}} \\right) $ | $ \\olsi{x} \\pm 2.575 \\cdot \\left( \\frac{s}{\\sqrt{n}} \\right); \\text{if} \\, \\sigma \\, \\text{is unknown,} \\, n \\ge30 $ |  \n",
    "\n",
    "#### MARGIN OF ERROR (ME)\n",
    "\n",
    "One of the main caveats of sample results is that they vary from sample to sample, so you shouldn’t report statistical results without also including the margin of error. \n",
    "\n",
    "The __margin of error__ is the amount by which you expect the sample results to vary from sample to sample. A large margin of error means you expect the sample results to change a lot (not a good thing), and a small margin of error means you don’t expect the sample results to change much with repeated sampling (a good thing).\n",
    "In this course, you tackle problems involving the margin of error for the sample mean or for the sample proportion. \n",
    "\n",
    "__Quantitative Data:__ If you want to determine the average value for the population, you use the sample mean, plus or minus the margin of error for your sample mean.\n",
    "\n",
    "__Categorical Data (yes/no, or success/failure):__ If you want to determine the percentage in the population with a certain characteristic, you use the sample proportion, plus or minus the margin of error for your sample proportion.\n",
    "\n",
    "The margin of error formulas look different on the surface in each situation, but they each contain three important components: the $Z^*$ value, The standard deviation of the population, and the sample size. \n",
    "\n",
    "Note: the $Z^*$ value is a number from the standard normal distribution and reflects the number of standard errors you need to add/ subtract in order to get the confidence level you want. For example, to be 95 percent $(95\\%)$ confident, you need to add/subtract about $ 2.0 \\cdot $ standard error ($1.96$ to be exact) because of the Empirical Rule.\n",
    "\n",
    "The specific formula for **__margin of error__ for the __sample mean__** is: $ \\pm \\left( Z_{\\alpha / 2} \\right) ^* \\cdot \\left( \\frac{\\sigma}{\\sqrt{n}} \\right) $\n",
    "\n",
    "The specific formula for **__margin of error__ for the __sample proportion__** is: $ \\pm \\left( Z_{\\alpha/2} \\right) ^* \\cdot \\left( \\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}} \\right) $\n",
    "\n",
    "##### INTERPRETING MARGIN OF ERROR CORRECTLY\n",
    "\n",
    "Margin of error measures the level of precision in your sample results. Precision represents consistency, but it doesn’t necessarily promise on-target results (they may be systematically over or under by a certain amount each time). In other word, margin of error doesn’t measure the amount of bias that may come with the sample results. Most types of bias are extremely difficult to measure, so you should make an effort minimize bias wherever possible.  \n",
    "\n",
    "<HR/>"
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    "### EXAMPLES/PROBLEMS &mdash; MISSION 10\n",
    "\n",
    "#### MARGIN OF ERROR (ME)\n",
    "\n",
    "##### CALCULATE MARGIN OF ERROR (ME)  \n",
    "$ \\newcommand{\\olsi}[1]{\\,\\overline{\\!{#1}}} % overline short italic $ \n",
    "\n",
    "*__Note:__*  \n",
    "\n",
    "> The specific formula for the **__margin of error__ of the __sample mean__** is: $ \\text{ME} = \\pm (Z_{\\alpha/2})^* \\cdot \\left( \\frac{\\sigma}{\\sqrt{n}} \\right) $  \n",
    "> The specific formula for the **__margin of error__ of the __sample proportion__ is: $ \\text{ME} = \\pm (Z_{\\alpha/2})^* \\cdot \\left(\\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}} \\right) $  \n",
    "\n",
    "1. Assume that a sample is used to estimate a population proportion p. Find the margin of error M.E. that corresponds to a sample of size 100 with 20% successes at a confidence level of 95%.  \n",
    "\n",
    "> $ \\text{ME} =  \\pm 1.96 \\cdot \\sqrt{\\frac{0.20 \\cdot 0.80}{100}} $  \n",
    "> $ \\text{ME} =  \\pm 1.96 \\cdot \\sqrt{\\frac{0.16}{100}} $  \n",
    "> $ \\text{ME} =  \\pm 1.96 \\cdot 0.04 $  \n",
    "> $ \\text{ME} =  \\pm 0.0784 $  \n",
    "\n",
    "\n",
    "\n",
    "2. We wish to estimate what percent of adult residents in a certain county are parents. Out of 400 adult residents sampled, 100 had kids. Based on this, construct a 98% confidence interval for the proportion, p, of adult residents who are parents in this county.  \n",
    "> Give your answers as decimals, to two places  \n",
    "> *__With 98\\% confidence:__*  \n",
    ">> $ \\olsi{p} = \\hat{p} \\pm (Z_{\\alpha/2})^2 \\cdot \\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}} $  \n",
    ">> $ \\olsi{p} = 0.25 \\pm 2.327 \\cdot \\sqrt{\\frac{0.25 \\cdot 0.75)}{400}} $  \n",
    ">> $ \\olsi{p} = 0.25 \\pm 0.05 $  \n",
    ">> $ \\olsi{p} = ( 0.20, 0.30 ) $  \n",
    ">> $ \\olsi{p} = ( 20\\%, 30\\% ) $  \n",
    "\n",
    "\n",
    "__Sample size:__  \n",
    "> $ \\text{ME} = \\pm (Z_{\\alpha/2})^* \\cdot \\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}} $  \n",
    "> $ \\frac{\\text{ME}}{Z} = \\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}} $  \n",
    "> $ \\frac{\\text{ME}^2}{Z^2} = \\frac{\\hat{p}(1-\\hat{p})}{n} $  \n",
    "> $ n = \\frac{Z^2}{ME^2} \\cdot \\left[\\hat{p} \\cdot (1-\\hat{p})\\right] $  \n",
    "\n",
    "__Note:__ If $\\hat{p}$ is unknown, what value should be assumed for $\\hat{p}$?\n",
    "\n",
    "3. A political candidate has asked you to conduct a poll to determine what percentage of people support her. If the candidate only wants a 4% margin of error at a 90% confidence level, what size of sample is needed?\n",
    "\n",
    "> If $\\hat{p}$ is unknown, what value should be assumed for $\\hat{p}$?  \n",
    "> $ n = \\frac{Z^2 \\cdot \\left[ \\hat{p} \\cdot (1-\\hat{p})\\right]}{ME^2} $  \n",
    "> Since the sample size, $n$, is proportional to the product $\\left[\\hat{p} \\cdot (1-\\hat{p})\\right]$:  \n",
    ">> We examine the value of $\\left[\\hat{p} \\cdot (1-\\hat{p})\\right]$ (which is a quadratic)  \n",
    ">> For $ \\hat{p} = 0.5, \\; \\left[\\hat{p} \\cdot (1-\\hat{p})\\right] = 0.25$ (its maximum value)  \n",
    ">>> $ \\lim_{\\hat{p} \\to 0} \\left[\\hat{p} \\cdot (1-\\hat{p})\\right] = 0$  \n",
    ">>> $ lim_{\\hat{p} \\to 1} \\left[\\hat{p} \\cdot (1-\\hat{p})\\right] = 0 $  \n",
    ">  \n",
    "> $\\therefore$ the maximum value of $n$ will also occur at $\\hat{p}=0.5$,  \n",
    "> Making $\\hat{p}=0.5$ the safest assumption when $\\hat{p}$ is unknown. \n",
    "> \n",
    "> Give your answer in whole people.\n",
    ">\n",
    "> $ n = \\frac{Z^2}{ME^2}  \\cdot \\left[\\hat{p} \\cdot (1-\\hat{p})\\right] $  \n",
    "> $ n = \\frac{1.645^2}{0.04^2}  \\cdot 0.25$  \n",
    "> $ n = \\frac{0.6765}{0.0016}$  \n",
    "> $ n = 423 $  \n",
    "\n",
    "#### APPLICATION: SPORTS CARS \n",
    "\n",
    "Sports car owners in a town complain that their cars are judged differently from family-style cars at the state vehicle inspection station. Previous records indicate that 30% of all passenger cars fail the inspection on the first time through. In a random sample of 150 sports cars, 60 failed the inspection on the first time through. Is there sufficient evidence to indicate the percentage (proportion) of first failures for sports cars is higher than the percentage (proportion) for all passenger cars?\n",
    "\n",
    "1. What is the parameter proportion ($p$) in this setting? (Description only)  \n",
    "> The proportion of sports car owners who fail the inspection the first time (p).  \n",
    "2. What is the sample proportion ($\\hat{p}$) in this setting?  \n",
    "> $ \\hat{p} = \\frac{60}{150} = 0.40 = 0.40\\% $\n",
    "\n",
    "3. Use the sample results to give a 99% confidence interval for $p$\n",
    "> *__With 99\\% confidence:__*  \n",
    ">> $ \\olsi{p} = \\hat{p} \\pm (Z_{\\alpha/2})^2 \\cdot \\left( \\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}} \\right) $  \n",
    ">> $ \\olsi{p} = 0.40 \\pm 2.575 \\cdot \\left( \\sqrt{\\frac{0.4 \\cdot 0.6)}{150}} \\right) $  \n",
    ">> $ \\olsi{p} = 0.40 \\pm 0.103 $  \n",
    ">> $ \\olsi{p} = ( 0.297, 0.503 ) $  \n",
    ">> $ \\olsi{p} = ( 29.7\\%, 50.3\\% ) $  \n",
    "\n",
    "4. Interpret the above confidence interval.  \n",
    "> *__With 99\\% confidence, the mean first pass inspection failure rate for sports cars is between 29.7\\% and 50.3\\%.__*  \n",
    "\n",
    "5. Based on 99\\% confidence interval, is there sufficient evidence to indicate the percentage (proportion) of first failures for sports cars is higher than the percentage (proportion) for all passenger cars?  \n",
    "> *__No, based on the provided data, the mean first pass inspection failure rate for sports cars could be the same as that of the larger population of cars.__*\n",
    "\n",
    "#### APPLICATION: DEFECTIVE RATE  \n",
    "\n",
    "A company claims that the defective rate of their product is 0.05. A sample of size 1000 and observe 70 of them are defective. Based on the sample result, is there sufficient evidence to show that the defective rate is not 0.05 (as company’s claim) based on 90% confidence Interval?\n",
    "\n",
    "1. Use the sample results to give a 90% confidence interval for the product defective rate.  \n",
    "> *__With 90\\% Confidence:__*\n",
    ">> $ \\olsi{p} = \\hat{p} \\pm (Z_{\\alpha/2})^2 \\cdot \\left( \\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}} \\right) $  \n",
    ">> $ \\olsi{p} = 0.07 \\pm 1.645 \\cdot \\left( \\sqrt{\\frac{0.07 \\cdot 0.93)}{1000}} \\right) $  \n",
    ">> $ \\olsi{p} = 0.07 \\pm 0.0133 $  \n",
    ">> $ \\olsi{p} = ( 0.0567, 0.0833 ) $  \n",
    ">> $ \\olsi{p} = ( 5.67\\%, 8.33\\% ) $  \n",
    "\n",
    "2. Interpret the above confidence interval.  \n",
    "> *__With 99\\% confidence, the mean product defective rate is between 5.67\\% and 8.33\\%.__*  \n",
    "\n",
    "3. Based on the above 90% confidence interval, do you believe the defective rate is 0.05 (as the company’s claim)? Justify your answer.  \n",
    "> *__No, based on the provided sample and with 90\\% confidence, the product defective rate is greater than the claimed 5.0\\%__*  \n",
    "\n",
    "#### APPLICATION: JURY SELECTION\n",
    "\n",
    "In the case of Green v. Pink. It was found that during a period of 15 years in Statopia, 5000 people were selected for grand jury duty, and 60% of them have special ability of Air.  \n",
    "\n",
    "1. Use the above data to construct a 95% confidence interval estimate of the proportion of grand jury members who had Air ability. What is Margin of Error?  \n",
    "> *__With 95\\% confidence:__*  \n",
    ">> $ \\olsi{p} = \\hat{p} \\pm (Z_{\\alpha/2})^2 \\cdot \\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}} $  \n",
    ">> $ \\olsi{p} = 0.60 \\pm 1.96 \\cdot \\sqrt{\\frac{0.6 \\cdot 0.4}{5000}} $  \n",
    ">> $ \\olsi{p} = 0.60 \\pm 0.0136 $  \n",
    ">> $ \\olsi{p} = ( 0.5864, 0.6136 ) $  \n",
    ">> $ \\olsi{p} = ( 58.64\\%, 61.36\\% ) $  \n",
    ">> $\\text{ME} = 0.0136 = 1.36\\%$  \n",
    "\n",
    "2. Given that among the people eligible for jury duty 63.1% of them had Air ability, does it appear that the jury selection process was somehow biased against Air ability? Please justify your answer based on the above 95% confidence interval.  \n",
    "> *__Yes, since the actual occurrence of ability Air in the general population lies outside the 95% confidence bands of our estimate, we can say with 95% confidence that the evidence points to the existence of a bias against people with the Air ability in the jury selection process.__*\n",
    "\n",
    "<HR/>\n",
    "<HR/>"
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    "## MISSION 11 &mdash; \n",
    "\n",
    "### SUMMARY NOTES &mdash; MISSION 11\n",
    "\n",
    "#### CONFIDENCE INTERVALS FOR THE DIFFERENCE BETWEEN TWO POPULATION MEANS\n",
    "\n",
    "We use a confidence interval for the difference between two population means under two conditions:  \n",
    "1. The variable you are dealing with is quantitative (eg., height, weight, exam score).\n",
    "2. __These two populations are independent__ and you wish to estimate the difference between the two population means. \n",
    "\n",
    "$ \\newcommand{\\olsi}[1]{\\,\\overline{\\!{#1}}} % overline short italic $ \n",
    "\n",
    "<!---\n",
    "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n",
    "t Distribution Block\n",
    "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n",
    "-->\n",
    "\n",
    "<div class=\"alert alert-block alert-info\">\n",
    "<p style=\"text-decoration:underline;\">\n",
    "    <b><em>Using the $t$ Distribrution (Theoretical)</em></b></p>  \n",
    "$\\text{Sample Sizes: } n_1,n_2$ <br/>\n",
    "$\\text{Sample Means: } \\olsi{x_1},\\olsi{x_2}$ <br/>\n",
    "$\\text{Critical Value: } (t_{\\alpha/2})^* $ (from $t$ table)\n",
    "    <blockquote style=\"text-align:left;margin-top: 0;\">\n",
    "        $ df = \\text{min}(n_1-1,n_2-1) $ </blockquote> \n",
    "$(1-\\alpha) \\; \\text{CI:} \\; \\; (\\olsi{x_1}-\\olsi{x_2}) \\; \\pm \\; (t_{\\alpha/2})^* \\cdot \\sqrt{\\frac{s_1^2}{n_1}+\\frac{s_2^2}{n_2}} $ <br/>    \n",
    "</div>\n",
    "\n",
    "\n",
    "<!---\n",
    "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n",
    "Z Distribution Block\n",
    "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n",
    "-->\n",
    "\n",
    "<div class=\"alert alert-block alert-info\">\n",
    "<p style=\"text-decoration:underline;\">\n",
    "    <b><em>Using the $Z$ Distribrution (Practical)</em></b></p>  \n",
    "$\\text{Sample Sizes: } n_1,n_2$ <br/>\n",
    "$\\text{Sample Means: } \\olsi{x_1},\\olsi{x_2}$ <br/>\n",
    "$\\text{Critical Value: } (Z_{\\alpha/2})^* $ (from the Standard Normal Distribution table)\n",
    "    <blockquote style=\"text-align:left;margin-top: 0;\">\n",
    "        $ \\text{If sample size is large enough: } n_1 \\ge 30, n_2 \\ge 30 $ </blockquote>\n",
    "    \n",
    "$(1-\\alpha) \\; \\text{CI:} \\; \\; (\\olsi{x_1}-\\olsi{x_2}) \\; \\pm \\; (Z_{\\alpha/2})^* \\cdot \\sqrt{\\frac{s_1^2}{n_1}+\\frac{s_2^2}{n_2}} $ <br/>    \n",
    "</div>\n",
    "\n",
    "- $90\\%$ __Confidence Interval__$(\\alpha=0.10)$; __Critical Value__$=(Z_{0.1/2})^*=(Z_{0.05})^*=1.645$\n",
    "> $ (\\olsi{x_1}-\\olsi{x_2}) \\; \\pm \\; 1.645 \\cdot \\sqrt{\\frac{s_1^2}{n_1}+\\frac{s_2^2}{n_2}} $  \n",
    "\n",
    "- $95\\%$ __Confidence Interval__$(\\alpha=0.05)$; __Critical Value__$=(Z_{0.05/2})^*=(Z_{0.025})^*=1.96$\n",
    "> $ (\\olsi{x_1}-\\olsi{x_2}) \\; \\pm \\; 1.96 \\cdot \\sqrt{\\frac{s_1^2}{n_1}+\\frac{s_2^2}{n_2}} $  \n",
    "\n",
    "- $99\\%$ __Confidence Interval__$(\\alpha=0.01)$; __Critical Value__$=(Z_{0.01/2})^*=(Z_{0.005})^*=2.575$\n",
    "> $ (\\olsi{x_1}-\\olsi{x_2}) \\; \\pm \\; 2.575 \\cdot \\sqrt{\\frac{s_1^2}{n_1}+\\frac{s_2^2}{n_2}} $  \n",
    "\n",
    "#### CONFIDENCE INTERVALS FOR THE DIFFERENCE BETWEEN TWO POPULATION PROPORTIONS\n",
    "\n",
    "We use a confidence interval estimate for the difference between two population proportions under three conditions:  \n",
    "1. The variable that you are dealing with is categorical (e.g., gender, political affiliation, opinion [agree, disagree, no opinion] or marital status).\n",
    "2. You want to estimate the difference between the two population proportions.\n",
    "3. The sample sizes are large enough.\n",
    "\n",
    "<!---\n",
    "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n",
    "Z Distribution Block\n",
    "%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n",
    "-->\n",
    "\n",
    "<div class=\"alert alert-block alert-info\">\n",
    "<p style=\"text-decoration:underline;\">\n",
    "    <b><em>Using the $Z$ Distribrution (Practical)</em></b></p>  \n",
    "$\\text{Sample Sizes: } n_1,n_2$ <br/>\n",
    "$\\text{Sample Proportions: } \\hat{p_1}=\\frac{x_1}{n_1},\\hat{p_2}=\\frac{x_2}{n_2}$\n",
    "    <blockquote style=\"text-align:left;margin-top: 0;\"> (sample proportion of $x$ successes in a sample of size $n$)</blockquote>\n",
    "$\\text{Critical Value: } (Z_{\\alpha/2})^* $ (from the Standard Normal Distribution table)\n",
    "    <blockquote style=\"text-align:left;margin-top: 0;\">\n",
    "        <b>If sample size is large enough:</b> <br/>\n",
    "        $n_i p_i \\ge 5 \\; \\text{for} \\, i=1,2 \\;\n",
    "        \\land \\; n_i (1-p_i) \\ge 5 \\; \\text{for} \\, i=1,2) $\n",
    "    </blockquote>\n",
    "\n",
    "$(1-\\alpha) \\; \\text{CI:} \\; \\; (\\hat{p_1}-\\hat{p_2}) \\; \\pm \\; (Z_{\\alpha/2})^* \\cdot \\sqrt{\\frac{\\hat{p_1}(1-\\hat{p_1})}{n_1}+\\frac{\\hat{p_2}(1-\\hat{p_2})}{n_2}}$  <br/>\n",
    "</div>\n",
    "    \n",
    "- $90\\%$ __Confidence Interval__$(\\alpha=0.10)$; __Critical Value__$=(Z_{0.1/2})^*=(Z_{0.05})^*=1.645$\n",
    "> $ (\\hat{p_1}-\\hat{p_2}) \\; \\pm \\; 1.645 \\cdot \\sqrt{\\frac{\\hat{p_1}(1-\\hat{p_1})}{n_1}+\\frac{\\hat{p_2}(1-\\hat{p_2})}{n_2}}$  \n",
    "\n",
    "- $95\\%$ __Confidence Interval__$(\\alpha=0.05)$; __Critical Value__$=(Z_{0.05/2})^*=(Z_{0.025})^*=1.96$\n",
    "> $ (\\hat{p_1}-\\hat{p_2}) \\; \\pm \\; 1.96 \\cdot \\sqrt{\\frac{\\hat{p_1}(1-\\hat{p_1})}{n_1}+\\frac{\\hat{p_2}(1-\\hat{p_2})}{n_2}}$  \n",
    "\n",
    "- $99\\%$ __Confidence Interval__$(\\alpha=0.01)$; __Critical Value__$=(Z_{0.01/2})^*=(Z_{0.005})^*=2.575$\n",
    "> $ (\\hat{p_1}-\\hat{p_2}) \\; \\pm \\; 2.575 \\cdot \\sqrt{\\frac{\\hat{p_1}(1-\\hat{p_1})}{n_1}+\\frac{\\hat{p_2}(1-\\hat{p_2})}{n_2}}$  \n",
    "\n",
    "<HR/>"
   ]
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    "### EXAMPLES/PROBLEMS &mdash; MISSION 11\n",
    "\n",
    "#### APPLICATIONS FOR THE DIFFERENCE BETWEEN TWO POPULATIONS\n",
    "\n",
    "<HR/>\n",
    "\n",
    "##### __Application: Are Blue and Green students different in ability in Statopia?__\n",
    "$ \\newcommand{\\olsi}[1]{\\,\\overline{\\!{#1}}} % overline short italic $  \n",
    "__Mean or Proportion?__  \n",
    "__One Sample or Two Samples?__  \n",
    "__Which Formula is Appropriate?__\n",
    "\n",
    "In a recent study of Statopian students, __63 percent__ of a __sample of 100__ Blue students have the “Water” ability. A __sample of 120__ Green students shows __60\\%__ possess the “Water” ability.\n",
    "\n",
    "1. Find a $99\\%$ confidence interval for the difference in the percentages of students who possess the water ability for these two populations. What is the Margin of Error (ME)?  \n",
    "> __Mean or Proportion?__ *The use of percentages indicates that this is a PROPORTION problem*  \n",
    "> __One Sample or Two Samples?__ *Independent samples were taken of the Blue and Green studen populations, therefore this is a Two Sample problem.*  \n",
    "> __Which Formula is Appropriate?__ *A __$99\\%$ Confidence Interval__ is requested for __a proportion problem__ with __two samples,__ therefore, the appropriate formula is:*  \n",
    "> $ 99\\% \\, \\text{CI:} \\; \\; (\\hat{p_1}-\\hat{p_2}) \\; \\pm \\; 2.575 \\cdot \\sqrt{\\frac{\\hat{p_1}(1-\\hat{p_1})}{n_1}+\\frac{\\hat{p_2}(1-\\hat{p_2})}{n_2}}$  \n",
    ">  \n",
    "> We will consider the sample of the Blue student poulation to be sample 1, therefore:  \n",
    "> $\\hat{p_1} = 63\\% = 0.63 \\; \\text{and} \\; n_1 = 100$  \n",
    ">  \n",
    "> We will consider the sample of the Green student poulation to be sample 2, therefore:  \n",
    "> $\\hat{p_2} = 60\\% = 0.60 \\; \\text{and} \\; n_2 = 120$  \n",
    ">\n",
    "> $ 99\\% \\, \\text{CI:} \\; \\; (0.63-0.60) \\; \\pm \\; 2.575 \\cdot \\sqrt{\\frac{0.63 \\cdot 0.37}{100}+\\frac{0.60 \\cdot 0.40}{120}}$  \n",
    "> $ 99\\% \\, \\text{CI:} \\; \\; 0.03 \\; \\pm \\; 2.575 \\cdot 0.0658 $  \n",
    "> $ 99\\% \\, \\text{CI:} \\; \\; 0.03 \\; \\pm \\; 0.1695 $  \n",
    "> $ 99\\% \\, \\text{CI:} \\; \\; ( -0.1395, 0.1995) $\n",
    ">  \n",
    "> $\\text{ME}$ *(the quantity following the $\\pm$)* $= 0.1695$  \n",
    "\n",
    "2. Interpret the above confidence interval.  \n",
    "> *__With 99\\% Confidence, the mean difference in the proportions of Blue and Green students who possess the Water ability kies between $-0.1395$ and $0.1995$.__\n",
    "\n",
    "3. Is there any significant difference in proportion of Water ability between Blue and Green students based on above confidence interval? Justify your answer using above Confidence Interval.\n",
    "> *__The evidence does NOT support there being a significant difference between the occurrence of the Water ability in Blue and Green students since it is not possible to say with confidence that the ability occurs more frequently in either Blue or Green students.__*  \n",
    "\n",
    "<HR/>\n",
    "\n",
    "##### __Application: Change of Skull Size?__\n",
    "\n",
    "__Mean or Proportion?__  \n",
    "__One Sample or Two Samples?__  \n",
    "__Which Formula is Appropriate?__\n",
    "\n",
    "The summary of maximum breadths of samples of male Egyptian skulls from 4000 B.C. and 150 A.D. is provided below.  \n",
    "* 4000 BC: sample size=49, sample mean: 129, sample standard deviation: 6.3\n",
    "* 150 AD: sample size=36, sample mean: 131, sample standard deviation: 6.0    \n",
    "\n",
    "1. Construct a 99% confidence interval for the difference in the skull sizes. What is Margin of Error?  \n",
    "> __Mean or Proportion?__ *The use of qunatitative mean values indicates that this is a __MEAN__ problem*  \n",
    "> __One Sample or Two Samples?__ *Independent samples were taken at two different times (4000 BC and 150 AD), therefore this is a __Two Sample__ problem.*  \n",
    "> __Which Formula is Appropriate?__ *A __$99\\%$ Confidence Interval__ is requested for __a mean problem__ with __two samples.__ The sizes of the samples are large enough to permit use of the standard normal ($Z$) distribution. However, I am choosing to use the $t$ distribution to illustrate its use. Therefore, the appropriate formula is:*  \n",
    "> $ 99\\% \\; \\text{CI:} \\; \\; (\\olsi{x_1}-\\olsi{x_2}) \\; \\pm \\; (t_{\\alpha/2})^* \\cdot \\sqrt{\\frac{s_1^2}{n_1}+\\frac{s_2^2}{n_2}} $  \n",
    ">  \n",
    "> We will consider the 4000 BC sample to be sample 1, therefore:  \n",
    "> $\\olsi{x_1} = 129, \\text{ and } n_1 = 49$  \n",
    ">  \n",
    "> We will consider the 150 AD sample to be sample 2, therefore:  \n",
    "> $\\olsi{x_2} = 131 \\text{ and } n_2 = 36$\n",
    ">\n",
    "> $\\text{df} = \\min{(n_1-1,n_2-1)} = \\min{(48,35)} = 35$  \n",
    ">> The $t$ table for CL=$99\\%$ only gives us values for $n=30 \\; (2.750)$ and for $n=40 \\; (2.704)$, so we take the value midway those two: \n",
    "> \n",
    "> $ ( t_{\\alpha/2})^* = 0.5 * (2.750 + 2.704) = 2.727$ (the comparable $(Z_{\\alpha/2})^*$ is $2.575$)  \n",
    ">  \n",
    "> $99\\% \\, \\text{CI:} \\; \\; (\\olsi{x_1}-\\olsi{x_2}) \\; \\pm \\; 2.727 \\cdot \\sqrt{\\frac{s_1^2}{n_1}+\\frac{s_2^2}{n_2}} $  \n",
    "> $ 99\\% \\, \\text{CI:} \\; \\; (129-131) \\; \\pm \\; 2.727 \\cdot \\sqrt{\\frac{6.3^2}{49}+\\frac{6.0^2}{36}}$  \n",
    "> $ 99\\% \\, \\text{CI:} \\; \\; -2.0 \\; \\pm \\; 2.727 \\cdot \\sqrt{0.81+1.0}$  \n",
    "> $ 99\\% \\, \\text{CI:} \\; \\; -2.0 \\; \\pm \\; 2.727 \\cdot 1.3454$  \n",
    "> $ 99\\% \\, \\text{CI:} \\; \\; -2.0 \\; \\pm \\; 3.6688 $  \n",
    "> $ 99\\% \\, \\text{CI:} \\; \\; ( -5.6688, 1.6688) $\n",
    ">  \n",
    "> $\\text{ME}$ *(the quantity following the $\\pm$)* $= 3.6688$  \n",
    "\n",
    "2. Change in skull sizes over time suggest interbreeding with people from other regions. Use the confidence interval constructed in part a to determine whether the skull sizes appear to have changed from 4000 B.C. to 150 A.D. Justify your answer using above Confidence Interval.  \n",
    "> *__The evidence does NOT support there being a significant in male Egyptian skull breadth between 4000 BC abd 150 AD since it is not possible to say with confidence that the skull breadth is greater or lessor in either sample.__*  \n",
    "\n",
    "<HR/>\n",
    "\n",
    "##### __Should We planning on using Cis for Two population means for …..?__\n",
    "\n",
    "Suppose a group of 100 people is on one weight loss plan for one month and then switches to another weight loss plan for one month. Suppose you want to compare the two plans, and you decide to compare the average weight loss for the first month to the second month.\n",
    "* Should we  plan on using a 95% confidence interval for the average difference or not? Explain.\n",
    "\n",
    "> *__I think the design of the experiment is flawed and that a comparison based on Confidence Intervals could be misleading. In this particular experimental design there are likely to be confounding variables. The primary one being that most strategies for weight loss cause the body to adapt by reducing the basal metabolic rate as time passes. Therefore, it is very likely that the second plan would unfairly be penalized just because the subjects have been on another plan for a month before starting and reduced basal metabolic rate would reduce the results of that second plan. A better design would be to divide the group into two and have each half try an assigned plan simultaneously, that would at least eliminate the bias resulting from reduced basal metabolic rate. At that point a two mean CI estimate might be useful in comparing the results of the two plans.__*\n",
    "\n",
    "<HR/>\n",
    "<HR/>"
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