11 - Interval Estimation

General Info

• sample cannot tell exact population mean $\mu$ → use sample mean to predict interval of confidence that population mean will lie in → confidence interval

note	sample	population
mean	$\bar{x}$	$\mu$
std	$s$	$\sigma$

Normal distribution of Sample mean

• $\overline{X_n}$ normally distributed with mean $\mu$ and std $\sigma /\sqrt{n}$ (Practical Interpretation of Limit Theorems)
  • standard normal random var / test statistic: $Z=\frac{\overline{X_n}-\mu }{\sigma /\sqrt{n}}$, $Z\sim N(0,1^2).$ (tag normal random var using z-test)
  • eg. 95% confidence interval → find $a$ in $P(-a\le Z\le a)=0.95$ → $a\approx 1.96$

Background Info for Confidence Interval

• to find the confidence interval containing $\mu$ with prob $1-\alpha$, find a such that:
  • toExplore why it has to be 1-alpha, why dont use the confidence level directly? refer to the normal dis graph for understanding?
  • $P(-a\le Z\le a)=1-\alpha \to P(Z\le -a)=\frac{\alpha }{2}$
  • $a$ is then labelled $z_{\frac{\alpha }{2}}$, determined for any $\alpha$ using tech

Confidence Interval for mean of known std

Calculating CI for mean of known std

• For interval containing $\mu$ with prob $1-\alpha$, sample mean $\bar{x}$
  • sample point estimate - margin of error $\le$ population point est $\ge$ sample point estimate - margin of error
  • $CI=\bar{x}-z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}\le \mu \le \bar{x}+z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$
  • sample mean, if unavailable: $=AVERAGE(input,range)$
  • Confidence level of the interval: $(1-\alpha )\cdot 100\%$
  • $(1-\alpha )$: confidence coefficient
  • margin of error: $z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$,
  • $=CONFIDENCE.NORM(\alpha ,\sigma ,n)$
  • width of confidence interval: $2\cdot z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$
  • $\uparrow$ sample size = $\downarrow$ width of interval (CI width is smaller as we have a larger sample size)
  • $\uparrow$ confidence level = $\uparrow$ width of interval
    • correlation between confidence level & width of interval

Determining sample size

• using confidence intervals to estimate population means, $\downarrow$ width of interval = $\uparrow$ sample size
  • $n={\left(\frac{2\times z_{\frac{\alpha }{2}}\sigma }{w}\right)}^2$
  • $\sigma$: population std, $n$: sample size

Confidence Interval for mean of unknown std

Calculating CI for mean of unknown std

• when $\sigma$ unknown, use same sample to estimate both $\mu$ and $\sigma$
• when using sample standard deviation $s$ to estimate $\sigma$, interval estimate for population mean is based on the t-distribution
• Interval estimate of a population mean, $\sigma$ unknown:
  • $\bar{x}-t_{\alpha /2}\frac{s}{\sqrt{n}}\le \mu \le \bar{x}+t_{\alpha /2}\frac{s}{\sqrt{n}}$
  • s: sample std
  • Confidence level of the interval: $(1-\alpha )\cdot 100\%$
  • $(1-\alpha )$: confidence coefficient
  • $t_{\alpha /2}$: find t-value from t-distribution with $n-1$ degree of freedom (Excel: T.INV($\frac{1-\alpha }{2}$; df)), take the positive result
  • $n$: sample size, $n\ge 30$ to use this expression
  • margin of error: $t_{\alpha /2}\frac{s}{\sqrt{n}}$ CONFIDENCE.T

Calculating T-distribution in EXCEL

• calculating t-distribution: $T.DIST(x,df,cumulative)$: find probability value for a given t-value
  • x : test statistic, or t-value
  • df: n - 1
  • cumulative is a logical value that determines the form of the function. if true, return cdf, else, return pdf

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Confidence Interval for Proportion

Calculating CI for Proportion

• true population proportion $p$ is unknown → taking a sample to obtain sample proportion $\bar{p}$ to estimate $p$
• $\bar{p}=X/n$
  • $\bar{p}$: sample proportion
  • $X$: number of successes in the sample
  • n: sample size
• $X\sim B(n,p),EX=np,{\sigma }_x=\sqrt{np(1-p)}$
  → $E\bar{p}=p$, $\sigma (\bar{p})=\sqrt{\frac{p(1-p)}{n}}$
• By Central Limit Theorem, $\bar{p}$ is approximately normally distributed if
  • both $np\ge 5\mathrm{and}n(1-p)\ge 5.$
• Confidence Interval for p:
  • $\bar{p}-z_{\frac{\alpha }{2}}\sqrt{\frac{\bar{p}(1-\bar{p})}{n}}\le p\le \bar{p}+z_{\frac{\alpha }{2}}\sqrt{\frac{\bar{p}(1-\bar{p})}{n}}$
  • margin of error: $z_{\frac{\alpha }{2}}\sqrt{\frac{\bar{p}(1-\bar{p})}{n}}$
  • width of confidence interval: $2\cdot z_{\frac{\alpha }{2}}\sqrt{\frac{\bar{p}(1-\bar{p})}{n}}$

Determining Sample size

• for $z_{\frac{\alpha }{2}}$ confidence interval, width $w$, preliminary proportion $\bar{p}=p^{\ast }$, sample size is
  • $n={\left(\frac{2\times z_{\frac{\alpha }{2}}}{w}\right)}^2{p}^{\ast }(1-p^{\ast })$
  • if dont have $\bar{p}$, choose $p^{\ast }=0.5$, as it is worst case, and maximizes $p^{\ast }(1-p^{\ast })$

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Test a claim using confidence interval (a, b)

• Test claim using a confidence interval (a, b)

Popular confidence level

Confidence level	$\alpha$	$z_{\frac{\alpha }{2}}$
90%	0.1	1.645
95%	0.05	1.960
98%	0.02	2.326
99%	0.01	2.576

• find $z_{\frac{\alpha }{2}}$ by using Inverse Normal (casio) with area = $\alpha /2$