Confidence Intervals (Part 3)
Sometimes an experiment is designed in such a way that each item in one sample is paired with an item in the other. Let \(D_1, D-2, ..., D_n\) be a small random sample \((n \leq 30)\) of differences of the paired data. If the population of differences is approximately normal, the \(100(1 - \alpha) \%\) confidence interval for the mean difference \(\mu_D\) is given as:
\[\overline{D} \pm t _{n-1, \alpha/2} \frac{s_D}{\sqrt{n}}\]where \(s_D\) is the sample standard deviation of \(D_1, D_2, ..., D_n\). If the sample size is large, the confidence interval is given as:
\[\overline{D} \pm z _{\alpha/2} \frac{s_D}{\sqrt{n}}\]#### Confidence Intervals for the Variance and Standard Deviation of a Normal Population :Confidence interval for the population variance \(\sigma^2\) (when the population is normal) is based on sample variance \(s^2\) and a probability distribution known as chi-square distribution. To be more specific, if \(X_1, X_2, ..., X_n\) is a random sample from a normal population with variance \(\sigma^2\), the sample variance can be given as:
\[s^2 = \frac{1}{n-1} \sum _{i=1}^{n} (X_i - \overline{X})^2\]The quantity
\[\frac{(n-1)s^2}{\sigma^2} = \frac{\sum _{i=1}^{n} (X_i - \overline{X})^2}{\sigma^2}\]follows a chi-square distribution with degree of freedom \(n-1\) and is denoted as \(\chi _{n-1}^2\). The plot of chi-square distribution for various degree of freedoms is shown below. It should be noted that chi-square distribution are right-skewed and the value of chi-square statistic is always positive. Due to asymmetric nature of the chi-square distribution, the upper and lower conficence bound is computed by two different quantities. To find the upper and lower bounds for a \(100(1 - \alpha) \%\) confidence interval for a variance, the values that cut off areas of \(\frac{\alpha}{2}\) in the right and left tails of the chi-square probability density curve are used. The calculation of confidence interval is shown by an example below.
from scipy.stats import chi2
import matplotlib.pyplot as plt
fig = plt.figure(figsize=(10,8))
ax = fig.add_subplot(111)
df=1
x = np.linspace(chi2.ppf(0.01, df), chi2.ppf(0.99, df), 1000)
ax.plot(x, chi2.pdf(x, df), 'r-', lw=2, alpha=0.6, label='DF: '+str(df))
df=5
x = np.linspace(chi2.ppf(0.01, df), chi2.ppf(0.99, df), 1000)
ax.plot(x, chi2.pdf(x, df), 'b-', lw=2, alpha=0.6, label='DF: '+str(df))
df=10
x = np.linspace(chi2.ppf(0.01, df), chi2.ppf(0.99, df), 1000)
ax.plot(x, chi2.pdf(x, df), 'g-', lw=2, alpha=0.6, label='DF: '+str(df))
df=20
x = np.linspace(chi2.ppf(0.01, df), chi2.ppf(0.99, df), 1000)
ax.plot(x, chi2.pdf(x, df), 'y-', lw=2, alpha=0.6, label='DF: '+str(df))
ax.set_ylim(0, 0.30)
ax.set_xlim(0, 50)
ax.set_xlabel('X')
ax.set_ylabel('PDF')
ax.set_title("PDF Plot (chi-squared distribution)")
ax.legend()
ax.grid()
plt.show()
Example: A simple random sample of 15 pistons is selected from a large population whose diameters are known to be normally distributed. The sample standard deviation of the piston diameters is s = 2.0 mm. Find a 95% confidence for the population variance \(\sigma^2\) and population standard deviation \(\sigma\).
Sol: The degree of freedom of chi-square distribution is \(n-1 = 14\). From chi-square table, the upper and lower cut off points can be found as \(\chi _{14, .975} = 5.63\) and \(\chi _{14, .025} = 26.12\), i.e.
\[5.63 < \frac{(n-1)s^2}{\sigma^2} < 26.12\]Solving this we get, \(2.144 < \sigma^2 < 9.948\). The confidence interval for the population standard deviation \(\sigma\) can be computed by taking the square root: \(1.464 < \sigma < 3.154\).
The calculation of confidence interval based on chi-square distribution requires that the population is normal. If the population distribution deviates from normal, the calculated confidence interval can be misleading.
#### Prediction Intervals :A prediction interval is the interval that is likely to contain the value of an item sampled from a population at a future time. Let \(\overline{X}\) be the sample mean of \(n\) samples from a normally distributed population. It will have a normal distribution which can be given by \(\overline{X} \sim N(\mu, \sigma^2/n)\). Suppose, at a future point, a value \(Y\) is observed from the population. The distribution of \(Y\) will be same as the population distribution, \(Y \sim N(\mu, \sigma^2)\). Hence, the difference \(Y - \overline{X}\) will be normally distributed with mean 0 and variance \(\sigma^2(1 + 1/n)\). i.e.
\[\frac{Y - \overline{X}}{\sigma\sqrt{1 + 1/n}} \sim N(0, 1)\]The population standard deviation \(\sigma\) can be estimated as the sample standard deviation \(s\) and hence the prediction interval can be given as:
\[\overline{X} \pm z _{\alpha/2} \times s\sqrt{1 + \frac{1}{n}}\]For a smaller sample size, Student’s t distribution is used and the prediction interval can be given as:
\[\overline{X} \pm t _{n-1, \alpha/2} \times s\sqrt{1 + \frac{1}{n}}\]One-sided prediction interval can be found similarly:
\[Lower \ bound = \overline{X} - t _{n-1, \alpha} \times s\sqrt{1 + \frac{1}{n}}\]\[Upper \ bound = \overline{X} + t _{n-1, \alpha} \times s\sqrt{1 + \frac{1}{n}}\]#### Reference :https://www.mheducation.com/highered/product/statistics-engineers-scientists-navidi/M0073401331.html