| Distribution | +PMF | +Mean \(E(X)\) | +Variance \(\text{Var}(X)\) | +
|---|---|---|---|
| Bernoulli(p) | +\(P(X=1) = p, P(X=0) = 1-p\) | +\(p\) | +\(p(1-p)\) | +
| Binomial(n, p) | +\(P(X=k) = \binom{n}{k} p^k +(1-p)^{n-k}\) | +\(np\) | +\(np(1-p)\) | +
| Geometric(p) | +\(P(X=k) = (1-p)^{k-1} p\) | +\(\frac{1}{p}\) | +\(\frac{1-p}{p^2}\) | +
| Poisson(λ) | +\(P(X=k) = \frac{\lambda^k +e^{-\lambda}}{k!}\) | +\(\lambda\) | +\(\lambda\) | +
| Distribution | +Mean \(E(X)\) | +Variance \(\text{Var}(X)\) | +|
|---|---|---|---|
| Uniform(a, b) | +\(f(x) = \frac{1}{b-a}\) | +\(\frac{a+b}{2}\) | +\(\frac{(b-a)^2}{12}\) | +
| Normal(μ, σ²) | +\(f(x) = \frac{1}{\sigma \sqrt{2\pi}} +e^{-\frac{(x-\mu)^2}{2\sigma^2}}\) | +\(\mu\) | +\(\sigma^2\) | +
| Exponential(λ) | +\(f(x) = \lambda e^{-\lambda x}, x \geq +0\) | +\(\frac{1}{\lambda}\) | +\(\frac{1}{\lambda^2}\) | +
Problem: A test for a disease is 99% accurate when +the disease is present and has a 1% false positive rate. If the disease +occurs in 2% of the population, what is the probability a person has the +disease given they test positive?
+Solution:
+- Let \(D\) = Disease, \(D^c\) = No Disease, \(T^+\) = Positive Test.
+\[
+P(D | T^+) = \frac{P(T^+ | D)P(D)}{P(T^+)}
+\]
+Where \(P(T^+) = P(T^+ | D)P(D) + P(T^+ |
+D^c)P(D^c)\):
+\[
+P(T^+) = (0.99)(0.02) + (0.01)(0.98) = 0.0198 + 0.0098 = 0.0296
+\] \[
+P(D | T^+) = \frac{(0.99)(0.02)}{0.0296} = \frac{0.0198}{0.0296} \approx
+0.668
+\] Answer: 66.8% chance they have the disease
+given a positive test.
Problem: A company’s fleet of cars has NOx emissions
+\(N(0.03, 0.02^2)\). If a sample of 25
+cars is taken:
+1. What is the standard error of the sample mean?
+2. What is the probability that the sample mean exceeds 0.045?
Solution:
+1. Standard Error:
+\[
+\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{0.02}{\sqrt{25}} = 0.004
+\]
Answer:
+- Standard error = 0.004
+- Probability = 0.00009 (very unlikely).
Problem: A light bulb has a lifetime with \(\sigma = 40\). A sample of 30 bulbs has an
+average lifetime of 780 hours.
+1. Construct a 96% confidence interval.
+2. What sample size is needed to ensure the margin of error is 9
+hours?
Solution:
+1. Use the formula \(\bar{x} \pm z^*
+\frac{\sigma}{\sqrt{n}}\):
+- \(z^*\) for 96%: \(z^* = 2.05\) (from Z-table).
+- Standard error:
+\[
+ \text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{40}{\sqrt{30}} \approx
+7.30
+ \]
+- CI:
+\[
+ 780 \pm 2.05 \cdot 7.30 \Rightarrow (765, 795)
+ \]
Answer:
+1. CI: (765, 795)
+2. Required sample size: 84
Problem: A plant reports average commute time is +32.6 minutes. A sample of 60 workers has \(\bar{x} = 34.5\), \(\sigma = 6.1\). Test at \(\alpha = 0.05\) whether the true mean +exceeds 32.6.
+Solution:
+1. Hypotheses:
+- \(H_0: \mu = 32.6\), \(H_a: \mu > 32.6\)
Test statistic:
+\[
+Z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{34.5 -
+32.6}{\frac{6.1}{\sqrt{60}}} = \frac{1.9}{0.787} \approx 2.41
+\]
Find p-value: From Z-table, \(P(Z > +2.41) \approx 0.0080\).
Conclusion: \(p = 0.008 < +0.05\), so reject \(H_0\).
Answer: There is sufficient evidence to conclude the +true mean exceeds 32.6 minutes.
+Problem: A researcher measures the performance of +students under 3 teaching methods (A, B, C). The mean performance scores +and ANOVA results are summarized below:
+| Source | +SS | +df | +MS | +F | +p-value | +
|---|---|---|---|---|---|
| Between | +45.2 | +2 | +22.6 | +8.3 | +0.002 | +
| Within | +54.6 | +27 | +2.02 | ++ | + |
| Total | +99.8 | +29 | ++ | + | + |
Solution:
+1. Hypotheses:
+- \(H_0: \mu_A = \mu_B = \mu_C\) (all
+group means are equal).
+- \(H_a\): At least one mean
+differs.
Conclusion: There is significant evidence that at +least one teaching method’s mean performance differs from the +others.
+Problem: A sample of 30 bulbs has \(\bar{x} = 780\), \(\sigma = 40\). Construct a 95% confidence +interval for the true mean.
+Solution:
+- \(n = 30\), \(\sigma = 40\), \(\bar{x} = 780\), \(z^* = 1.96\) (for 95%).
+- Standard Error:
+\[
+ \text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{40}{\sqrt{30}} \approx
+7.30
+ \]
+- CI:
+\[
+ 780 \pm 1.96 \cdot 7.30
+ \]
+\[
+ 780 \pm 14.31 \Rightarrow (765.69, 794.31)
+ \]
Answer: \((765.7, +794.3)\).
+Problem: A company claims their machines fill +bottles with \(\mu = 500\) ml. A sample +of 36 bottles has \(\bar{x} = 495\) and +\(\sigma = 10\). Test \(H_0: \mu = 500\) vs \(H_a: \mu < 500\) at \(\alpha = 0.05\).
+Solution:
+1. Hypotheses:
+- \(H_0: \mu = 500\), \(H_a: \mu < 500\).
Test statistic:
+\[
+Z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{495 -
+500}{\frac{10}{\sqrt{36}}} = \frac{-5}{1.67} \approx -3.00
+\]
Find p-value: From Z-table, \(P(Z < +-3.00) \approx 0.0013\).
Compare to \(\alpha = +0.05\):
+Conclusion: There is significant evidence that the +mean bottle fill is less than 500 ml.
+Problem: Given the following data:
+| \(x\) | +\(y\) | +
|---|---|
| 1 | +2 | +
| 2 | +4 | +
| 3 | +5 | +
| 4 | +4 | +
| 5 | +5 | +
Solution:
+1. First, compute means:
+\[
+ \bar{x} = \frac{1+2+3+4+5}{5} = 3, \quad \bar{y} =
+\frac{2+4+5+4+5}{5} = 4
+ \]
+Compute \(b\):
+\[
+ b = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i -
+\bar{x})^2} = \frac{(1-3)(2-4) + (2-3)(4-4) + \dots}{(1-3)^2 + (2-3)^2 +
+\dots} = 0.6
+ \]
Answer: Regression line: \(\hat{y} = 2.2 + 0.6x\).
+The goal of linear regression is to model the +relationship between a predictor variable \(x\) (independent variable) and a response +variable \(y\) (dependent +variable).
+The least squares regression line minimizes the sum
+of the squared residuals (vertical distances between observed and
+predicted values).
+The line is given by:
+\[
+\hat{y} = a + bx
+\]
+Where:
+- \(b\) is the
+slope:
+\[
+ b = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i -
+\bar{x})^2}
+ \]
+- \(a\) is the
+y-intercept:
+\[
+ a = \bar{y} - b\bar{x}
+ \]
Example Calculation:
+Suppose you have the following data:
+| \(x\) | \(y\) | |———|———| | 1 | 2 | | 2 | 3 | | 3 | 5
+|
Compute means:
+\[
+\bar{x} = \frac{1 + 2 + 3}{3} = 2, \quad \bar{y} = \frac{2 + 3 + 5}{3} =
+3.33
+\]
Compute \(b\):
+\[
+b = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}
+\]
Compute \(a\):
+\[
+a = \bar{y} - b\bar{x} = 3.33 - (1.33)(2) = 0.67
+\]
Final Regression Line:
+\[
+\hat{y} = 0.67 + 1.33x
+\]
Example: From the regression line \(\hat{y} = 0.67 + 1.33x\):
+- Slope \(1.33\): For each 1-unit
+increase in \(x\), \(y\) increases by 1.33.
+- Intercept \(0.67\): When \(x = 0\), \(y\) is predicted to be 0.67.
To test the significance of the regression slope (\(\beta\)):
+ANOVA compares the means of three or more groups to +determine if at least one group mean is significantly different.
+The F-statistic compares the variation between
+groups to the variation within groups:
+\[
+F = \frac{\text{MS}_{\text{between}}}{\text{MS}_{\text{within}}}
+\]
Where:
+- Mean Square Between (MSB): Measures variation between
+group means.
+\[
+ \text{MSB} = \frac{\text{SS}_{\text{between}}}{k-1}
+ \]
+- Mean Square Within (MSW): Measures variation within
+groups.
+\[
+ \text{MSW} = \frac{\text{SS}_{\text{within}}}{n-k}
+ \]
+- Degrees of Freedom:
+- Between groups: \(df_{\text{between}} = k -
+1\)
+- Within groups: \(df_{\text{within}} = n -
+k\)
+- Total: \(df_{\text{total}} = n -
+1\)
If the ANOVA test rejects \(H_0\), +post-hoc tests (e.g., Tukey’s HSD) identify which pairs +of group means differ significantly.
+Problem: Three fertilizers are tested for crop yield +(in bushels per acre). The data is summarized as follows:
+| Group | +Sample Size (\(n_i\)) | +Mean (\(\bar{x}_i\)) | +Variance (\(s_i^2\)) | +
|---|---|---|---|
| Fertilizer A | +10 | +20 | +4 | +
| Fertilizer B | +10 | +25 | +5 | +
| Fertilizer C | +10 | +30 | +6 | +
Solution:
+1. Hypotheses:
+- \(H_0: \mu_A = \mu_B = \mu_C\)
+- \(H_a:\) At least one mean
+differs.