Stats Exam 2

Probability Theory

The mathematical study of chance that describes the randomness of a random phenomenon.

Random Phenomenon

A situation where individual outcomes are uncertain, but there is a regular pattern of outcomes in a very large number of repetitions.

Probability of an Event $A$ ($P(A)$)

The proportion of times the event would occur in a very long series of repetitions (in the long run).

Sample Space ($S$)

The set of all possible outcomes of a random phenomenon or process.

Event

A subset of a sample space made up of one or more outcomes.

Probability Rule 1 (Range of Values)

The probability of any event must satisfy $0 \le P(\text{event}) \le 1$, where $0$ indicates the event never occurs and $1$ indicates it occurs every time.

Probability Rule 2 (Sum of Outcomes)

The sum of the probabilities of all possible individual outcomes in the sample space must equal $1$ ($P(S) = 1$).

Disjoint (Mutually Exclusive) Events & Addition Rule

Events that share no outcomes and cannot occur together, calculated as: $P(A \text{ or } B) = P(A) + P(B)$

General Addition Rule for Non-Disjoint Events

$$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$$

Complement Rule

The probability that an event does not occur is given by: $P(A^c) = 1 - P(A)$

Discrete Probability Model

A model with a finite or countable sample space where each individual outcome has a distinct known probability.

Continuous Probability Model

A model where the sample space contains an infinite number of possible outcomes within an interval, and probabilities are measured as areas under a density curve.

Probability of an Exact Single Value in a Continuous Distribution

Always equal to $0$ (e.g., $P(X = 10.5) = 0$).

Total Area Under Any Density Curve

Always exactly equal to $1$ (or $100\%$).

Uniform Distribution Density Formula

For a flat continuous distribution: $\text{Height} = \frac{1}{\text{max} - \text{min}}$

Normal Curve Properties

A symmetric, single-peaked, bell-shaped density curve completely defined by its mean ($\mu$) and standard deviation ($\sigma$).

Mean ($\mu$) of a Normal Curve

The center line and peak of symmetry of the distribution.

Standard Deviation ($\sigma$) of a Normal Curve

The horizontal distance from the center mean to the inflection points where the curve changes curvature from concave down to concave up.

68–95–99.7% Rule (Empirical Rule)

In a normal distribution:

• Approximately $68\%$ of observations fall within $\mu \pm 1\sigma$
• Approximately $95\%$ fall within $\mu \pm 2\sigma$
• Approximately $99.7\%$ fall within $\mu \pm 3\sigma$

$z$-Score Formula for an Individual Observation

$z = \frac{x - \mu}{\sigma}$, measuring the number of standard deviations $x$ lies above or below the mean $\mu$.

Standard Normal Distribution Parameters and Notation

$N(0, 1)$, with mean $\mu = 0$ and standard deviation $\sigma = 1$.

Finding Area to the Right of $z$

$P(Z > z) = 1 - P(Z < z)$, or by symmetry, $P(Z > z) = P(Z < -z)$

Finding Area Between Two $z$-Scores ($z_1$ and $z_2$)

$$P(z_1 \le Z \le z_2) = P(Z \le z_2) - P(Z \le z_1)$$

Backward Normal Calculation Formula for an Individual Observation

$$x = \mu + z\sigma$$

Parameter vs. Statistic

A parameter is a fixed, typically unknown number describing a population. A statistic is a number calculated from sample data used to estimate a parameter.

Notation for Sample vs. Population (Mean and Standard Deviation)

• Mean: Sample is $\bar{x}$; Population is $\mu$
• Standard Deviation: Sample is $s$; Population is $\sigma$

Law of Large Numbers

As sample size ($n$) increases, the sample mean ($\bar{x}$) gets closer and closer to the true population mean ($\mu$).

Sampling Distribution of $\bar{x}$

The distribution of values taken by the sample mean across all possible samples of the same size $n$ from the same population.

Mean and Standard Deviation of the Sampling Distribution of $\bar{x}$

• Mean: $\mu_{\bar{x}} = \mu$
• Standard Deviation: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

Central Limit Theorem (CLT)

For large sample sizes ($n \ge 30$), the sampling distribution of $\bar{x}$ is approximately normal, regardless of the population distribution shape.

$z$-Score Formula for a Sample Mean ($\bar{x}$)

$$z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$$

Backward Calculation Formula for Sample Mean ($\bar{x}$)

$$\bar{x} = \mu + z\left(\frac{\sigma}{\sqrt{n}}\right)$$

Point Estimate vs. Interval Estimate

A point estimate is a single number estimating a parameter (e.g., $\bar{x}$); an interval estimate provides a range of plausible values formulated as $\text{Estimate} \pm \text{Margin of Error}$.

One-Sample $z$ Confidence Interval Formula for $\mu$

$$\bar{x} \pm z^*\left(\frac{\sigma}{\sqrt{n}}\right)$$

Conditions for a One-Sample $z$ Confidence Interval for $\mu$

1. Simple Random Sample (SRS)
2. Normal population or large sample size ($n \ge 30$)
3. Known population standard deviation ($\sigma$)

Common Critical Values ($z^*$)

• $90\%$ confidence: $z^* = 1.645$
• $95\%$ confidence: $z^* = 1.960$
• $98\%$ confidence: $z^* = 2.326$
• $99\%$ confidence: $z^* = 2.576$

Correct Interpretation of a $95\%$ Confidence Interval

We are $95\%$ confident that the true population mean ($\mu$) falls within the calculated interval boundaries.

Effect of Increasing Confidence Level ($C$)

$z^*$ increases, which increases the margin of error, making the interval wider and reducing precision.

Effect of Increasing Sample Size ($n$) on Confidence Intervals

$\frac{\sigma}{\sqrt{n}}$ decreases, which decreases the margin of error, making the interval narrower and increasing precision.

Sample Size Determination Formula for Estimating $\mu$

$n = \left(\frac{z^* \sigma}{m}\right)^2$, where $m$ is the desired margin of error (always rounded up to the next whole integer).

Goal of Hypothesis Testing (Significance Testing)

To assess the strength of evidence provided by sample data against a null hypothesis regarding a population parameter.

Null Hypothesis ($H_0$)

The claim of 'no difference' or 'no effect' being tested, always written with an equality sign: $H_0: \mu = \mu_0$

Alternative Hypothesis ($H_a$)

The claim about the population parameter that researchers seek evidence in favor of, formulated using $>$, $<$, or $\neq$.

One-Sided vs. Two-Sided Hypotheses

• One-sided: Directs interest toward one direction ($H_a: \mu > \mu_0$ or $H_a: \mu < \mu_0$)
• Two-sided: Directs interest toward any difference ($H_a: \mu \neq \mu_0$)

Test Statistic for a One-Sample $z$-Test of $\mu$

$z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}}$, which measures the distance of $\bar{x}$ from $\mu_0$ in standard error units.

$P$-Value Definition

The probability, calculated under the assumption that $H_0$ is true, of observing a test statistic as extreme as or more extreme than the calculated value.

Decision Rule for Significance Tests

• Reject $H_0$ if $P\text{-value} \le \alpha$
• Fail to reject $H_0$ if $P\text{-value} > \alpha$

$P$-Value Calculation Formulas by Test Direction

• Right-tailed ($H_a: \mu > \mu_0$): $P(Z \ge z)$
• Left-tailed ($H_a: \mu < \mu_0$): $P(Z \le z)$
• Two-tailed ($H_a: \mu \neq \mu_0$): $2 \times P(Z \ge |z|)$

Statistically Significant at Level $\alpha$

An outcome so unlikely under $H_0$ that the resulting $P$-value is less than or equal to the significance level $\alpha$

Practice: Uniform Distribution Calculation

A concrete cure time $X$ is uniformly distributed between $45$ and $75$ minutes. Find the height of the curve, and calculate the probability that a specific batch takes between $50$ and $62$ minutes to cure: $P(50 \le X \le 62)$.

1. Height of the curve:
   $\text{Height} = \frac{1}{\text{max} - \text{min}} = \frac{1}{75 - 45} = \frac{1}{30}$
2. Width of the interval:
   $\text{Width} = 62 - 50 = 12$
3. Probability:
   $\text{Area} = \text{Height} \times \text{Width} = \left(\frac{1}{30}\right) \times 12 = \frac{12}{30} = 0.40$
   Therefore, $P(50 \le X \le 62) = 0.40$ (or $40\%$).

Practice: General Addition Rule for Non-Disjoint Events

At a facility, $35\%$ of workers wear hard hats with visors (Event $A$), $40\%$ wear high-visibility vests (Event $B$), and $15\%$ wear both. What is the probability that a randomly chosen worker wears a visor or a vest: $P(A \text{ or } B)$?

1. Identify event relationship:
   The events are not disjoint because workers can wear both ($P(A \text{ and } B) = 0.15$).
2. Apply General Addition Rule:
   $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$
   $P(A \text{ or } B) = 0.35 + 0.40 - 0.15 = 0.60$
   Therefore, $P(A \text{ or } B) = 0.60$ (or $60\%$).

Practice: Forward Normal Distribution Calculation

The compressive strength of a batch of concrete blocks is normally distributed with mean $\mu = 3200\,\text{psi}$ and standard deviation $\sigma = 150\,\text{psi}$. What is the probability that a randomly selected block has a compressive strength greater than $3425\,\text{psi}$?

1. Calculate the z-score:
   $z = \frac{x - \mu}{\sigma} = \frac{3425 - 3200}{150} = \frac{225}{150} = 1.50$
2. Find cumulative area:
   From the standard normal table, the area to the left of $z = 1.50$ is $0.9332$.
3. Calculate right-tail probability:
   $P(X > 3425) = 1 - P(Z < 1.50) = 1 - 0.9332 = 0.0668$
   Therefore, the probability is $0.0668$ (or $6.68\%$).

Practice: Backward Normal Calculation (Percentile to Observation)

Steel rebar yield strength follows $N(60\,\text{ksi}, 2.5\,\text{ksi})$. What minimum yield strength represents the top $10\%$ of strongest rebar?

1. Find cumulative percentile:
   The top $10\%$ corresponds to the $90^{\text{th}}$ percentile, meaning $90\%$ ($0.9000$) of the area lies to the left.
2. Find critical z-value:
   Looking up $0.9000$ in the standard normal table yields $z \approx 1.285$.
3. Apply backward calculation formula:
   $x = \mu + z \sigma = 60 + (1.285 \times 2.5) = 60 + 3.21 = 63.21\,\text{ksi}$
   Therefore, the minimum yield strength is $63.21\,\text{ksi}$.

Practice: Sampling Distribution of the Sample Mean ($\bar{x}$)

Soil sample moisture content has mean $\mu = 18\%$ and standard deviation $\sigma = 4.2\%$. If a technician tests a random sample of $n = 36$ soil cores, what is the probability that the average moisture content $\bar{x}$ is less than $17\%$?

1. Determine sampling parameters:
   $\mu<em>{\bar{x}} = \mu = 18\%$$\sigma</em>{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{4.2}{\sqrt{36}} = \frac{4.2}{6} = 0.70\%$
2. Calculate the z-score:
   $z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} = \frac{17 - 18}{0.70} = \frac{-1.00}{0.70} \approx -1.43$
3. Find lower-tail probability:
   $P(\bar{x} < 17) = P(Z < -1.43) = 0.0764$
   Therefore, the probability is $0.0764$ (or $7.64\%$).

Practice: Constructing and Interpreting a Confidence Interval for $\mu$

A safety inspector samples $n = 64$ ladders and finds a sample mean failure load of $\bar{x} = 312\,\text{lbs}$. Assume population standard deviation $\sigma = 24\,\text{lbs}$. Construct and interpret a $95\%$ confidence interval for the true mean failure load $\mu$.

1. Check conditions:
   Simple random sample, large sample ($n = 64 \ge 30$), and known $\sigma = 24\,\text{lbs}$.
2. Identify critical value:
   For $95\%$ confidence, $z^* = 1.960$.
3. Calculate margin of error:
   $m = z^* \left(\frac{\sigma}{\sqrt{n}}\right) = 1.960 \left(\frac{24}{\sqrt{64}}\right) = 1.960 \times 3 = 5.88\,\text{lbs}$
4. Construct interval:
   $\bar{x} \pm m = 312 \pm 5.88 = (306.12\,\text{lbs}, 317.88\,\text{lbs})$
5. Interpretation:
   We are $95\%$ confident that the true population mean ladder failure load $\mu$ falls between $306.12\,\text{lbs}$ and $317.88\,\text{lbs}$.

Practice: Sample Size Determination for Estimating $\mu$

An estimator wants to estimate the true mean hourly wage of framing carpenters within a margin of error $m = \$1.25$ with $99\%$ confidence. Assume $\sigma = \$4.50$. How many carpenters must be sampled?

1. Formula and critical value:
   For $99\%$ confidence, critical value is $z^* = 2.576$.
   $n = \left(\frac{z^* \sigma}{m}\right)^2$
2. Calculate sample size:
   $n = \left(\frac{2.576 \times 4.50}{1.25}\right)^2 = \left(\frac{11.592}{1.25}\right)^2 = (9.2736)^2 \approx 85.999$
3. Rounding rule:
   Always round up to the next integer:
   $n = 86\text{ carpenters}$

Practice: One-Sided Hypothesis Test for $\mu$

Bolts have a specified mean shear strength of $\mu_0 = 400\,\text{psi}$. A random sample of $n = 49$ bolts yields $\bar{x} = 406\,\text{psi}$. Assume $\sigma = 14\,\text{psi}$. Test at $\alpha = 0.05$ whether the new alloy has higher mean strength than $400\,\text{psi}$.

1. Hypotheses:
   $H<em>0: \mu = 400\,\text{psi}$$H</em>a: \mu > 400\,\text{psi}$
2. Significance level:
   $\alpha = 0.05$
3. Calculate test statistic:
   $z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{406 - 400}{\frac{14}{\sqrt{49}}} = \frac{6}{2} = 3.00$
4. Find P-value:
   $P\text{-value} = P(Z \ge 3.00) = 0.0013$
5. Conclusion:
   Since $P\text{-value} = 0.0013 \le 0.05$, reject $H_0$. There is statistically significant evidence at the $\alpha = 0.05$ level that the mean shear strength is greater than $400\,\text{psi}$.

Practice: Two-Sided Hypothesis Test for $\mu$

Drywall boards have a specified target thickness of $\mu_0 = 0.500\,\text{inches}$ with $\sigma = 0.020\,\text{inches}$. A sample of $n = 36$ boards shows $\bar{x} = 0.493\,\text{inches}$. Test at $\alpha = 0.05$ whether the mean thickness differs from $0.500\,\text{inches}$.

1. Hypotheses:
   $H<em>0: \mu = 0.500\,\text{inches}$$H</em>a: \mu \neq 0.500\,\text{inches}$
2. Significance level:
   $\alpha = 0.05$
3. Calculate test statistic:
   $z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} = \frac{0.493 - 0.500}{\frac{0.020}{\sqrt{36}}} = \frac{-0.007}{0.00333} = -2.10$
4. Find P-value:
   Because the test is two-sided, double the tail probability:
   $P(Z \le -2.10) = 0.0179$
   $P\text{-value} = 2 \times 0.0179 = 0.0358$
5. Conclusion:
   Since $P\text{-value} = 0.0358 \le 0.05$, reject $H_0$. There is statistically significant evidence at the $\alpha = 0.05$ level that the mean thickness differs from $0.500\,\text{inches}$.