# Least-Squares Estimates for Simple Linear Regression The model is: \[ Y = \beta_ + \beta_1 x + \epsilon \] ## Given Data - \(\sum x = 20\) - \(\sum Y = 50\) - \(\sum x^2 = 120\) - \(\sum xY = 180\) - \(n = 5\) ## Formulas The least-squares estimates for \(\beta_1\) and \(\beta_\) are: \[ \hat{\beta}_1 = \frac{\sum xY - \frac{1}{n}(\sum x)(\sum Y)}{\sum x^2 - \frac{1}{n}(\sum x)^2} \] \[ \hat{\beta}_ = \bar{Y} - \hat{\beta}_1 \bar{x} \] where: - \(\bar{x} = \frac{\sum x}{n}\) - \(\bar{Y} = \frac{\sum Y}{n}\) ## Step 1: Calculate Means \[ \bar{x} = \frac{20}{5} = 4 \] \[ \bar{Y} = \frac{50}{5} = 10 \] ## Step 2: Calculate \(\hat{\beta}_1\) \[ \hat{\beta}_1 = \frac{180 - \frac{1}{5}(20)(50)}{120 - \frac{1}{5}(20^2)} \] Calculate each term: - \(\frac{1}{5}(20)(50) = 4 \times 50 = 200\) - \(\frac{1}{5}(20^2) = 4 \times 20 = 80\) So: \[ \hat{\beta}_1 = \frac{180 - 200}{120 - 80} = \frac{-20}{40} = -.5 \] ## Step 3: Calculate \(\hat{\beta}_\) \[ \hat{\beta}_ = 10 - (-.5)(4) = 10 + 2 = 12 \] ## Final Answers - \(\boxed{\hat{\beta}_ = 12}\) - \(\boxed{\hat{\beta}_1 = -.5}\)

# Least-Squares Estimates for Simple Linear Regression The linear regression model is defined as: \[ Y = \beta_0 + \beta_1 x + \epsilon \] ## Given Data - \(\sum x = 20\) - \(\sum Y = 50\) - \(\sum x^2 = 120\) - \(\sum xY = 180\) - \(n = 5\) ## Formulas for Estimates The least-squares estimates for \(\beta_0\) and \(\beta_1\) are calculated using the following formulas: \[ \hat{\beta}_1 = \frac{\sum xY - \frac{1}{n}(\sum x)(\sum Y)}{\sum x^2 - \frac{1}{n}(\sum x)^2} \] \[ \hat{\beta}_0 = \bar{Y} - \hat{\beta}_1 \bar{x} \] Where: - \(\bar{x} = \frac{\sum x}{n}\) - \(\bar{Y} = \frac{\sum Y}{n}\) ## Step 1: Calculate Means Calculating the means of \(x\) and \(Y\): \[ \bar{x} = \frac{20}{5} = 4 \] \[ \bar{Y} = \frac{50}{5} = 10 \] ## Step 2: Calculate \(\hat{\beta}_1\) Substituting the values into the formula for \(\hat{\beta}_1\): \[ \hat{\beta}_1 = \frac{180 - \frac{1}{5}(20)(50)}{120 - \frac{1}{5}(20^2)} \] ### Calculate Each Term 1. For \(\frac{1}{5}(20)(50)\): \[ = \frac{1000}{5} = 200 \] 2. For \(\frac{1}{5}(20^2)\): \[ = \frac{400}{5} = 80 \] ### Final Calculation of \(\hat{\beta}_1\) \[ \hat{\beta}_1 = \frac{180 - 200}{120 - 80} = \frac{-20}{40} = -0.5 \] ## Step 3: Calculate \(\hat{\beta}_0\) Using the value of \(\hat{\beta}_1\) to find \(\hat{\beta}_0\): \[ \hat{\beta}_0 = 10 - (-0.5)(4) = 10 + 2 = 12 \] ## Final Answers - \(\hat{\beta}_0 = 12\) - \(\hat{\beta}_1 = -0.5\) \(\boxed{\hat{\beta}_0 = 12}\) \(\boxed{\hat{\beta}_1 = -0.5}\)

# Dice Game: Maximums and Wagers ## Overview In this game, two players roll three fair six-sided dice each, recording the maximum of their rolls. The player with the higher maximum wins, and the loser pays an amount equal to that higher maximum. If both maxima are equal, no money is exchanged. ### Variables Defined - \( X \): Maximum of three dice for a player. - \( Y \): Wager paid (0 if tie, otherwise the higher maximum). ## Distribution of \( X \) ### Cumulative Distribution Function (CDF) For independent dice, the cumulative distribution function for the maximum is given by: \[ P(X \leq k) = P(\text{all dice} \leq k) \] Since each die has: \[ P(\text{die} \leq k) = \frac{k}{6} \] For three dice: \[ P(X \leq k) = \left(\frac{k}{6}\right)^3 \] ### Probability Mass Function (PMF) The PMF is calculated as: \[ P(X = k) = P(X \leq k) - P(X \leq k - 1) \] Thus, \[ P(X = k) = \left(\frac{k}{6}\right)^3 - \left(\frac{k-1}{6}\right)^3 \] ### Simplified PMF Calculation Calculating the PMF yields: \[ P(X = k) = \frac{3k^2 - 3k + 1}{216} \] ### Values of \( k \) | \( k \) | \( P(X = k) \) | |---------|---------------------| | 1 | \( \frac{1}{216} \) | | 2 | \( \frac{7}{216} \) | | 3 | \( \frac{19}{216} \)| | 4 | \( \frac{37}{216} \)| | 5 | \( \frac{61}{216} \)| | 6 | \( \frac{91}{216} \)| ## Inversion Method for Generating \( X \) To simulate \( X \) from \( U(0,1) \): 1. Let \( U \sim U(0,1) \). 2. Compute \( k = \lceil 6U^{1/3} \rceil \). 3. Return \( X = k \). ## Distribution of \( Y \) ### Support for \( Y \) The support for \( Y \) is \( \{0, 1, 2, 3, 4, 5, 6\} \). ### Tie Probability \[ P(Y = 0) = \sum_{k=1}^{6} P(X = k)^2 = \frac{2297}{7776} \] ### Wager Probability for \( k \geq 2 \) For \( k \geq 2 \): \[ P(Y = k) = 2 P(X = k) P(X \leq k - 1) \] ### Numerical Values for \( Y \) | \( y \) | \( P(Y = y) \) | |---------|-----------------------| | 0 | \( \frac{2297}{7776} \) | | 1 | 0 | | 2 | \( \frac{7}{23328} \) | | 3 | \( \frac{19}{2916} \) | | 4 | \( \frac{37}{864} \) | | 5 | \( \frac{122}{729} \) | | 6 | \( \frac{11375}{23328} \)| ## Inversion Method for Generating \( Y \) To simulate \( Y \): 1. Let \( U \sim U(0,1) \). 2. Compute cumulative probabilities \( G(y) \): - \( G(0) = P(Y = 0) \) - \( G(2) = G(0) + P(Y = 2) \) - \( G(3) = G(2) + P(Y = 3) \) - Continue until \( G(6) \). 3. Return the smallest \( y \) such that \( G(y) \geq U \). ## Summary of Results ### Maximum \( X \) - Support: \( \{1, 2, 3, 4, 5, 6\} \) - PMF: \[ P(X = k) = \frac{3k^2 - 3k + 1}{216} \] - Generation Formula: \[ X = \lceil 6 U^{1/3} \rceil \] ### Wager \( Y \) - Support: \( \{0, 1, 2, 3, 4, 5, 6\} \) - Tie Probability: \[ P(Y = 0) = \frac{2297}{7776} \] - Wager Probability: \[ P(Y = k) = 2 P(X = k) P(X \leq k - 1), \quad k \geq 2 \] - Generation Method: Compare \( U \) with cumulative probabilities of \( Y \).

# Simple Linear Regression Analysis ## Model Definition The simple linear regression model is expressed as: \[ Y = \beta_0 + \beta_1 x + \epsilon \] ## Given Data - Number of observations \( n = 5 \) - \( \sum x = 20 \) - \( \sum Y = 50 \) - \( \sum x^2 = 120 \) - \( \sum xY = 180 \) ## Definitions The least-squares estimates minimize the sum of squared errors. The formulas used for estimating \(\beta_0\) and \(\beta_1\) are: \[ \hat{\beta}_1 = \frac{n \sum xY - (\sum x)(\sum Y)}{n \sum x^2 - (\sum x)^2} \] \[ \hat{\beta}_0 = \bar{Y} - \hat{\beta}_1 \bar{x} \] Where: - \( \bar{x} = \frac{\sum x}{n} \) - \( \bar{Y} = \frac{\sum Y}{n} \) ## Calculation Steps ### Step 1: Calculate Sample Means 1. Calculate \( \bar{x} \): \[ \bar{x} = \frac{20}{5} = 4 \] 2. Calculate \( \bar{Y} \): \[ \bar{Y} = \frac{50}{5} = 10 \] ### Step 2: Compute the Slope Estimate (\(\hat{\beta}_1\)) #### Numerator Calculation: \[ n \sum xY - (\sum x)(\sum Y) = 5(180) - 20(50) = 900 - 1000 = -100 \] #### Denominator Calculation: \[ n \sum x^2 - (\sum x)^2 = 5(120) - 20^2 = 600 - 400 = 200 \] #### Slope Estimate: \[ \hat{\beta}_1 = \frac{-100}{200} = -0.5 \] ### Step 3: Compute the Intercept (\(\hat{\beta}_0\)) \[ \hat{\beta}_0 = \bar{Y} - \hat{\beta}_1 \bar{x} = 10 - (-0.5)(4) = 10 + 2 = 12 \] ## Final Answers Summary - **Least-Squares Estimates:** - \(\hat{\beta}_0 = 12\) - \(\hat{\beta}_1 = -0.5\) - **Estimated Regression Line:** \[ \hat{Y} = 12 - 0.5x \] ### Interpretation This regression line indicates that for each one-unit increase in \( x \), the predicted value of \( Y \) decreases by 0.5 units, starting from an intercept of 12 when \( x = 0 \).

# Cam Follower Analysis ## Given Information - Lift of follower: \( h = 30 \, \text{mm} = 0.03 \, \text{m} \) - Motion: Simple Harmonic Motion (SHM) during rise - Angle of rise: \( \beta = 180^\circ = \pi \, \text{rad} \) - Cam speed: \( N = 600 \, \text{rpm} \) ## Objectives - Find the **maximum follower velocity**. - Find the **maximum follower acceleration**. - Explain **why high acceleration causes wear**. ## Definitions and Concepts Used For SHM cam motion during rise through angle \( \beta \): ### Displacement \[ y = \frac{h}{2} \left(1 - \cos\frac{\pi \theta}{\beta}\right) \] ### Velocity \[ v = \frac{dy}{dt} = \frac{h \pi \omega}{2 \beta} \sin\frac{\pi \theta}{\beta} \] ### Maximum Velocity Maximum velocity occurs when \( \sin(\cdot) = 1 \): \[ v_{\max} = \frac{h \pi \omega}{2 \beta} \] ### Acceleration \[ a = \frac{h \pi^2 \omega^2}{2 \beta^2} \cos\frac{\pi \theta}{\beta} \] ### Maximum Acceleration Maximum acceleration occurs when \( \cos(\cdot) = 1 \): \[ a_{\max} = \frac{h \pi^2 \omega^2}{2 \beta^2} \] ## Step-by-Step Solution ### Step 1: Convert Cam Speed to Angular Speed 1. Convert \( N \) to revolutions per second (rps): \[ N = 600 \, \text{rpm} = 10 \, \text{rps} \] 2. Calculate angular speed \( \omega \): \[ \omega = 2 \pi N = 2 \pi \times 10 = 20\pi \, \text{rad/s} \] ### Step 2: Calculate Maximum Velocity Using the formula for maximum velocity: \[ v_{\max} = \frac{h \pi \omega}{2 \beta} \] Substituting the known values: \[ v_{\max} = \frac{0.03 \times \pi \times 20\pi}{2\pi} = \frac{0.03 \times 20\pi^2}{2\pi} = 0.03 \times 10\pi = 0.3\pi \, \text{m/s} \] Calculating numerical value: \[ v_{\max} \approx 0.942 \, \text{m/s} \; (\approx 942 \, \text{mm/s}) \] ### Step 3: Calculate Maximum Acceleration Using the formula for maximum acceleration: \[ a_{\max} = \frac{h \pi^2 \omega^2}{2 \beta^2} \] Substituting the known values: \[ a_{\max} = \frac{0.03 \times \pi^2 \times (20\pi)^2}{2\pi^2} = \frac{0.03 \times \pi^2 \times 400\pi^2}{2\pi^2} = 0.015 \times 400\pi^2 = 6\pi^2 \, \text{m/s}^2 \] Calculating numerical value: \[ a_{\max} \approx 59.2 \, \text{m/s}^2 \] ### Step 4: Explanation of Wear due to High Acceleration High acceleration produces large inertia forces, which increases: - Contact pressure - Friction - Vibration - Surface fatigue These factors contribute to wear between the cam and the follower. ## Final Answers Summary - **Maximum follower velocity**: \( 0.942 \, \text{m/s} \) (≈ 942 mm/s) - **Maximum follower acceleration**: \( 59.2 \, \text{m/s}^2 \) (≈ \( 5.92 \times 10^4 \, \text{mm/s}^2 \)) - **Reason for wear**: High acceleration increases inertia forces, contact stress, friction, and vibration.

# Thermal Insulation Analysis ## Given Information - Thermal conductivity of insulation: \( k = 0.05 \, \text{W/m} \cdot \text{K} \) - Heat transfer coefficient: \( h = 10 \, \text{W/m}^2 \cdot \text{K} \) - Geometry: Cylindrical pipe with insulation ## Objectives 1. Find the **critical radius of insulation**. 2. Explain the **significance of the critical radius**. 3. Determine whether **adding insulation always reduces heat loss**. ## Definitions and Concepts Used In a cylindrical body, heat loss to surrounding air occurs through: 1. **Conduction** through insulation (depends on thickness and \( k \)). 2. **Convection** from the outer surface (depends on surface area and \( h \)). When insulation is added to a pipe, two opposing effects occur: - Increasing insulation thickness increases conduction resistance, which reduces heat loss. - Increasing insulation thickness also increases the outer surface area, which increases convection heat loss. At a particular radius, these two effects balance, leading to maximum heat loss. This radius is known as the **critical radius of insulation**. ### Critical Radius Formula For a cylinder, the critical radius \( r_c \) is given by: \[ r_c = \frac{k}{h} \] ## Step-by-Step Solution ### Step 1: Calculate Critical Radius Substituting the given values: \[ r_c = \frac{k}{h} = \frac{0.05}{10} = 0.005 \, \text{m} \] ### Step 2: Convert to Millimeters \[ r_c = 0.005 \, \text{m} = 5 \, \text{mm} \] ## Detailed Significance of Critical Radius When insulation is first added to a bare pipe (small radius): 1. **Outer Surface Area Increases Rapidly**: The outer surface area increases as insulation is added. - The convective heat transfer increases because: \[ Q = hA(T_s - T_\infty) \] Where \( A = 2\pi r L \) increases with radius. 2. **Conduction Resistance Initially Small**: With small thickness, the conduction resistance through insulation is still relatively small. Consequently, the initial effect is: - **Increase in convection effect** > **Increase in conduction resistance**, leading to increased heat loss. As insulation thickness continues to increase: - **Conduction resistance** becomes significant, reducing heat flow through insulation. This effect becomes dominant, and thus: - **Beyond the critical radius**, heat loss decreases. The critical radius serves as the boundary between: - Region where insulation increases heat loss. - Region where insulation decreases heat loss. ## Whether Adding Insulation Always Reduces Heat Loss **No, not always.** - If the outer radius of the insulated pipe is less than the critical radius, adding insulation increases heat loss. - If the outer radius exceeds the critical radius, adding insulation decreases heat loss. For most practical pipes, the pipe radius is greater than the critical radius, meaning insulation typically reduces heat loss. However, for small wires, cables, and thin tubes, insulation may initially increase heat loss. ## Final Answers Summary - **Critical radius of insulation**: \( 0.005 \, \text{m} = 5 \, \text{mm} \) - **Significance**: It is the radius at which heat loss from the cylinder becomes maximum due to the balance between increased convection area and conduction resistance. - **Insulation Effect**: Adding insulation does not always reduce heat loss; it increases heat loss up to the critical radius and reduces it beyond that.

# Ungrouped Data Analysis ## Given Data A group of 8 students scored the following on a short test worth 20 points: - Scores: \( 20, 14, 15, 16, 20, 18, 16, 10 \) ## Objectives 1. Calculate **Arithmetic Mean**. 2. Find the **Median**. 3. Determine the **Mode**. 4. Calculate the **Third Quartile (Q3)**. 5. Find the **Decile (D1)**. 6. Calculate the **Standard Deviation (S)**. ## Step-by-Step Calculations ### Step 1: Arithmetic Mean The arithmetic mean (\( \bar{x} \)) is calculated as follows: \[ \bar{x} = \frac{\sum x}{n} \] Where: - \( \sum x = 20 + 14 + 15 + 16 + 20 + 18 + 16 + 10 = 129 \) - \( n = 8 \) Calculating the mean: \[ \bar{x} = \frac{129}{8} = 16.125 \] ### Step 2: Median To find the median, first order the scores: - Ordered Scores: \( 10, 14, 15, 16, 16, 18, 20, 20 \) Since there are 8 scores (even number), the median is the average of the 4th and 5th scores: \[ \text{Median} = \frac{16 + 16}{2} = 16 \] ### Step 3: Mode The mode is the score that appears most frequently: - Ordered Scores: \( 10, 14, 15, 16, 16, 18, 20, 20 \) The mode is \( 16 \) and \( 20 \) (bimodal). ### Step 4: Third Quartile (Q3) To find \( Q3 \): 1. Identify the position of \( Q3 \): \[ Q3 = \frac{3(n+1)}{4} = \frac{3(8+1)}{4} = \frac{27}{4} = 6.75 \] 2. The 6th and 7th values in the ordered list are \( 18 \) and \( 20 \): \[ Q3 = 18 + 0.75(20 - 18) = 18 + 1.5 = 19.5 \] ### Step 5: First Decile (D1) To find \( D1 \): 1. Identify the position of \( D1 \): \[ D1 = \frac{n+1}{10} = \frac{8+1}{10} = 0.9 \] Since \( D1 \) is 0.9, it rounds to the first score in the ordered list: \[ D1 = 10 \] ### Step 6: Standard Deviation (S) 1. Calculate the variance (\( S^2 \)): \[ S^2 = \frac{\sum (x - \bar{x})^2}{n} \] Calculating each squared difference: \[ \begin{align*} (20 - 16.125)^2 & = 14.265625 \\ (14 - 16.125)^2 & = 4.515625 \\ (15 - 16.125)^2 & = 1.265625 \\ (16 - 16.125)^2 & = 0.015625 \\ (20 - 16.125)^2 & = 14.265625 \\ (18 - 16.125)^2 & = 3.515625 \\ (16 - 16.125)^2 & = 0.015625 \\ (10 - 16.125)^2 & = 37.265625 \\ \end{align*} \] Sum of squared differences: \[ \sum (x - \bar{x})^2 = 14.265625 + 4.515625 + 1.265625 + 0.015625 + 14.265625 + 3.515625 + 0.015625 + 37.265625 = 75.5 \] Calculating variance: \[ S^2 = \frac{75.5}{8} = 9.4375 \] Calculating standard deviation (\( S \)): \[ S = \sqrt{9.4375} \approx 3.07 \] ## Final Answers Summary - **Arithmetic Mean**: \( 16.125 \) - **Median**: \( 16 \) - **Mode**: \( 16 \) and \( 20 \) (bimodal) - **Third Quartile (Q3)**: \( 19.5 \) - **First Decile (D1)**: \( 10 \) - **Standard Deviation (S)**: \( 3.07 \)

# Ungrouped Data Analysis ## Given Data The scores obtained by a group of 8 students on a short test worth 20 points are: - Scores: \( 20, 14, 15, 16, 20, 18, 16, 10 \) ## Objectives 1. Calculate **Arithmetic Mean**. 2. Find the **Median**. 3. Determine the **Mode**. 4. Calculate the **Third Quartile (Q3)**. 5. Compute the **Quartile Deviation (DQ)**. 6. Calculate the **Sample Variance (S²)**. 7. Calculate the **Sample Standard Deviation (S)**. ## Step-by-Step Calculations ### Step 1: Arrange Data in Ascending Order - Ordered Scores: \( 10, 14, 15, 16, 16, 18, 20, 20 \) ### Step 2: Calculate Arithmetic Mean The arithmetic mean (\( \bar{x} \)) is calculated as follows: \[ \bar{x} = \frac{\sum x}{n} \] Where: - \( \sum x = 20 + 14 + 15 + 16 + 20 + 18 + 16 + 10 = 129 \) - \( n = 8 \) Calculating the mean: \[ \bar{x} = \frac{129}{8} = 16.125 \] ### Step 3: Find the Median Since \( n = 8 \) (even), the median is the average of the 4th and 5th values: \[ \text{Median} = \frac{16 + 16}{2} = 16 \] ### Step 4: Determine the Mode The mode is the score that appears most frequently: - The scores \( 16 \) and \( 20 \) appear twice. Thus, the mode is: \[ \text{Mode} = 16 \text{ and } 20 \quad (\text{bimodal}) \] ### Step 5: Calculate Third Quartile (Q3) Using the position formula for \( Q3 \): \[ Q3 \text{ position} = \frac{3(n+1)}{4} = \frac{3(8+1)}{4} = \frac{27}{4} = 6.75 \] To find \( Q3 \): - The 6th value is \( 18 \) - The 7th value is \( 20 \) Interpolating: \[ Q3 = 18 + 0.75(20 - 18) = 18 + 1.5 = 19.5 \] ### Step 6: Calculate Quartile Deviation (DQ) \[ DQ = \frac{Q3 - Q1}{2} \] First, calculate \( Q1 \): \[ Q1 \text{ position} = \frac{n+1}{4} = \frac{8+1}{4} = 2.25 \] Interpolating: - The 2nd value is \( 14 \) - The 3rd value is \( 15 \) \[ Q1 = 14 + 0.25(15 - 14) = 14 + 0.25 = 14.25 \] Calculating \( DQ \): \[ DQ = \frac{19.5 - 14.25}{2} = \frac{5.25}{2} = 2.625 \] ### Step 7: Calculate Sample Variance (S²) Using the shortcut formula: \[ S^2 = \frac{\sum x^2 - \frac{(\sum x)^2}{n}}{n-1} \] Calculating \( \sum x^2 \): \[ \sum x^2 = 20^2 + 14^2 + 15^2 + 16^2 + 20^2 + 18^2 + 16^2 + 10^2 = 400 + 196 + 225 + 256 + 400 + 324 + 256 + 100 = 2157 \] Now, calculating \( S^2 \): \[ S^2 = \frac{2157 - \frac{(129)^2}{8}}{7} = \frac{2157 - \frac{16641}{8}}{7} \] Calculating: \[ \frac{16641}{8} = 2080.125 \] Now substituting: \[ S^2 = \frac{2157 - 2080.125}{7} = \frac{76.875}{7} \approx 10.9821 \] ### Step 8: Calculate Sample Standard Deviation (S) \[ S = \sqrt{S^2} \approx \sqrt{10.9821} \approx 3.31 \] ## Final Answers Summary - **Arithmetic Mean**: \( 16.125 \) - **Median**: \( 16 \) - **Mode**: \( 16 \) and \( 20 \) (bimodal) - **Third Quartile (Q3)**: \( 19.5 \) - **Quartile Deviation (DQ)**: \( 2.625 \) - **Sample Variance (S²)**: \( 10.98 \) - **Sample Standard Deviation (S)**: \( 3.31 \)

# One-Sample t-Test Analysis ## Given Data A sample of 25 students' sleep hours: - Scores: - 4 (1 time) - 5 (2 times) - 6 (8 times) - 7 (8 times) - 8 (6 times) **Significance Level:** - \( \alpha = 0.05 \) ## Hypotheses - **Null Hypothesis (\( H_0 \))**: \( \mu = 7 \) (true mean hours of sleep is 7) - **Alternative Hypothesis (\( H_A \))**: \( \mu < 7 \) (true mean hours of sleep is less than 7) ### Test Type - **Left-tailed Test**. ## Model - Population standard deviation (\( \sigma \)) is unknown, so we will use a one-sample t-test. ### Degrees of Freedom \[ df = n - 1 = 25 - 1 = 24 \] ## Mechanics ### Step 1: Compute Sample Mean 1. **Calculate Total Score**: \[ \sum x = (4)(1) + (5)(2) + (6)(8) + (7)(8) + (8)(6) = 4 + 10 + 48 + 56 + 48 = 166 \] 2. **Calculate Sample Mean**: \[ \bar{x} = \frac{\sum x}{n} = \frac{166}{25} = 6.64 \] ### Step 2: Compute Sample Variance 1. **Calculate \( \sum x^2 \)**: \[ \sum x^2 = (4^2)(1) + (5^2)(2) + (6^2)(8) + (7^2)(8) + (8^2)(6) = 16 + 50 + 288 + 392 + 384 = 1130 \] 2. **Compute Sample Variance**: \[ s^2 = \frac{\sum x^2 - \frac{(\sum x)^2}{n}}{n-1} = \frac{1130 - \frac{166^2}{25}}{24} \] \[ \frac{166^2}{25} = 1102.24 \] Now substitute back: \[ s^2 = \frac{1130 - 1102.24}{24} = \frac{27.76}{24} \approx 1.157 \] 3. **Calculate Sample Standard Deviation**: \[ s = \sqrt{s^2} \approx \sqrt{1.157} \approx 1.076 \] ### Step 3: Calculate Test Statistic \[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{6.64 - 7}{1.076/\sqrt{25}} \] \[ t = \frac{-0.36}{1.076/5} = \frac{-0.36}{0.2152} \approx -1.67 \] ### Step 4: Critical Value For \( \alpha = 0.05 \) (left-tailed) with \( df = 24 \): \[ t_{\text{critical}} = -1.711 \] ## Decision - Since \( t = -1.67 > -1.711 \), we **fail to reject \( H_0 \)**. ## Final Answer - **Test Statistic**: \( t = -1.67 \) - **Critical Value**: \( t_{\text{critical}} = -1.711 \) ### Conclusion There is not enough evidence at the 5% significance level to conclude that the mean sleep time is under 7 hours.

# One-Sample t-Test Summary ## Given Data A sample of 25 students' sleep hours was analyzed. The frequency of scores is as follows: - 4: 1 time - 5: 2 times - 6: 8 times - 7: 8 times - 8: 6 times **Significance Level:** - \( \alpha = 0.05 \) ## Hypotheses - **Null Hypothesis (\( H_0 \))**: \( \mu = 7 \) (true mean hours of sleep is 7) - **Alternative Hypothesis (\( H_A \))**: \( \mu < 7 \) (true mean hours of sleep is less than 7) ### Test Type - **Left-tailed Test**. ## Mechanics 1. **Sample Mean**: - \( \bar{x} = 6.64 \) 2. **Sample Variance**: - \( s^2 \approx 1.157 \) 3. **Sample Standard Deviation**: - \( s \approx 1.076 \) 4. **Test Statistic**: - \( t \approx -1.67 \) 5. **Critical Value**: - \( t_{\text{critical}} = -1.711 \) ## Conclusion - **Decision**: Since \( t = -1.67 > -1.711 \), we **fail to reject \( H_0 \)**. ### Final Answer Summary - **Test Statistic**: \( t \approx -1.67 \) - **Critical Value**: \( t_{\text{critical}} = -1.711 \) - Conclusion: There is not enough evidence at the 5% significance level to conclude that the mean sleep time is under 7 hours.