University:
Jackson College
Course:
MAT-010 | Basic Mathematics
Academic year:

2024
Views:

221

Pages:

4
Author:

Bennett Wheeler

Question 1 Is bivariate normal distribution an assumption for the Pearson Correlation? Answer 1 Without making any assumptions, we can calculate the Pearson correlation coefficient for a given set of data if a linear relationship between two variables is detected. Additionally, this allows us to determine the strength of the linear association between the variables. However, if we want to test the significance of the correlation value and determine if the data supports a meaningful correlation, the assumption of normality is necessary for constructing the test statistics. Bivariate normality is particularly important when performing significance tests on the correlation coefficient. Question 2 ( 𝜇 1 − 𝜇 2 ) (μ 1 −μ 2 ) for two normal distributions. Obtain the appropriate point estimator. Answer 1 Step 1 Point Estimator: Point estimators use sample data to estimate population parameters. They provide an approximate value of the parameter, often by using a sample mean or difference in sample means for independent populations. Step 2 For two normal distributions with means 𝜇 1 μ 1 and 𝜇 2 μ 2 : 𝜇 1 − 𝜇 2 : mean difference μ 1 −μ 2 :mean difference Thus, the point estimator for the difference in means is the difference between the sample means, 𝑋 ˉ 1 − 𝑋 ˉ 2 X ˉ 1 − X ˉ 2 . Question 3 Using the daily high and low temperatures at Chicago's O'Hare International Airport for a year, a meteorologist modeled a scatterplot relating 𝑦 = high temperature y=high temperature to 𝑥 = low temperature x=low temperature, both in degrees Fahrenheit. The regression line calculated is 𝑦 ^ = 16.6 + 1.02 𝑥 y ^ =16.6+1.02x with 𝜎 = 6. 6 ∘ σ=6.6 ∘ F. About what percent of days with a low temperature of 4 0 ∘ 40 ∘ F have a high temperature greater than 7 0 ∘ 70 ∘ F? Answer 1 Step 1 Given: 𝑦 ^ = 16.6 + 1.02 ( 40 ) = 57.4 y ^ =16.6+1.02(40)=57.4 The average high temperature on days with a low temperature of 4 0 ∘ 40 ∘ F is 57. 4 ∘ 57.4 ∘ F, with 𝜎 = 6.6 σ=6.6. Step 2 Calculate the z-score: 𝑧 = 70 − 57.4 6.6 ≈ 1.90 z= 6.6 70−57.4 ≈1.90 Using the z-table: 𝑃 ( 𝑍 > 1.90 ) = 1 − 0.9713 = 0.0287 P(Z>1.90)=1−0.9713=0.0287 Thus, about 2.87% of days with a low of 4 0 ∘ 40 ∘ F have a high temperature greater than 7 0 ∘ 70 ∘ F. Question 4 c) According to the 2 Standard Deviation Rule, data values more than 2 standard deviations from the mean are considered "unusually high" or "unusually low." Question 5 Would you expect distributions of the following variables to be uniform, unimodal, or bimodal? Symmetric or skewed? Explain why. a) Ages at a Little League game b) Number of siblings of people in your class c) Pulse rates of college-age males d) Number of times each face of a die shows in 100 tosses Answer 1 a) Unimodal and skewed: Most participants at a Little League game are children, creating a peak at lower ages, with a skew towards higher ages for adults. b) Unimodal and skewed: Most individuals have few siblings, creating a peak at low numbers, skewed right as sibling count increases. c) Unimodal and symmetric: In a healthy population, pulse rates cluster around a central value, following a bell shape. d) Uniform: Each die face has equal probability, leading to approximately uniform results across 100 tosses.