Statistics Answer Note #25

Question 1 The dataset shoes how far, on average, a person in Illinois drove each year from 1960 to 1003. You want to calculate the correlation between year and distance. What correlation coefficient should you use and why? Answer To calculate the correlation between year and distance, the most appropriate correlation coefficient to use would be Pearson's correlation coefficient (r). This is because we are dealing with a linear relationship between two continuous variables, year and distance, and Pearson's correlation coefficient is specifically designed to measure the strength and direction of linear relationships. Before we dive into the calculations, let's first understand what Pearson's correlation coefficient is and how it can be interpreted. Pearson's r ranges from -1 to 1 and indicates the strength and direction of the linear relationship between two variables. A positive r value indicates a positive linear relationship, meaning as one variable increases, the other tends to also increase. On the other hand, a negative r value indicates a negative linear relationship, meaning as one variable increases, the other tends to decrease. A value of 0 indicates no linear relationship between the two variables. The formula for calculating Pearson's correlation coefficient is: r = (sum of (x-x̅)(y-ȳ)) / √((sum of (x-x̅)^2) * (sum of (y-ȳ)^2)) Where: - x and y are the variables in our dataset (year and distance) - x̅ and ȳ are the means of x and y, respectively. To calculate this, we need to first find the means of year and distance. Let's assume that the data is already cleaned and organized into a table with the years in one column (x) and the average distance driven in Illinois in the corresponding year in the second column (y). Step-by-Step Breakdown: 1. Calculate the means: - x̅ = sum of all x values / total number of x values - For example, if our dataset contains the years 1960-2003, we would have 44 years. So x̅ = (1960+1961+1962+...+2003)/44 = 1982 - ȳ = sum of all y values / total number of y values - For our sample dataset, let's say the average distance driven in Illinois was 10,000 miles per year. So ȳ = (10000+10000+10000+...+10000)/44 = 10,000 2. Calculate the sum of (x-x̅)(y-ȳ): - This can be done by subtracting x̅ from each x value and y̅ from each y value, then multiplying them together and adding them all up. - For example, for the first row in our table (1960, 8000), the calculation would be: (1960 - 1982)*(8000 - 10000) = -422000 3. Calculate the sum of (x-x̅)^2 and the sum of (y-ȳ)^2: - Similar to step 2, this can be done by subtracting the means from each value, squaring the results, and adding them all up. - For example, for the first row in our table, the calculations would be: (1960 - 1982)^2 = 484 and (8000 - 10000)^2 = 40000 4. Plug the values into the formula: - r = (-422000) / √((484)*(40000)) = -0.873 5. Interpret the result: - The result of -0.873 indicates a strong negative linear relationship between year and distance. This means that as the years increase, the distance driven in Illinois tends to decrease. This could be due to advancements in technology, changes in transportation modes, or other factors. Question 2 This dataset contains four variables: Exam_Score, Hours_Studied, Anxiety, and ACT_Test. Exam_Score represents the final exam score, while Hours_Studied indicates the time spent studying. Anxiety is a self-reported measure of anxiety about the final grade, ranging from 0 to 100. ACT_Test reflects the score on the ACT college entrance exam. This dataset allows for analysis of how study habits, anxiety levels, and standardized test scores influence final exam outcomes. What correlation coefficient should you use and why? Answer The most appropriate correlation coefficient to use in this scenario would be the Pearson correlation coefficient, also known as the Pearson's r. This coefficient measures the linear relationship between two continuous variables, and is suitable for analyzing the relationship between variables such as Hours_Studied, Anxiety, and ACT_Test with the final Exam_Score. Question 3 A random sample of 42 college graduates revealed that they worked an average of 5.5 years on the job before being promoted. The sample standard deviation was 1.1 years. Using the 0.99 degree of confidence, what is the confidence interval for the population mean? Answer Based on the given data, we can calculate the confidence interval for the population mean using the following formula: Confidence Interval = Sample mean ± (critical value) x (standard deviation of the sample)/√(sample size) Before we proceed with the calculation, let's first understand the concept of confidence interval. A confidence interval is a range of values within which we can be relatively certain that the true population mean lies. The level of confidence is determined by the critical value, which is based on the desired confidence level. In this case, the confidence level is 0.99, meaning we can be 99% confident that the true population mean falls within the calculated interval. Now, let's calculate the confidence interval using the given data. Step 1: Identify the critical value To find the critical value, we need to refer to the t-distribution table. Since we have a sample size of 42, and a confidence level of 0.99, we need to look for the column corresponding to 41 degrees of freedom (df= n-1) and 0.995 significance level (0.99+0.005). The critical value for this is 2.708. Step 2: Calculate the standard error The standard error is the standard deviation of the sample divided by the square root of the sample size. In this case, it is (1.1/√42) = 0.169 Step 3: Calculate the upper and lower limits for the confidence interval The upper limit = sample mean + (critical value x standard error) = 5.5 + (2.708 x 0.169) = 5.93 The lower limit = sample mean - (critical value x standard error) = 5.5 - (2.708 x 0.169) = 5.07 Therefore, the 99% confidence interval for the population mean is 5.07 to 5.93 years. Question 4 One way we found the coefficients to the regression equation, 𝑦𝑦=𝑦𝑦+𝑦𝑦𝑦𝑦 was to use orthogonal projections with: � 𝑦𝑦 𝑦𝑦 �=(𝑦𝑦𝑦𝑦𝑦𝑦)−1𝑦𝑦𝑦𝑦𝑦𝑦 Let 𝑦𝑦 and 𝑦𝑦 be defined below: 𝑦𝑦=� 1 𝑦𝑦0 1 𝑦𝑦1 1 𝑦𝑦2 � 𝑦𝑦=� 𝑦𝑦0 𝑦𝑦1 𝑦𝑦2 � Show that the values of vector [𝑦𝑦 𝑦𝑦]𝑦𝑦 are the same as the values of 𝑦𝑦 and 𝑦𝑦 that you get solving the system of equations that minimize the sum of squared errors. 𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦=0 𝑦𝑦𝑦𝑦 𝑦𝑦𝑦𝑦=0 Where 𝑦𝑦(𝑦𝑦,𝑦𝑦)=�𝑦𝑦0−(𝑦𝑦+𝑦𝑦𝑦𝑦0)�2+�𝑦𝑦1−(𝑦𝑦+𝑦𝑦𝑦𝑦1)�2+�𝑦𝑦2−(𝑦𝑦+ 𝑦𝑦𝑦𝑦2)�2 . Answer In order to understand how the values of vector [𝑦𝑦 𝑦𝑦]𝑦 are the same as the values of 𝑦𝑦 and 𝑦𝑦 obtained through minimizing the sum of squared errors, it is helpful to understand the concept of orthogonal projections and how they are used to find regression coefficients. Orthogonal projection is a method used to measure the vertical distance between a point and a straight line. In regression analysis, it is used to find the best fit line that minimizes the error between the observed data and the predicted values. This line is known as the regression line or the line of best fit. The equation 𝑦𝑦=𝑦𝑦+𝑦𝑦𝑦𝑦 represents the true relationship between the independent variable 𝑦𝑦 and the dependent variable 𝑦𝑦. However, in real data, there is usually some error present in the observed values of 𝑦𝑦. The aim of regression analysis is to find the values of 𝑦𝑦 and 𝑦𝑦 that minimize the total error. To do this, we use the method of least squares. This method involves minimizing the sum of squared errors between the observed values of 𝑦𝑦 and the predicted values from the regression line. This is where the 𝑦𝑦(𝑦𝑦,𝑦𝑦) comes in, as it represents the total sum of squared errors. The equation for 𝑦𝑦(𝑦𝑦,𝑦𝑦) can also be written as 𝑦𝑦(𝑦𝑦,𝑦𝑦)=∑(𝑦𝑦𝑦−𝑦𝑦−𝑦𝑦𝑦𝑦𝑦)^2, where 𝑦𝑦𝑦 is the observed value of 𝑦𝑦 for the 𝑦th data point. Now, to find the values of 𝑦𝑦 and 𝑦𝑦 that minimize 𝑦𝑦(𝑦𝑦,𝑦𝑦), we take the partial derivatives of 𝑦𝑦(𝑦𝑦,𝑦𝑦) with respect to 𝑦𝑦 and 𝑦𝑦 and set them equal to 0, as given in the question. This gives us a system of equations that can be solved to obtain the values of 𝑦𝑦 and 𝑦𝑦 that minimize the sum of squared errors. Now, coming back to the orthogonal projections method, we can see that the equation Question 5 College students are randomly selected and arranged in groups of three. The random variable x is the number in the group who say that they take one or more online courses. Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviation. If a probability distribution is not given, identify the requirements that are not satisfied. x P(x) 0 0.101 1 0.354 2 0.400 3 0.145 Answer In this scenario, it seems that a probability distribution is given. A probability distribution is a function that assigns probabilities to the possible outcomes of a random variable. In this case, the random variable x represents the number of students in a group of three who say they take one or more online courses. The given probabilities for each possible value of x (0, 1, 2, and 3) add up to 1, which is a requirement for a probability distribution. To find the mean of a probability distribution, we multiply each possible value of x by its corresponding probability and then sum them up. Thus, the mean (μ) can be calculated as: μ = (0)(0.101) + (1)(0.354) + (2)(0.400) + (3)(0.145) = 0.985 To find the standard deviation, we need to calculate the variance first. The variance (σ²) of a probability distribution is given by: σ² = Σ (x - μ)² * P(x) Where Σ denotes the summation symbol, x is the possible values of the random variable, μ is the mean, P(x) is the corresponding probability. For this given distribution, the variance can be calculated as: σ² = [(0-0.985)²(0.101)] + [(1-0.985)²(0.354)] + [(2-0.985)²(0.400)] + [(3-0.985)²(0.145)] = 0.677 And the standard deviation (σ) is the square root of the variance, thus: σ = √0.677 = 0.823 Therefore, the mean and standard deviation of this given probability distribution is 0.985 and 0.823 respectively.