Questions and Answers #1 Modeling Data Distributions

Questions and Answers Sheet 1 Modeling Data Distributions Question #1 A random sample of 𝑛1 = 14 winter days in Denver gave a sample mean pollution index 𝑥1 = 43 . Previous studies show that 𝜎1 = 19 . For Englewood (a suburb of Denver), a random sample of 𝑛2 = 12 winter days gave a sample mean pollution index of 𝑥2 = 37 . Previous studies show that 𝜎2 = 13 . Assume the pollution index is normally distributed in both Englewood and Denver. (a) State the null and alternate hypotheses. 𝐻0 : 𝜇1 = 𝜇2 ; 𝜇1 > 𝜇2 𝐻0 : 𝜇1 < 𝜇2 ; 𝜇1 = 𝜇2 𝐻0 : 𝜇1 = 𝜇2 ; 𝜇1 < 𝜇2 𝐻0 : 𝜇1 = 𝜇2 ; 𝜇1 ≠ 𝜇2 (b) What sampling distribution will you use? What assumptions are you making? The Student's t. We assume that both population distributions are approximately normal with known standard deviations. The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations. The standard normal. We assume that both population distributions are approximately normal with known standard deviations. The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations. (c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate. (Test the difference 𝜇1 − 𝜇2 . Round your answer to two decimal places.) (d) Find (or estimate) the P-value. (Round your answer to four decimal places.) (e) Based on your answers in parts (i)-(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \alpha? At the 𝛼 = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. At the 𝛼 = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant. At the 𝛼 = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the 𝛼 = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant. (f) Interpret your conclusion in the context of the application. Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver. Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver. Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver. Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver. (g) Find a 99% confidence interval for 𝜇1 − 𝜇2 . (Round your answers to two decimal places.) lower limit upper limit (h) Explain the meaning of the confidence interval in the context of the problem. Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver. Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver. Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver. Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver. Answer: Step 1 Hey there! Thank you for posting the question. Since your question has more than 3 parts, we are solving the first 3 parts for you, according to our policy. If you need help with any of the other parts, please re-post the question and mention the part you want to be answered. Step 2 a) The null and alternative hypotheses are shown below: Null hypothesis: 𝐻0 : 𝜇1 = 𝜇2 Alternative hypothesis: 𝐻1 : 𝜇1 ≠ 𝜇2 b) Here, the assumptions are both population distributions are approximately normal with known standard deviations. Therefore, the sampling distribution will use here is the standard normal. Thus, the correct option is the standard normal. We assume that both population distributions are approximately normal with known standard deviations. c) The test statistic is obtained as follows: 𝑥1 −𝑥2 𝑧= 2 2 𝜎 𝜎 √ 1+ 2 𝑛1 𝑛2 𝑧= 43−37 2 2 √19 +13 14 12 = 0.95 Thous, the value of the test statistics is 0.95. Question #2 This exercise requires the use of a graphing calculator or computer programmed to do numerical integration. The normal distribution curve, which models the distributions of data in a wide range of applications, is given by the function 1 (𝑥 − 𝜇)2 /(2𝜎 2 ) 𝑝(𝑥) = 𝜎𝑒 − √2𝜋 where 𝜋 = 3,14159265 ... and \sigma and \mu are constants called the standard deviation and the mean, respectively. Its graph (for 𝜎 = 1 and 𝜇 = 2) is shown in the figure. +∞ With  = 5 and 𝜇 = 0, approximate ∫0 𝑝(𝑥)𝑑𝑥 .(Round your answer to four decimal places.) Answer: Step 1 Here, 𝑥 ∼= 𝑁(𝑀, 𝜎 2 ) +∞ To find ∫0  𝑝(𝑥)𝑑𝑥 , 10 here  ( x) = ( x ) is the PDF of x  + 0  ( x)dx = P[ x  0] 𝑥 ∼ 𝑁(𝑀, 𝜎 2 ) ⇒ 𝑥−𝑀 𝜎 ∼ 𝑁(0,1) 𝑥 For 𝑀 = 0, 𝜎 = 5𝑀 = ∼ 𝑁(0,1) 5 1 e( x −  )2 / 2 2 , x  R  2K 𝑥 Then, 𝑃[𝑥 > 0] = 𝑃 [ > 0] + 𝑃[𝑦 > 0] 5 = 1 − 𝑃[𝑦 ≤ 0] = 1 − 𝜙(0) = 1−0⋅5= 0⋅5 Step 2 Hence,the required value is 0.5. Question #3 Find the amount that results from the given investment. $700 invested at 4\% compounded daily after a period of 4 years: After 4 years, the investement results in how much $? Answer: Step 1 Compound interest: In compound interest, interest is added back to the principal sum so that interest is earned on that added during the next compounding period. That is, compound interest will give an interest on the interest. The interest payments will change in the time period in which the initial sum of money stays in the bank or with the 𝑏 ⇒ 𝑒𝑟. The general formula for compound interest is, 𝐴 = 𝑃 ⋅ (1 + 𝑟/𝑛)𝑛𝑡 Where: A is the future value of the investment loan including the loan, P is the principle amount, r is the annual interest rate in decimals, n is the number of times interest is compounded per year, t is the time of years the money is invested or borrowed. Step 2 Find the investment results after 4 years: The invested amount is $700 at 4% compound daily interest. The aim is to know the investment results after 4 years. Here, The principal amount is 𝑃 = $700 Annual interest rate is 𝑟 = 4% = 0.04, The number of times interest is compounded per year is daily. That is, 𝑛 = 365. The number of years the money is invested or time period is 𝑡 = 4 years. The investment result after 4 years is obtained as $821.4504 from the calculation given below: 𝑟 𝑛𝑡 𝐴 = 𝑃 × (1 + ) = 700 × (1 + = 700 × (1 + 𝑛 0.04 365×4 ) 365 0.04 1.460 365 ) = 821.4504 Step 3 Answer: The investment result after 4 years is $821.4504. Question #4 MODELING REAL LIFE. You mix 0.25 cup of juice concentrate for every 2 cus of water to make 18 cups of juice. How much juice concentrate do you use? How much water do you use? You use\square cups of juice concentrate and\square cups of water . Answer: Step 1 From given, every 2 cups of water has a mix of 0.25 cups of juice concentration. 0.25𝑐𝑢𝑝𝑠 1 = cups 2𝑐𝑢𝑝𝑠 8 1 So, each cup of juice has cups of juice concentration. 8 Step 2 Obtain the juice concentration for 18 cups. 1 18 8 8 18 × ( ) = = 9 4 = 2.25 cups of juice concentration 18 − 2.25 = 15.75 cups of water So, you use 2.25 cups of juice concentration and 15.75 cups of water. Question #5 In a normal distribution, a data value located 1 standard deviations below the mean has Standard Score: 𝑧 = In a normal distribution, a data value located 2.2 standard deviations above the mean has Standard Score: 𝑧 = In a normal distribution, the mean has Standard Score: 𝑧 = Answer: Step 1 Introduction: For a normally distributed random variable, X with mean \mu, and standard deviation 𝜎, the z-score for the value x is given as follows: 𝑥−𝜇 𝑧= . 𝜎 Step 2 1.The z-score gives the number of standard deviations that a raw score or data value lies above or below the mean. The numerical value of the z-score gives the distance of the data value from the mean in terms of number of standard deviations, and the sign of the z-score gives the direction with respect to the mean; if the data value lies above the mean, then the z-score is positive; if the data value lies below the mean, then the zscore is negative. Here, the data value is given to be 1 standard deviation below the mean. The term “1 standard deviation” indicates that the numerical value of the z-score must be 1, whereas the word “below” indicates that the sign of the z-score should be negative. Thus, the required z-score is –1. Step 3 b.Here, the data value is said to be 2.2 standard deviations above the mean. The term “2.2 standard deviations” indicates that the numerical value of the z-score must be 2.2, whereas the word “above” indicates that the sign of the z-score should be positive. Thus, the required z-score is 2.2. Step 4 c.The z-score for the mean of the data values, \mu, is calculated below, by substituting 𝑥 = 𝜇 in the formula for the z-score: 𝑥−𝜇 𝑧= = = 𝜇−𝜇 0 𝜎 𝜎 𝜎 = 0. Thus, the required z-score for the mean is 0. Question #6 a. Let 𝑋 ∼ 𝑁(0,1) and 𝐸 ∼ 𝑁(0,1) be independent, and let 𝑌 = 𝑋 + 𝛽𝐸 . Show that 𝑟𝑥𝑦 = 1 √𝛽 2 +1 b.Use the results of part (a) to generate bivariate samples (𝑥𝑖 , 𝑦𝑖 ) of size 20 with population correlation coefficients -.9, -.5, 0, .5, and .9, and compute the sample correlation coefficients. c. Have a partner generate scatterplots as in part (b) and then guess the correlation coefficients. Answer: Question #7 a. Let 𝑋 ∼ 𝑁(0,1) and 𝐸 ∼ 𝑁(0,1) be independent, and let 𝑌 = 𝑋 + 𝛽𝐸 . Show that 𝑟𝑥𝑦 = 1 √𝛽 2 +1 b.Use the results of part (a) to generate bivariate samples (𝑥𝑖 , 𝑦𝑖 ) of size 20 with population correlation coefficients -.9, -.5, 0, .5, and .9, and compute the sample correlation coefficients. c. Have a partner generate scatterplots as in part (b) and then guess the correlation coefficients. Answer: b)We are using R. to generate samples X and Y. (1) Let correlation between X and Y be -0.9. We will use R to find the value of 𝛽 . Then we will generate samples X and E from standard normal distribution. After that we will model sample Y. Question #8 Presenting data in the form of table. For the data set shown by the table, Solve, a) Create a scatter plot for the data. b) Use the scatter plot to determine whether an exponential function, a logarithmic function, or a linear function is the best choice for modeling the data. (If applicable, you will use your graphing utility to obtain these functions.) Intensity (wattd per meter 2 ) Loudness Level (decibels) 0.1(loud thunder) 110 1(rock concert, 2 yd from speakers) 120 10(jackhammer) 130 100(jet takeoff, 40 yd away) 140 Answer: Step 1 a) We can draw scatter plot using Excel as follows, Enter the data in Excel and select it. Go to insert ⇒ scatter ⇒ select 1st scatter plot ⇒ add trendline to the plot ⇒ Tick on display equation on plot and display R square on the plot. We get result as follows, Step 2 b) From the above scatter plot, we can observe that, the regression equation for logarithmic model is, 𝑦̂ = 4.3429 ln(𝑥) + 120 and 𝑅2 = 1 Since, the value of 𝑅2 = 1 for Logarithmic function, the logarithmic function is the best choice for modeling the data. Question #9 True/False a) It is appropriate to use A Normal approximation for a Binomial distribution with 𝑛 = 15 and 𝑝 = 0.2 b) Both the Hypergeometric and the Binomial distributions deal with dichotomous (binary) data. Answer: Step 1 a) To use Normal approximation for a Binomial distribution we must have np and 𝑛(1 − 𝑝) both greater than 5. For this problem 𝑛𝑝 = 15(0.2) = 3 < 5 So we can not use the Normal approximation. So the given statement is false. Step 2 b) For both Hypergeometric and Binomial distributions there are 2 possible outcomes: success and failure. For Hypergeometric distribution, the probability for success and failure is not the same in every trial. For Binomial distribution, the probability for success and failure is the same in every trial. So both the Hypergeometric and the Binomial distributions deal with dichotomous (binary) data. So the given statement is true. Question #10 A privately owned liquor store operates both a drive-n facility and a walk-in facility. On a randomly selected day, let X and Y, respectively, be the proportions of the time that the drive-in and walk-in facilities are in use, and suppose that the joint density function of these variables is 2  ( x + 2 y ) 0  1.0  y  1 f ( x, y ) =  3 0, else where a) Find the marginal density of X. b) Find the marginal density of Y. c) Find the probability that the drive-in facility is busy less than one-half of the time. Answer: Step 1 a) The marginal density of X is: 1 𝑓(𝑥) = ∫0 𝑓(𝑥, 𝑦)𝑑𝑦 1 2 = ∫ (𝑥 + 2𝑦)𝑑𝑦 0 3 = 2𝑥 3 1 1 4 ∫0 𝑑𝑦 + 3 ∫0 𝑦𝑑𝑦 1 2𝑥 1 4 𝑦 2 [𝑦]0 + [ ] 3 3 2 0 2𝑥 2 = + 2 3 2(𝑥+1) = , 0≤𝑥≤1 = 3 Step 2 b) The marginal density of Y is given below: 1 𝑓(𝑦) = ∫0 𝑓(𝑥, 𝑦)𝑑𝑥 12 = ∫0 (𝑥 + 2𝑦)𝑑𝑥 3 1 2 = ∫0 𝑥𝑑𝑥 + 3 2 𝑥2 1 = [ ] + 3 1 3 + = 2 0 4𝑦 3 1+4𝑦 3 4𝑦 4𝑦 3 3 1 ∫0 𝑑𝑥 [𝑥]10 , 0≤𝑦≤1 c) It is given that drive in facility is denoted by x, then 1 1 2 𝑃 (𝑋 < ) = ∫ 𝑓(𝑥)𝑑𝑥 2 0 =∫ 1 2 2(𝑥 0 + 1) 𝑑𝑥 3 1 2 2 = ∫ (𝑥 + 1)𝑑𝑥 3 0 1 1 2 2 2 2 = ∫ 𝑥𝑑𝑥 + ∫ 𝑑𝑥 3 0 3 0 1 2 𝑥2 2 2 1 = [ ] + [𝑥]20 3 2 0 3 2 1 2 1 = [ ]+ [ ] 3 4 3 2 1 1 = + 12 3 5 = 12 Therefore, the required probability is 0.4167.