1 Binomial Probability [15 points]

Transcription

1 Economics 250 Assignment 2 (Due November 13, 2017, in class) i) You should do the assignment on your own, Not group work! ii) Submit the completed work in class on the due date. iii) Remember to include the departmental cover page! 1 Binomial Probability [15 points] Suppose that the Real Estate Agent, Jack Mills, has five contacts, and he believes that for each contact the probability of making a sale is 0.4, and all contacts are independent. Let X be the number of sales in five contacts. i) What is the mean and standard deviation of X? [5 points] X follows a binomial distribution with mean µ X = np = 5(0.4) = 2, and standard deviation σ X = np(1 p) = ii) What is the probability that he makes between two and four sales (inclusive)? [5 points] P (2 X 4) = P (X = 2) + P (X = 3) + P (X = 4) = 5! 2!3! 0.42 (1 0.4) 3 + 5! 3!2! 0.43 (1 0.4) 2 + 5! 4!1! 0.44 (1 0.4) 1 = = iii) What is the probability that he makes at most four sales? [5 points] You could solve this problem by using P (X 4) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4), but it is more easy to use the fact that P (X 4) = 1 P (X = 5). 1

2 P (X 4) = 1 P (X = 5) = 1 5! 5!0! 0.45 (1 0.4) 0 = = Normal Distribution Approximation for Binomial Distribution [15 points] A car rental company has determined that the probability a car will need service work in any given month is 0.4, and the service requirement is independent event. The company has 1,000 cars. i) Use the normal approximation to calculate the probability that at most 385 cars will require service in a particular month. [5 points] Let X be the number of cars require service work in a particular month. Then X has a binomial distribution B(n, p) = B(1000, 0.4) with mean µ X = np = 400 and standard deviation σ X = np(1 p) = 240. The normal approximating probability is P (X 385) = P ( X µ X σ X = P (Z 0.7) = ) ii) Using software to calculate the exact Binomial probability in part i), you will get the probability value of Now, use the normal approximation with continuity correction to calculate the probability that at most 385 cars will require service in a particular month. [5 points] 2

3 The normal approximating probability with continuity correction improved in this way: P (X 385) = P (X 385.5) = P ( X µ X σ X = P (Z 0.4) = ) iii) Use the normal approximation to calculate the probability that at most 400 cars will require service in a particular month. [5 points] For this question, we don t need any calculation since the probability is the area under normal curve to the left of the mean. It is 0.5. You can confirm this result by: P (X 400) = P ( X µ X σ X = P (Z 0) = ) 3 Binomial Distribution [15 points] Recently, the U.S. Department of Education released a report on online learning stating that blended instruction, a combination of conventional face-to-face and online instruction, appears more effective in terms of student performance than conventional teaching. Suppose, for all students nationwide the population proportion who prefer this type of blended instruction is 85%. You decide to poll the incoming students at your institution to see if they prefer courses that blend face-to-face instruction with online components. i) If you had drawn an SRS of 400 incoming students, what are the mean and standard deviation of the sample proportion who prefer this type of blended instruction? [5 points] 3

4 µˆp = p = 0.85 p(1 p) 0.85(1 0.85) σˆp = = = n 400 ii) If you had drawn an SRS of 400 incoming students, what is the probability that the sample proportion would be less than the population proportion? [5 points] µˆp = p = 0.85 ˆp is approximately Normally distributed (since both np and n(1 p) are greater than 10) around center at p = 0.85 = 85%. So, the probability that the sample proportion would be less than the population proportion is 0.5. iii) If you had drawn an SRS of 400 incoming students, what is the probability that the sample proportion would be greater than 88%? [5 points] ˆp µˆp P (ˆp > 0.88) = P ( > σˆp = P (Z > 1.68) = 1 P (Z < 1.68) = = ) Sampling Distribution and Probability [15 points] The heights of South African men are normally distributed with a mean of 6 inches and a standard deviation of 4 inches. i) What is the probability that a randomly selected South African man is taller than 72 inches? [5 points] Denote the height as random variable X, then we have X N(µ, σ) = N(6, 4), 4

5 we are looking for P (X > 72) P (X > 72) = P ( X µ 72 6 > ) σ 4 = P (Z > 0.75) = 1 P (Z < 0.75) = = ii) If a random sample of three South African men were selected at random, what is the probability that the sample mean height is greater than 72 inches? [5 points] We are taking a random smaple with size n = 3. Denote the sample mean as X. Since the population has Normal distribution, so we have X N(µ, σ/ n) = N(6, 4/ 3). We are looking for P ( X > 72) P ( X > 72) = P ( X µ σ/ n = P (Z > 1.30) > / 3 ) = 1 P (Z < 1.30) = = iii) Now, suppose the heights of South African men are NOT normally distributed with a mean of 6 inches and a standard deviation of 4 inches. If a random sample of 200 South African men were selected at random, what is the probability that the sample mean height is greater than 6.5 inches? [5 points] We are taking a random smaple with size n = 200. Denote the sample mean as X. By Central Limit Theorem X N(µ, σ/ n) = N(6, 4/ 200), we are looking for P ( X > 6.5) P ( X > 6.5) = P ( X µ σ/ n = P (Z > 1.77) = 1 P (Z < 1.77) = = > / 200 )

6 5 Confidence Interval [15 points] Suppose that shopping times for customers at a local grocery store are normally distributed. A random sample of 16 shoppers in the local grocery store had a mean time of 25 minutes. Assume we do not know the population mean, µ, of the shopping times, but we know the standard deviation σ = 6 minutes. i) Find a 5% confidence interval for the population mean, µ. [5 points] Given information: sample size n = 16; sample mean x = 25; population standard deviation σ = 6. From Table A or Table D in textbook, we can find z = 1.6 for 5% confidence interval. Then, we can compute the margin of error m = z σ = ( ) n = 2.4. So, the 5% confidence interval for the unknown population 4 mean is ( x m, x + m) = (22.06, 27.4) ii) How can you interpret the calculated confidence interval? [5 points] Interpretation: If the population is repeatedly sampled and intervals are calculated in this fashion (this means: if we took many, many additional random samples with the same sample size and from each computed a 5% confidence interval for µ ), then in the long run approximately 5% of the intervals would contain the true value of the unknown population mean, µ. iii) If we want the margin of error for a 5% confidence interval to be 2, what is the sample size needed? [5 points] Margin of error is m = z σ n this implies n = size is ( z σ m) 2. So, the required sample 6

7 ( n = z σ ) ( ( )) = 1.6 = 35 m 2 6 Test of Significance [25 points] Suppose a stock price, X is normally distributed with a unknown population mean of µ and a population standard deviation of A random sample of observations had a sample mean of 1.5. Suppose that the significance level is pre-set at α = i) A researcher wishes to test the null hypothesis H 0 : µ = 2 against the alternative hypothesis H a : µ 2. Does the researcher reject the null at this level of significance? What is the power of the test when µ = 2.02?[10 points] The research is performing a two-tailed hypothesis test: Calculate the test statistic z: H 0 : µ = 2 H a : µ 2 z = x µ 0 σ/ n = / = 2.5 There are two ways to determine if the null hypothesis will be rejected or not. Method 1: using p-value For this two-tailed test, the p-value is p-value = 2P (Z z ) = 2P (Z 2.5) = The researcher will reject the null hypothesis at this significance level, since p-value = < α =

8 Method 2: comparing test statistic with critical value for α level For two-tailed test at α = 0.05 level, the critical value is z = 1.6 (this can be found from Table A in textbook). So, the null hypothesis will be rejected if test statistic is either greater than 1.6 or less than The researcher calculated the test statistic as z = 2.5 < 1.6. She will reject the null hypothesis at this significance level. The power to detect the alternative µ = 2.02 is the probability that H 0 will be rejected when, in fact, µ = We know the z-values less than 1.6 or greater than 1.6 would be viewed as significant at the 0.05% level. We can rewrite the upper rejection rule in terms of x: x / 1.6 For lower rejection rule in terms of x: x x x / 1.6 x x Now, we know that rejection can happen when either x or x We find that P ( x 2.032) = P ( x µ σ/ n = P (Z 0.6) = / ) P ( x 1.608) = P ( x µ σ/ n = P (Z 2.6) = / ) The power is the sum of their probabilities =

9 ii) If the researcher uses a significance level of α = 0.01 test, can she have the same conclusion as part i)? [5 points] The researcher will not reject the null hypothesis at this significance level (α = 0.01), since in this case p-value = > α = iii) Now, suppose the researcher wishes to test the null hypothesis H 0 : µ = 2 against the alternative hypothesis H a : µ < 2. Does the researcher reject the null at significance level α = 0.05? [5 points] The research is performing a one-tailed hypothesis test: Calculate the test statistic z: H 0 : µ = 2 H a : µ < 2 For this one-tailed test, the p-value is z = x µ 0 σ/ n = / = 2.5 p-value = P (Z < z) = P (Z < 2.5) = The researcher will reject the null hypothesis at this significance level α = 0.05, since p-value = < α = iv) If the researcher uses a significance level of α = 0.01 test (one-tailed test as part iii)), can she have the same conclusion as part iii)? [5 points] The researcher will also reject the null hypothesis at this significance level (α = 0.01), since in this case p-value = < α = 0.01.