(d) The estimated probability that a respondent answers Yes to both - PDF document

SOKOINE UNIVERSITY OF AGRICULTURE COLLEGE OF SOCIAL SCIENCES AND HUMANITIES DEPARTMENT OF POLICY PLANNING AND MANAGEMENT MASTER OF ARTS IN PROJECT MANAGEMENT AND EVALUATION M 602: STATISTICAL METHODS FOR PROJECT MANAGEMENT INSTRUCTOR: Prof. Kim Abel Kayunze ASSIGNMENT Suppose in a national household budget survey, 1,117 respondents in a region are asked, among other questions, whether they are members of an environmental group and whether they would be very willing to pay much higher prices to protect the environment, and the results are as tabulated below. Pay higher prices Responses Yes No Total Yes 30 66 96 Member of environmental No 88 933 1,021 group Total 118 999 1,117 Giving pertinent arguments/reasons, compute: (a) The estimated probability that a randomly selected respondent is a member of an environmental group (b) The estimated probability of being very willing to pay much higher prices to protect the environment, (i) if the respondent is a member of an environmental group, and (ii) if the respondent is not a member of an environmental group (c) The estimated probability that a respondent is both a member of an environmental group and very willing to pay much higher prices to protect the environment (d) The estimated probability that a respondent answers Yes to both questions or No to both questions

1.0. INTRODUCTION TO PROBABILITY 1.1.What is probability? For a particular possible outcome for a random phenomenon, the probability of that outcome is the proposition of times that the outcome would occur in a very long sequence of observations. 1.2.Basic Concepts of Probability 1.2.1. Experiment or a trial is the process of making an observation or recording a measurement under a given set of conditions. Note, an experiment is realized whenever the set of conditions is realized Example Consider the following experiment. Toss a coin and observe whether the upside of the coin is Head or Tail. Two events may be occurred: • H: Head is observed, • T: Tail is observed. 1.2.2. Events are outcomes of the experiment . 2.0.BASIC PROBABILITY RULES There are four probability rules which are most useful. Let P (A) denote the probability of a possible outcome or set of outcomes denoted by the letter A as shown below; 2.1. Rule Number One: P ( not A) = 1 – P(A) If you know the probability a particular outcome occurs, then the probability it does not occur is 1 minus that probability. By using the rule number one; If the event A can occur in Y A ways out of a total of X possible outcomes and equally likely outcomes, the probability that the event A will occur is given by P (A) = Y A / X P (A) = Probability of an event A to occur Y A = number of possibilities that an A can occur X = totally number of all equally possible outcomes 2.2. Rule Number Two: If A and B are distinct possible outcome ( with no overlap) , then P(A or B) = P(A)+P(B) Suppose you take a survey to estimate the population of people who believe that embryonic stem cell research should be banned by the federal Government .Let A represent your getting a sample proportion estimates that is much too below, being more than 0.10 below the population proportion .Let B represent your sample proportion estimates being much too high at least 0.10

above the population proportion. Example, P(A)=P(B)=0.03 , then the overall probability your sample proportion is in error by more than 0.10 ( without specifying the direction of error) is P(A or B) = P(A) + P(B) = 0.03 + 0.03 = 0.06 2.3. Rule Number Two: If A and B are possible outcomes , then P(A) and P(B) = P(A) x P(B, given A) Suppose the probability that a randomly selected American adult is married equals 0.56. Of those who are married, general social survey indicate that the probability a person reports being very happy when asked to choose among ( very happy , pretty happy, not too happy) is about 0.40, that is given you are married , the probability of being very happy is 0.40. So P (married and very happy) = P(married) x P9very happy, given married)=0.56 x 0.40 = 0.22 About 22% of the adults’ population is both married and very happy. NOTE: In some cases, A and B are independent in a sense that whether one occurs does not. That is, P (B, given A) = P (B) = P(B), so the previous rule simplifies. 2.4.Rule Number Four :If A and B are independent , then P(A and B)= P(A) x P(B) For example, the probability of a correct inference set is at 0.95.Suppose A represents an inference about men in the population of interest (such as a prediction about the proportion of men who vote for President) being correct. Then since these are independent samples and inferences, the probability that both inferences are correct is P (A and B) = P (A) x P (B) = 0.95 x 0.95 = 0.90

3.0. RESPONSES TO QUESTION Pay higher prices Responses Yes No Total Member of Yes 30 66 96 environmental No 88 933 1,021 group Total 118 999 1,117 3.1. a). The estimated probability that a randomly selected respondent is a member of an environmental group By using the rule number one; If the event A can occur in Y A ways out of a total of X possible outcomes and equally likely outcomes, the probability that the event A will occur is given by P (A) = Y A / X P (A) = Probability of an event A to occur Y A = number of possibilities that an A can occur X = totally number of all equally possible outcomes So the probability that the randomly selected respondent from the population will be a member of an environmental group will be calculated as follows; Pay higher prices Responses Yes No Total Yes 30 66 96 No 88 933 1,021 Total 118 999 1,117 Member of environmental group  The blue shaded row shows the total number of the members of an environmental group which is equal to 96 (number of possibilities/ ways that a random selected respondent from the population will be a member of an environmental group)  The green shaded area shows the total number of population/ household (total number of all/ equally possible outcomes).

Hence; P (A) = 96/ 1117 = 0.086 Therefore; the estimated probability that a randomly selected respondent is a member of an environmental group is 0.086 or 8.6% 3.1. b). The estimated probability of being very willing to pay much higher prices to protect the environment, (i) if the respondent is a member of an environmental group, and (ii) if the respondent is not a member of an environmental group b) i. Pay higher prices Responses Yes No Total Member of Yes 30 66 96 environmental No 88 933 1,021 group Total 118 999 1,117  People who are willing to pay much higher prices to protect the environment and also is the member of environmental group (intersection) are 30  So the estimated probability that the randomly selected respondent from the population he/she is willing to pay much higher prices to protect the environment but the respondent is a member of an environmental group P (A) = Y A / X 30/ 96 =0.3125 b) ii. Pay higher prices Responses Yes No Total Member of Yes 30 66 96 environmental No 88 933 1,021 group Total 118 999 1,117  People willing to pay much higher prices to protect the environment and he/she is not the member of environmental group (intersection) are 88.

 So the estimated probability that the randomly selected respondent from the population he/she is willing to pay much higher prices to protect the environment but he/she is not a member of an environmental group will P (A) = Y A / X 88/ 1021 =0.08619 3.1.c). The estimated probability that a respondent is both a member of an environmental group and very willing to pay much higher prices to protect the environment Pay higher prices Responses Yes No Total Member of Yes 30 66 96 environmental No 88 933 1,021 group Total 118 999 1,117  The shaded area are the respondents who are the members of the environmental group and they are willing to pay high much prices to protect the environment  So the estimated probability that a randomly selected respondent from the total respondents will be both a member of an environmental group and very willing to pay much higher prices to protect the environment We will use rule number one P (A) = Y A / X 30/1117 =0.027 Apply Rule number two P(A and B ) = P (A) x P (B, given A) P (A) = 96/ 1117 = 0.086 P (B, given A) = Y A / X 30/ 96 =0.3125 P (A and B) = 0.086 x 0.3125 = 0.027