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### assignment4_2006.ans_mk

Course: PBAF 527, Fall 2009
School: Washington
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Word Count: 2032

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527: PBAF Quantitative Methods Winter 2004 Assignment 4 Answer Sheet Part I In the entire population of water supplies in WA, the concentration of fluoride is known to be approximately normally distributed, with mean 3 ppm and standard dev 1 ppm. a. This question reviews our use of the cumulative normal probability distribution where the random variable X is the concentration of fluoride in the water supply. We...

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527: PBAF Quantitative Methods Winter 2004 Assignment 4 Answer Sheet Part I In the entire population of water supplies in WA, the concentration of fluoride is known to be approximately normally distributed, with mean 3 ppm and standard dev 1 ppm. a. This question reviews our use of the cumulative normal probability distribution where the random variable X is the concentration of fluoride in the water supply. We want to calculate the probability of a concentration greater than 2 ppm: P(X > 2) = P(Z > (2-3)/1) = P(Z > -1) = .5 + P(-1< Z < 0) = .5 + .3413 = 0.841 Thus, we can say that about 84% of the water supplies will have a concentration of greater than 2 ppm of fluoride. Next, what is the chance that a random water supply contains 1 to 2 ppm? We want: P(1 < X < 2) = P(X<2) - P(X<1). Well, P(X < 2) = 0.1587 from above. P(X < 1) = P(Z < (1-3)/1) = P(Z < -2) = 0.0228. And 0.1587 - 0.0228 = 0.1359. Thus, about 13.6% of the water supplies will have a concentration between 1 and 2 ppm of fluoride. b. Now we need: P(X1 > 2 and X2 > 2) (remember that the probability of independent events both occurring is the product of their individual probabilities) = P(X1>2)P(X2>2) = (0.841)(0.841) = 0.707, thus about 70.7% of the time two randomly selected water supplies will both contain more than 2 ppm. Next we calculate P(1 < X1 < 2 and 1 < X2 < 2) = P(1 < X1 < 2)P(1 < X2 < 2) = (0.1359)(0.1359) = 0.0185. In both cases in part b. the probabilities are a lot lower than the probability that one random water supply will satisfy the condition given, since it is less likely to happen that two random draws from the population will both satisfy the condition. c. Now we average the two concentrations and create a new random variable: = (X1+X2)/2. has a new mean and standard deviation which can be calculated according to the rules of expectation from about a linear combination of independent random variables. E[X+Y]=E[X]+E[Y], etc. The mean is E[] = E[(X1 + X2)/2] = (1/2)(E[X1] + E[X2]) = (1/2)(3 + 3) = 3. The var is V[] = V[(X1 + X2)/2] = (1/4)(V[X1]+V[X2]) = (1/4)(2V[X]) = 0.5, or it equals V(X)/n = = .5. So, just as we have seen in class the sample mean is on average equal to the population mean, but has less variance than random selections from the population. The standard deviation is the square root of the variance, so it equals .707. Finally, P(>2)=1-P(<2)=1-P(Z<(2-3)/0.707)=1-P(Z<-1.414) =1-0.0793=0.9207 And P(1<<2)= P(<2) - P(<1) = 0.0793-0.0023=0.077 The probability that the average of two random water supply concentrations is over 2 ppm is 92.07%, higher than for one water supply since we expect the average to fall close to the mean concentration of 3 ppm. In other words, the chance that the average of the two would fall below 2 ppm (away from the mean) is less than it would be for a single draw. The probability that the average will be between 1 and 2 ppm is only about 7.7% since once again the average of the two supplies would be expected to fall close to the mean of 3 ppm. d. Calculate mean and standard deviation for a random sample of 10 water supplies. E[] = 3 and V[] = 2/n = 1/10 = 0.1. Thus SD[] = 0.32. Wow, the standard deviation and variance for the mean of 10 samples are small!! P(>2) = 1-P(Z<(2-3)/0.32) = 1-P(Z<-3.13). P(Z<-3.13) is basically off of the charts, i.e., it is very close to 0, thus 1-P(Z<-3.13) is very close to 1. This makes sense because the average concentration in 10 samples should be extremely close to the population mean which is 3 ppm (included in the range that is above 2 ppm). And furthermore, there is an extremely tiny probability of an average concentration for 10 samples of between 1 and 2 ppm for the same reason. e. If you know that a sampling distribution where n=10 has a mean of 3ppm and a standard deviation (standard error) of 1ppm, there are a few things you can say about the population mean: = 3 since the mean of a sampling distribution is equal to the population mean. = 1ppm * 10 =3.16 If the sample size is 40, but the sampling distribution still has a standard error of 1ppm, the standard deviation of the population must be larger. We can solve for it: Std .Error = ... so = Std.Error n n = 1 40 = 6.32 Therefore, if the sample size gets bigger while the standard error of the sampling distribution stays the same size, it must mean that the population standard deviation grew. Increases in sample sizes reduce the standard error of sampling distributions, but the population standard deviation also has an important effect on the width of sampling distributions. Moving from Populations to Samples 2 Part II 1a. We want the 95 percent confidence interval. We are given the mean and the sample standard deviation. n=21. So, we need to use the t-distribution because the population SD is unknown and the sample size is small. If the CI is 95% then =5%=.05 and /2=.025. t/2=2.086 with 20 degrees of freedom (n-1=21-1=20 df ). We want this confidence interval: P( x - 2.086SE < < x + 2.086SE)= .95 s 3.22 ^ = = .70 Next, we'll calculate the estimate of the standard error. SE = n 21 Then, we'll plug in SE estimate and sample mean: P( x - 2.086SE < < x + 2.086SE)= .95 =P[52.6 - (2.086)(.7) < < 52.6 + (2.086)(.7)]=P[51.14< < 54.06]=.95 so, x 2.086SE=[51.14,54.06] 1b. The mean pulse rate for all U.S. adult males is 72 heartbeats per minute. 95 percent of the time, the mean heart rate for U.S. adult males who jog at least 15 miles per week will fall in the range of 51.14 to 54.06 beats per minute. Only 5% percent of the time will it fall out of the range. So, we can say that it appears that jogging 15 miles per is associated with a reduced heat rate in adult males. 1c. We need to assume that the heart rate is normally distributed in the population. 2. We want to know if these sample sizes are large enough to construct a confidence interval for the true proportion of employees at each fitness level exhibiting signs of stress. First, we need to think about how close we need to be to "true". Do we need to be within .01? .02? .03? What would meet our needs? Second, will a 95% confidence interval provide us with enough confidence or should the interval be larger, say 99%? Likely being 99% confident is fine. In order to figure out the sample size, ^^ we need an estimate of the variance. The table below presents that estimate, pq Fitness Level Poor Average Good Sample Size 242 212 95 Proportion with Signs of Stress ^ p .155 .133 .108 .845 .867 .892 ^ q ^^ pq (Var) .131 .115 .096 ^^ pq (est. SD) .362 .339 .310 2a. To check to see if each subsample is large enough to use the large sample formula for the confidence ^ ^ ^ interval, we need to check for each that p 3 p (1 - p) / n does not include 0 or 1. ^ ^ ^ Poor: - p 3 p(1 - p) / n =.155-3* sqrt(.362/242) =.085 ^ ^ ^ p + 3 p (1 - p ) / n = .155+3 *sqrt (.362/242) = .225 Average: ^ ^ ^ p - 3 p(1 - p) / n =.133-3* sqrt(.339/212) =.063 ^ ^ ^ p + 3 p (1 - p ) / n = .133+3 *sqrt (.339/212) = .203 3 Good: ^ ^ ^ p - 3 p(1 - p) / n =.108-3* sqrt(.310/295) =.0125 ^ ^ ^ p + 3 p (1 - p ) / n = .108+3 *sqrt (.310/95) = .204 So, each of the samples is large enough given that the sample proportion does not fall within 3 standard errors of 0 or 1. Notice that the smallest sample (for the good sample) is getting close to being too small at n=95. 2b. ^ ^ P( p poor -1.96SE < ppoor< p poor + 1.96SE)= .95 =P[.155 - (1.96)( ^ so, p poor .131 ) < ppoor < .155 + (1.96)( 242 1.96SE=[.109,.201] .131 )]=P[.109<ppoor< .201]=.95 242 ^ ^ P( paverage -1.96SE < paverage< paverage + 1.96SE)= .95 .115 ) < paverage < .133 + (1.96)( 212 ^ so, paverage 1.96SE=[.087,.179] ^ ^ P( p good -1.96SE < pgood< p good + 1.96SE)= .95 =P[.133 - (1.96)( =P[.108 - (1.96)( ^ so, p good .096 ) < pgood < .108 + (1.96)( 95 1.96SE=[.046,.170] .115 )]=P[.087<p1< .179]=.95 212 .096 )]=P[.046<pgood< .170]=.95 95 2c. We are 95 percent confident (given our sample sizes for each) group that, of those with poor fitness levels, between 11% and 20% will show signs of stress. Of those with average fitness levels between 9% and 18% will show signs of stress. Of those with good fitness levels between 5 and 17 % will exhibit signs of stress. These confidence intervals overlap, leading use to question whether there are real differences in the rates for the groups. However, all these sample sizes differ, the distance from the mean differs, and these results do depend on the sample size. If we increased the sample size for those in good health, the 95% confidence interval would decrease in size. Therefore, although they are all 95% confidence intervals, it is difficult to compare them. 2d. We'd like to get the sample size to be 95% and within .01. So, we use our equation: 2 z / 2 pq , where z/x=1.96 and D=.01 pq is calculated in the table above. n= D2 So 1.96 2 (.096) n= = 3688 This is a larger sample size due to the smaller distance we can tolerate 2 .01 from the mean. 4 Part III There are no unique answers for these problems so this answer key will walk through a generic set for those presented in the assignment sheet. 1. The hypothesis was that ecologically minded commuters would have lower than average personal wage earnings because higher wage earners would use their income to "buy" convenience (parking, cars, etc.) 2. Below is the printout of COMPARE MEANS. Report 2003 HOUSEHOLD TOTAL INCOME Ecocommute_Yes_1 .00 1.00 Total Mean 73961.8999 62056.0214 71865.3201 N 3495 747 4242 Std. Deviation 78869.16008 58247.24095 75776.43231 To construct a confidence interval about the proportion of cases in the ecological commuters category we need to calculate an estimate: The point estimate is just 747/4242 = 0.176 (17.6% of people use public transportation). To get the standard deviation, S, estimate for a proportion, we will first find the variance, then take the square root. Var(proportion) = p(1-p)/n = .176*(1-.0176)/4242 = .0000342 and its square root equals 0.006. N here is 4242 because the mean is a proportion of the 4242 cases. Since this is a large sample, we just use the normal tables. For a 99% confidence interval (0.5% in each tail) we find that Z = 2.58 (or 2.576 to be exact). Now, we have all of the tools, the sample proportion (),and its standard deviation (s) P((P-2.58*S< p <P+2.58*S) = 0.99 P((0.176-2.58(0.006))< p < (0.176+2.58(0.006))) = 0.99 P(0.16 < p < .19) = 0.99 So I can say with a 99% confidence level that between 16% and 19% of households in WA are commuting ecologically. 3. To find confidence intervals for the mean outcome for each category: get SE for each category (REMEMBER THAT SPSS GIVES YOU THE SE BUT ON A QUIZ YOU MIGHT BE ASKED TO CALCULATE IT YOURSELF!) For non-eco commuters = 0 SE=S/n = 78869/3495 = 1334 For eco-commuters = 1 SE = S/n = 58247/747 = 2131 get t-s...

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