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This file contains worksheets for each of the major problems we will discuss in ISDS 2710 Discrete (Counting… How many?) Binomial (Fill in the basic idea here.) Hypergeometric (Fill in the basic idea here.) Poisson (Fill in the basic idea here.) Continuous (Measuring… How much?) Uniform (Fill in the basic idea here.) Exponential (Fill in the basic idea here.) Normal (Fill in the basic idea here.) sampling sample means (Fill in the basic idea here.) sample proportions (Fill in the basic idea here.) confidence intervals (Fill in the basic idea here.) means σ known (Fill in the basic idea here.) σ unknown (Fill in the basic idea here.) proportions Hypothesis testing (Fill in the basic idea here.) BACK TO MAIN BINOMIAL WORKSHEET Problem description: (Put a brief summary of your problem here.) BASIC IDEA: Parameters: n =#trials 4 p = probability of success on each trial 0.2 x = #successes of interest 1 Computation f(x) = probability of exactly x successes 0.4096 F(x) = probability of no more than x successes 0.8192 E(x) = expected #successes =n*p 0.8 This is the expected #fries I expect to sell given 8 customers. variance = n*p*(1-p) 0.64 stdev 0.8 CV 100% NOTE: My standard is very large! Discussion Once you have calculated your results, I want two statements 1. A summary of the results 2. A discussion of how you will use this for the problem at hand Example#1 BINOMIAL WORKSHEET Problem description: Suppose that I've been collecting data at Chick-fil-A I notice that 15% of all customers buy french fries suppose that 8 customers will come in today. (we'll keep it small!) How many fries will we sell? BASIC IDEA: Success= buying french fries; trial = one customer Parameters: n =#trials 8 p = probability of success on each trial 0.15 x = #successes of interest 1 Computation f(x) = probability of exactly x successes 0.384693 F(x) = probability of no more than x successes 0.657183 E(x) = expected #successes =n*p 1.2 variance = n*p*(1-p) 1.02 stdev 1.00995 CV 0.841625 Let's calculate the probabilities for each possibility #fries sold Probability 0 0.272 Make sure you understand the formulas. 1 0.385 Create a column graph of this data. 2 0.238 3 0.084 4 0.018 5 0.003 6 0.000 7 0.000 8 0.000 let's also calculate cumulative probabilities: #fries sold 0 0.272 Make sure you understand the formulas. 1 or less 0.657 Create a column graph of this data. 2 or less 0.895 3 or less 0.979 4 or less 0.997 5 or less 1.000 6 or less 1.000 7 or less 1.000 8 or less 1.000 Questions: 1. Why is it generally more useful to know the cumulative probability, rather than the probability for a single value? 2. Why is this even more evident as the number of trials increases? GOING ONLINE Find an article based on the theme, which gives some suggestion of trials and successes. Using data from the article, if available, (and possibly scaled down), set up your own Binomial. Be prepared to share in the discussion. BACK TO MAIN (I've just put in some sample #s.) t to sell given 8 customers. own Binomial. BACK TO MAIN HYPERGEOMETRIC WORKSHEET Problem description: (Put a brief summary of your problem here.) BASIC IDEA: Parameters: BUCKET_INFO Pull N = #items in total 8 n =#items pulled n1 = #type 1 items 5 x1 = #type 1 pulled n2 = #type 2 items 3 x2 = #type 2 pulled Computation f(x1,x2) = probability of pulling out x1 type 1 and x2 type 2 NOTE: It is very easy to have more th 0.428571 Each type adds an addition combina in the numerator Discussion Once you have calculated your results, I want two statements 1. A summary of the results 2. A discussion of how you will use this for the problem at hand Example#1 HYPERGEOMETRIC WORKSHEET Suppose we want to look at the possible outcomes when we have 12 people in the class suppose that 8 have previously taken an online class, and 4 have not. If we pick a random group of 4 people, what is the likelihood that 3 members of the group have been online before? BASIC IDEA: Parameters: BUCKET_INFO Pull N #students In the class 12 n #in the group n1 #who have been online previously 8 x1 #online in grp. n2 #who have no prior online experience 4 x2 #not_on in grp. Computation f(3,1) = probability of pulling out 3 online students, and 1 with no online experience 0.452525 Discussion 1. A summary of the results The chance of such a group is 45% 2. A discussion of how you will use this for the problem at hand Having a group made up of a good # of online people is likely. (To think about: How is this relevant for this class?) GOING ONLINE Find an article based on the theme, which gives some suggestion of different types of items Using data from the article, if available, (and possibly scaled down), set up your own Hypergeometric Be prepared to share in the discussion. BACK TO MAIN =#items pulled = 4 = #type 1 pulled = 3 = #type 2 pulled = 1 very easy to have more than 2 types adds an addition combination term n online before? #in the group = 4 #online in grp. = 3 #not_on in grp. = 1 pergeometric POISSON WORKSHEET suppose I'd like to know the probability of 24 arrivals in 3 days, when I expect 12/day BASIC IDEA: Parameters: Expected # arrivals in a period of length one day 12 µ = scaled expectation, if necessary, for the time period of interest 36 x=#arrivals of interest 24 Computation f(x)= likelihood of x arrivals in the time period f(24)= 0.008394 F(x) = likelihood of x or fewer arrivals in the time period (Again, consider why it is usually more helpfu F(24)= 0.022446 F(x) than f(x).) Expected # arrrivals 12 variance 12 standard deviation 3.464102 CV 29% NOTE: For the Poisson, we will have a large s Discussion Once you have calculated your results, I want two statements 1. A summary of the results 2. A discussion of how you will use this for the problem at hand Example#1 POISSON WORKSHEET Suppose that the University has collected information, and has noticed that on average, 160 new students sign up for online p each semester. What is the likelihood that no more than 175 new students join in the fall? BASIC IDEA: Parameters: Expected # new students each semester 160 µ = scaled expectation, if necessary, for the time period of interest 160 x=#arrivals of interest 175 Computation f(175)= likelihood of exactly 175 f(175)= 0.015238 F(175) = likelihood of 175 or less F(175)= 0.888642 NOTE: This also tells me the likelihood of more than 175 new students: = 1-F(175) 0.111358 Discussion Once you have calculated your results, I want two statements 1. There is an 89% chance that enrollment will be 175 or less 2. this will help me plan for very large, or very small, classes the chance of a large class, 10%, is too large to be ignored. GOING ONLINE Find an article based on the theme, which gives some suggestion of "arrivals." Using data from the article, if available, (and possibly scaled down), set up your own Poisson Be prepared to share in the discussion. BACK TO MAIN (I've put in sample values) (=12/day*3days) sider why it is usually more helpful to know the Poisson, we will have a large standard deviation. 0 new students sign up for online programs Let's look at the probabilities 1. Create a column graph of the d for a few values of x 2. what can you say about the sh x f(x) 3. How does the mean relate to t MORE DETAIL===>! 0 0 10 0 20 0 30 0 40 0 50 0 60 0 70 0 80 9.72E-13 90 5.15E-10 100 9.01E-08 110 5.82E-06 120 0.000152 130 0.001729 140 0.009132 150 0.023658 160 0.031523 170 0.022516 180 0.008944 190 0.002041 200 0.000275 210 2.26E-05 own Poisson 220 1.15E-06 230 3.72E-08 240 7.8E-10 250 1.08E-11 260 1E-13 270 0 280 0 290 0 300 0 310 0 320 0 330 0 340 0 350 0 360 0 370 0 380 0 390 0 400 0 410 0 420 0 1. Create a column graph of the data. 2. what can you say about the shape? 3. How does the mean relate to the graph? UNIFORM WORKSHEET Description of problem: Suppose we know that the typical homeowner uses at least 100 gallons of water each day, but no more than 200 what is the likelihood of using between 125 and 140 gallons? BASIC IDEA: Parameters: a = minimum 100 b = maximum 200 lowerlimit 125 upperlimit 175 Computation f(lowerlimit,upperlimit)= likelihood of an outcome between the lowerlimit and upperlimit) f(125,175)= 0.5 E(outcome) = (a+b)/2 150 variance = (b-a)^2/12833.3333 std. dev 28.86751 CV 19% (since the CV is above 10% , our stdev is large) Discussion Once you have calculated your results, I want two statements 1. A summary of the results 2. A discussion of how you will use this for the problem at hand Example#1 UNIFORM WORKSHEET Description of problem: Suppose that I believe that an online student will spend between 1 and 8 hours per week on class material what is the likelihood of a student spending less that 3 hours on class? BASIC IDEA: Parameters: a = minimum 1 b = maximum 8 lowerlimit 1 upperlimit 3 Computation f(lowerlimit,upperlimit)= likelihood of an outcome between the lowerlimit and upperlimit) f(125,175)= 0.285714 E(outcome) = (a+b)/2 4.5 variance = (b-a)^2/124.083333 std. dev 2.020726 CV 45% (since the CV is well above 10% , our stdev is large) Discussion 1. 28%of students will spend no more than 3 hours on the class, if my estimates are correct. 2. I might need to encourage more online effort. I should also note that the standard deviation is very large; a lot GOING ONLINE Find an article based on the theme, which gives some suggestion of a continuous problem. Using data from the article, if available, (and possibly scaled down), set up your own uniform Be prepared to share in the discussion. BACK TO MAIN We might use the uniform, based on a sample. Suppose that we had surveyed Their daily usages are shown below 149 191 107 189 r each day, but no more than 200 167 164 113 164 151 126 127 159 134 134 160 190 MORE DETAIL 135 197 145 162 163 187 171 117 suppose we create a histogram of the data: Bins 100 120 140 160 180 200 Bin Frequency 100 0 120 10 15 140 3 Frequency 10 160 9 180 6 5 200 8 0 dev is large) More 0 100 100 Since there doesn't appear to be any real pattern, other than all bet likely outcomes. (Of course, larger samples might tell us more!) per week on class material our stdev is large) are correct. ndard deviation is very large; a lot of variability among student participation should be expected. own uniform e. Suppose that we had surveyed 36 customers 154 176 120 112 144 150 161 139 177 151 149 125 Histogram water usage/day Frequency 100 140 140 100 120120 160 160 More 200 180 200 180 More quantity(gallons) Bin any real pattern, other than all between 100 and 200, we assume equally r samples might tell us more!) EXPONENTIAL WORKSHEET Description of problem: suppose that we know that there is an average of 1.5 minutes between arrivals What is the likelihood the next arrival is less than 30 seconds(half a minute) later? BASIC IDEA: Parameters: µ 1.5 x 0.5 Computation F(x) = probability that the next arrival is within x time periods F(.5) 0.283469 E(outcome) =µ 1.5 variance =µ2 2.25 std. dev 1.5 CV 100% (since the CV is 100% , our stdev is large) (NOTE: for the exponential, the standard deviation will always be la Discussion Once you have calculated your results, I want two statements 1. A summary of the results 2. A discussion of how you will use this for the problem at hand Example#1 EXPONENTIAL WORKSHEET Description of problem: Suppose that I believe that an online student will send me an email question every 3 hours, on average. what is the likelihood of there being less than 1 hour between successive emails? BASIC IDEA: an arrival is an email message. Parameters: µ 3 x 1 Computation F(x) = probability that the next arrival is within x time periods F(1) 0.283469 E(outcome) =µ 3 variance =µ2 9 std. dev 3 CV 100% Note: for the Exponential, the CV is always 100%! Discussion 1. There is a 28% chance there will be less than an hour between successive emails. 2. It it likely that many emails will come in frequently, and only rarely will there be a long gap. GOING ONLINE Find an article based on the theme, which gives some suggestion of "arrivals." Using data from the article, if available, (and possibly scaled down), set up your own Exponential Be prepared to share in the discussion. BACK TO MAIN Do you remember that example we saw in the demo_numerical file MORE DETAIL the average time was 1.5 minutes Suppose that we had a larger sample, say 100 values: 1.09 0.59 0.20 4.04 1.97 1.81 0.93 1.94 0.06 0.79 0.54 2.03 0.48 2.51 0.59 0.62 0.66 1.14 1.74 2.61 1.03 2.85 0.43 0.65 0.15 0.82 0.28 0.07 0.17 2.72 3.14 3.45 0.77 2.63 0.13 2.03 8.48 0.06 5.41 0.66 we could calculate the mean 1.58 1. Create a histogram of the sample data, using 25-30 bins. andard deviation will always be large.) 1.2 ery 3 hours, on average. 1 0.8 0.6 0.4 0.2 0 1 2 3 1 2 3 be a long gap. own Exponential we saw in the demo_numerical file, about time between arrivals? ple, say 100 values: 2.13 0.18 0.67 4.72 2.60 2.74 0.92 1.26 1.12 1.26 0.94 0.62 2.65 2.18 2.58 1.91 0.17 0.33 0.67 0.87 3.40 0.46 5.04 1.61 2.08 3.54 0.26 5.40 0.23 1.27 2.33 0.21 0.36 0.05 0.56 1.13 0.02 0.29 0.34 1.51 1.71 1.23 2.92 2.03 1.82 0.07 1.76 1.23 2.63 1.97 0.36 0.85 1.43 1.51 4.03 7.50 0.18 0.29 0.32 0.91 And the standard deviation 1.561161 le data, using 25-30 bins. Histogram 1.5 Frequency 1 0.5 Frequency 0 Bin Series1 Series2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 NORMAL WORKSHEET Description of problem: Suppose we know that online students tend to surf the web an average of 9 hours/week, with a standard deviation of 5 hours. What is the likelihood of spending between 9 and 12 hours on the web? BASIC IDEA: Parameters: µ 9 σ 5 x 12 Computation steps 1. graph the points, and shade the area of interest (shaded in light blue) 2. calculate z = (x-µ)/σ = 0.6 3. Look up F(z) F(.6) = 0.2257 4. Shade in F(z) (shaded in red) 5. determine the probability of interest Since the blue shaded area and red shaded areas µ=9, σ=5 match up, the probability of interest is: 0.2257 6.Discussion Once you have calculated your results, I want two statements 1. A summary of the results 2. A discussion of how you will use this for the problem at hand Example#1 Description of problem: Suppose we know that online students tend to surf the web an average of 9 hours/week, with a standard deviation of 5 hours. Now, suppose we want to know the likelihood that a student surfs more than 12 hours/week. BASIC IDEA: Parameters: µ 9 σ 5 x 12 Computation steps 1. graph the points, and shade the area of interest (shaded in light blue) (Notice how this area differs from the previous example.) 2. calculate z = (x-µ)/σ = 0.6 3. Look up F(z) F(.6) = 0.2257 4. Shade in F(z) (shaded in red) µ=9, σ=5 5. determine the probability of interest Since the blue shaded area and red shaded areas complement, we subtract = .5-.2257= 0.2743 6.Discussion Once you have calculated your results, I want two statements 1. A summary of the results 2. A discussion of how you will use this for the problem at hand Example#2 Description of problem: Suppose we know that online students tend to surf the web an average of 9 hours/week, with a standard deviation of 5 hours. Now, suppose we want to know the likelihood that a student surfs less than 6 hours/week. BASIC IDEA: Parameters: µ 9 σ 5 x 6 Computation steps x=6 1. graph the points, and shade the area of interest x=6 (shaded in light blue) (Notice how this area differs from the previous example.) 2. calculate z = (x-µ)/σ = (Notice z is negative, but -0.6 remember, normal is 3. Look up F(z) symmetric around µ) F(.6) = 0.2257 4. Shade in F(z) (shaded in red) x=6 µ=9, σ=5 5. determine the probability of interest Since the blue shaded area and red shaded areas complement, we subtract = .5-.2257= 0.2743 6.Discussion Once you have calculated your results, I want two statements 1. A summary of the results 2. A discussion of how you will use this for the problem at hand Example#3 Description of problem: Suppose we know that online students tend to surf the web an average of 9 hours/week, with a standard deviation of 5 hours. Now, suppose we want to know the likelihood that a student surfs between 6 and 10 hours/week. NOTE: We have two x-values now, so we're going to deermine 2 z-values. BASIC IDEA: Parameters: µ 9 σ 5 x1 6 x2 10 Computation steps 1. graph the points, and shade the area of interest (shaded in light blue) 2. calculate z1 : = (x1-µ)/σ = -0.6 and z2: = (x2-µ)/σ = 0.2 x=6 µ=9, x=10 σ=5 3. Look up F(z1) F(.6) = 0.2257 and F(z2) F(.2) = 0.0793 4. Shade in F(z1) (shaded in red) and F(z2) (shaded in greeen) 5. determine the probability of interest since the blue-shaded area = (red-shaded area) + (green-shaded area) we add the probabilities = ..2257+ .0793 0.31 6.Discussion Once you have calculated your results, I want two statements 1. A summary of the results 2. A discussion of how you will use this for the problem at hand Example 4. - YOUR TURN SOMETHING TO THINK ABOUT Notice that the Standard normal table only includes z values up to 3.09 Notice the pattern of the probabilities. Why is it satisfactory to conclude that if z>3.09, then F(z)=.5 Description of problem: Suppose we know that online students tend to surf the web an average of 9 hours/week, with a standard deviation of 5 hours. Now, suppose we want to know the likelihood that a student surfs between 11 and 14 hours/week. BASIC IDEA: Parameters: µ 9 σ 5 x1 x2 Computation steps x=6 x=10 µ=9, 6.Discussion Once you have calculated your results, I want two statements 1. A summary of the results 2. A discussion of how you will use this for the problem at hand GOING ONLINE Find an article based on the theme, which gives some suggestion of a symmetric continuous variable. Using data from the article, if available, (and possibly scaled down), set up your own Normal Be prepared to share in the discussion. BACK TO MAIN Standard Normal STANDARD NORMAL TABLE 0 0.01 0.02 0 0 0.004 0.008 0.1 0.0398 0.0438 0.0478 0.2 0.0793 0.0832 0.0871 0.3 0.1179 0.1217 0.1255 0.4 0.1554 0.1591 0.1628 0.5 0.1915 0.195 0.1985 0.6 0.2257 0.2291 0.2324 0.7 0.258 0.2611 0.2642 0.8 0.2881 0.291 0.2939 0.9 0.3159 0.3186 0.3212 1 0.3413 0.3438 0.3461 1.1 0.3643 0.3665 0.3686 1.2 0.3849 0.3869 0.3888 1.3 0.4032 0.4049 0.4066 1.4 0.4192 0.4207 0.4222 1.5 0.4332 0.4345 0.4357 1.6 0.4452 0.4463 0.4474 1.7 0.4554 0.4564 0.4573 1.8 0.4641 0.4649 0.4656 1.9 0.4713 0.4719 0.4726 x=12 2 0.4772 0.4778 0.4783 2.1 0.4821 0.4826 0.483 2.2 0.4861 0.4864 0.4868 2.3 0.4893 0.4896 0.4898 2.4 0.4918 0.492 0.4922 2.5 0.4938 0.494 0.4941 2.6 0.4953 0.4955 0.4956 2.7 0.4965 0.4966 0.4967 2.8 0.4974 0.4975 0.4976 2.9 0.4981 0.4982 0.4982 hours/week. 3 0.4987 0.4987 0.4987 x=12 d 10 hours/week. x=10 nd 14 hours/week. x=10 continuous variable. 0.03 0.04 0.05 0.06 0.07 0.08 0.09 BACK TO NORMAL 0.012 0.016 0.0199 0.0239 0.0279 0.0319 0.0359 BACK TO SAMPLE MEANS 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753 BACK TO SAMPLE PROPORTIONS 0.091 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141 BACK TO CI_MEANS_σKNOWN 0.1293 0.1331 0.1368 0.1406 0.1443 0.148 0.1517 BACK TO CI_MEANS_σUNKNOWN 0.1664 0.17 0.1736 0.1772 0.1808 0.1844 0.1879 BACK TO CI_PROPORTIONS 0.2019 0.2054 0.2088 0.2123 0.2157 0.219 0.2224 BACK TO HT 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133 0.3238 0.3264 0.3289 0.3315 0.334 0.3365 0.3389 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621 0.3708 0.3729 0.3749 0.377 0.379 0.381 0.383 0.3907 0.3925 0.3944 0.3962 0.398 0.3997 0.4015 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319 0.437 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706 0.4732 0.4738 0.4744 0.475 0.4756 0.4761 0.4767 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817 0.4834 0.4838 0.4842 0.4846 0.485 0.4854 0.4857 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.489 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952 0.4957 0.4959 0.496 0.4961 0.4962 0.4963 0.4964 0.4968 0.4969 0.497 0.4971 0.4972 0.4973 0.4974 0.4977 0.4977 0.4978 0.4979 0.4979 0.498 0.4981 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.4986 0.4988 0.4988 0.4989 0.4989 0.4989 0.499 0.499 BACK TO NORMAL BACK TO SAMPLE MEANS BACK TO SAMPLE PROPORTIONS BACK TO CI_MEANS_σKNOWN BACK TO CI_MEANS_σUNKNOWN BACK TO CI_PROPORTIONS BACK TO HT SAMPLE MEANS WORKSHEET Description of problem: Suppose we know that a typical student spends an average of 20 hours/week on school work,σ=8 hours. Suppse that a survey of 10 online students showed an average of 23 hours/week. Is this a significant difference? BASIC IDEA: Parameters: µ population mean 20 n sample size σ population std. dev 8 xbar sample mean 23 Computation 1. Since this is a sampling problem, we first calculate standard deviation of sample distribution σn = σ/sqrt(n) 2.529822 2. Calculate z: = (xbar-µ)/σn 1.185854 3. Look up F(z) in standard normal table 0.383 4. Discuss, using the 40% rule µ=20, 1. Since F(z) < .40, we conclude that the difference is not significant 2. Online student work habits are not very different from regular students. GOING ONLINE Find an article based on the theme, which gives some suggestion of a sample Using data from the article, if available, (and possibly scaled down), set up your own sample means problem. Be prepared to share in the discussion. BACK TO MAIN GOTO STD NORMAL TABLE n school work,σ=8 hours. 10 µ=20, σ=2 xbar=12 own sample means problem. SAMPLE PROPORTIONS WORKSHEET Description of problem: Suppose we know that a typical student is responsible for 25% of his/her tuition, on average. Suppse that a survey of 10 online students showed that, on average, they are responsible for 36% of tuition. Is this a significant difference? BASIC IDEA: Parameters: p population proportion 0.25 pbar sample proportion 0.36 n sample size 10 Computation 1. Since this is a sampling problem, we first calculate standard deviation of sample distribution σn = sqrt(p*(1-p)/n) 0.136931 2. Calculate z: = (pbar-p)/σn 0.803326 3. Look up F(z) in standard normal table 0.2881 4. Discuss, using the 40% rule µ=20, 1. Since F(z) < .40, we conclude that the difference is not significant 2. Online student's contribution to tuition payments are not very different from regular students. GOING ONLINE Find an article based on the theme, which gives some suggestion of a sample Using data from the article, if available, (and possibly scaled down), set up your own sample proportions problem Be prepared to share in the discussion. BACK TO MAIN GOTO STD NORMAL TABLE , on average. sponsible for 36% of tuition. µ=20, σ=2 xbar=12 regular students. own sample proportions problem. CONFIDENCE INTERVAL - MEANS σ KNOWN WORKSHEET Description of problem: Suppose that we have surveyed 10 students, and found that the average courseload is 14 hours. Suppose we know that σ=4 hours. what is the 90% (2-sided) confidence interval? (Why 2-sided? In this case, we are concerned on both sides: if BASIC IDEA: Parameters: xbar sample mean 14 σ population std. dev 4 n sample size 10 CL confidence level 80% Computation 1. To start, we need to determine a z-value, based on the confidence level: σn = sqrt(p*(1-p)/n) #NUM! 2. Calculate z: = (pbar-p)/σn #NUM! 3. Look up F(z) in standard normal table 0.2881 4. Discuss, using the 40% rule µ=20, 1. Since F(z) < .40, we conclude that the difference is not significant 2. Online student's contribution to tuition payments are not very different from regular students. GOING ONLINE Find an article based on the theme, which gives some suggestion of a sample Using data from the article, if available, (and possibly scaled down), set up your own sample proportions problem Be prepared to share in the discussion. BACK TO MAIN GOTO STD NORMAL TABLE z-value TYPICAL CONFIDENCE LEVELS Confidence level 2-sided 80% 1.28 load is 14 hours. 90% 1.64 95% 1.96 99% 2.58 µ=20, σ=2 xbar=12 regular students. own sample proportions problem. z-value 1-sided 0.84 1.28 1.64 2.33 When σ is unknown, there are a couple fof slight changes that must be made to the worksheet. Copy over the worksheet for CI, when σ is known, and make the appropriate changes For a CI-proportions problem, there are a couple fof slight changes that must be made to the worksheet. Copy over the worksheet for CI, when σ is known, and make the appropriate changes e worksheet.