Baggage Check
Thousands of travelers pass through Guadalajara airport each day. Before leaving the airport, each passenger must pass through the Customs inspection area. Customs officials want to be sure that passengers do not bring illegal items into the country. But they do not have time to search every traveler’s luggage. Instead, they require each person to press a button. Either a red or a green bulb lights up. If the red light shows, the passenger will be searched by Customs agents. A green light means “go ahead.” Customs officers claim that the probability that the light turns green on any press of the button is 0.70. You observe 100 passengers at the Customs “stop-light.” Only 65 get a green light. Does this give you reason to doubt the Customs officials?
a.) Use your calculator to simulate 50 groups of 100 passengers activating the Customs light. Use the command randBin(100, 0.7, 50)/100 store in L1.List 1 will contain 50 values of , the proportion of the 100 passengers who got a green light.
b.) Sort L 1 in descending order. In how many of the 50 simulations did you obtain a value of that is less than or equal to 0.65? Do you believe the Customs agents?
c.) Describe the shape, center, and spread of the sampling distribution of for samples of n = 100 passengers.
d.) Use the sampling distribution from part (c) to find the probability of getting a sample proportion of 0.65 or less if p=0.7 is actually true. How does this compare with the results of your simulation in part (b)?
e.) Repeat parts (c) and (d) for samples of size n = 1000 passengers. Report your findings as compared to n = 100.
Resources:
Beth Chance's Reese's Pieces applet and sampling pennies www.rossmanchance.com/applets/index.html
David Lane's Sampling Distribution
www.ruf.rice.edu/~lane/stat_sim/sampling_dist/
Central Limit Theorem
www.whfreeman.com/tps3e
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a. When I did the simulation and put it in descending order I got
ReplyDelete.79 .78 .77 .76 .76 .75 .75 .74 .74 .74 .74 .74 .74 .74 .74 .73 .72 .71 .71 .71 .71 .71 .71 .7 .7 .69 .69 .69 .69 .69 .69 .69 .69 .69 .68 .68 .68 .68 .67 .66 .65 .64 .64 .64 .63 .61 .61 .6 .57
b. 9 are less than or equal to .65
a. i did the same thing and got
ReplyDelete.79 .78 .78 .77 .76 .76 .75 .75 .74 .74 .74 .74 .74 .74 .74 .74 .73 .73 .72 .71 .71 .71 .71 .71 .71 .7 .7 .69 .69 .69 .69 .69 .69 .69 .69 .68 .68 .68 .68 .67 .66 .65 .64 .64 .64 .63 .61 .61 .6 .57
b. i also got 9 are less then or equal to .65
b. i do believe the customs agent because when i calculated the average of my stimulation i got .7012 which is close to 70%. allison what was your average?
ReplyDeleteb. my average was .6994 which is close to 70% so i believe the customs agent too.
ReplyDeletec. the shape is pretty symetrical looking. i dont know about the center or spread.
Allison, where is the center of a symmetric distribution? The spread is your smallest value to your largest.
ReplyDeleteWhen I did RandBin(100,0.7,50)/100 and sorted the results in descending order, I got:
ReplyDelete.82 .79 .78 .77 .77 .76 .75 .75 .75 .75 .75 .75 .74 .74 .74 .73 .73 .73 .72 .72 .72 .71 .71 .7 .7 .7 .69 .69 .69 .69 .69 .69 .69 .68 .68 .67 .67 .67 .66 .66 .66 .66 .65 .64 .64 .64 .63 .63 .62 .6
b) 8 were less than or equal to .65. The average of my data was .7034, so I believe the customs agents.
c) The center is around .7. The spread is from .6 to .82
.78 .77 .76 .76 .75 .75 .74 .74 .74 .74 .74 .74 .74 .74
ReplyDelete.73 .72 .71 .71 .71 .71 .71 .71 .7 .7 .7 .7 .69 .69 .69
.69 .69 .69 .69 .69 .68 .68 .68 .68 .67 .67 .66 .65 .64
.64 .63 .61 .61 .6 .57 .56
I had 9 data below or equal to .65. I do believe the customs officer because the mean is .695 which is extremely close to .70, 70%.
When I graphed the simulation, I found that the data was very slightly skewed left. The spread is from .56 to .78 with its center around .69.
When I did the normalcdf ( .65, .7, .695, .049) I got .3614. So there is a probability of about 36.14 % for getting .65 or less.
A.)
ReplyDelete.79 .79 .77 .77 .76 .75 .75 .75 .75 .75 .75 .75 .74 .74 .73 .71 .71 .71 .71 .71 .71 .71 .7 .7 .7 .7 .7 .69 .68 .68 .68 .68 .68 .68 .67 .67 .67 .67 .65 .65 .64 .64 .64 .63 .62 .62 .6 .59 .59 .59
B.)
There were 12 numbers in my simulation that were less than or equal to .65. My average is .6944, so I believe the customs agent because this number rounds up to .7.
C.)
My graph looked symmetrical. The spread of my data is from .59 to .79. The center was around .71.
D.)
When I did normCDF, my result was .384. There is a 38.4% chance of getting a value equal to or less than .65.
the probability that p hat is less than .65 should be less than .384. Check your calculations. Remember rule of thumb tells you that this distribution is approx. normal
ReplyDeletec. my spread is .57 to .79 and my center was around .71.
ReplyDeleteA) My values are...
ReplyDelete.64 .75 .63 .68 .68 .67.75 .73 .75 .71 .7 .64
.65 .72 .75 .71 .7 .74 .69 .57 .73 .72 .72 .69 .68 .71 .74 .75 .67 .71 .7 .7 .65 .61 .69 .74 .73 .66 .71 .8 .75 .75 .68 .67 .69 .81 .75 .71 .74 .8
B) I put my list in ascending order but I did that after I typed the numbers and I didn't feel like retyping it. But I counted 8 times is the probability at or under .65. I believe the agents because there are only 8 instances where it would be probable that .65 and less percent of people are stopped. also my average is .7064 so for the most part, they are accurate.
C) My graph is quite symmetrical. The spread is .57-.81 and the center is at .71.
D) I am having trouble with this portion of the assignment!
eric, in part c you are saying it is symmetrical, is there a rule of thumb that allows you to use a normal dist? check that then draw curve and shade area you are looking for. Be sure to use correct standard deviation.
ReplyDeleteamber, could you show what you are entering in your calculator, I think your standard deviation is wrong, check text for formula
ReplyDeletems criscuolo i'm confused. i dont know what you mean?
ReplyDeleteEr yeah, I did not do the standard deviation right... at all. haha.
ReplyDelete.78 .77 .77 .76 .76 .75 .75 .75 .75 .75 .74 .73 .73 .73 .73 .72 .72 .72 .71 .71 .71 .71 .71 .7 .7 .7 .69 .69 .69 .69 .68 .68 .68 .68 .68 .68 .67 .67 .67 .66 .66 .66 .66 .66 .66 .65 .65 .65 .64 .63 .62
ReplyDeleteb. 6 of these numbers are less than or equal to .65
The average of my simulation was ..6968, which is very close to .7, so i believe the customs agent
c) center- .69, the data is symmetrical with a spread of .78-.62
d)..ill be back!
c. my center is .7 my datais pretty symetrical and my spread is .57-.79
ReplyDeletefocus on the rules of thumb. What do each of the rules allow you to do or assume? check your notes or text on standard deviations for proportions and sample means.
ReplyDeletems criscuolo i am trying to do part e (the randbin(1000, .7, 500)/1000 stored in list one) but it is taking forever. Am i typing it correctly into my calculator ? i'm not sure if it really takes this long
ReplyDeletea) .84 .8 .77 .76 .76 .76 .75 .75 .75 .75 .74 .74 .74 .74 .74 .73 .73 .73 .73 .72 .72 .72 .72 .71 .71 .71 .71 .71 .71 .7 .7 .7 .69 .69 .69 .69 .69 .68 .68 .68 .68 .67 .66 .65 .64 .64 .62 .61 .6 .58
ReplyDeleteb) 7 of my simulations were less than or equal to .65. i do believe the customs agent because i calculated the mean of my simulations to be .7078 which is very close to .7.
c) My distribution was slightly skewed right. It had a spread of .84 to .58 and a center of .71
d) i used normalcdf(.58,.65,.71) and found that the probability of getting a sample proportion less than or equal to .65 to be .0277.
ReplyDeleteNot sure if this is right...
e) when using a sample size of 1000, my data was much more reliable. the distribution was symmetric with a spread of 7.3 to 6.71 and a center of 7.0134.
ReplyDeletea)When I put my numbers in descending order I got the values
ReplyDelete.85 .79 .79 .79 .78 .77 .77 .77 .77 .76 .74 .74 .74 .74 .73 .73 .72 .71 .71 .71 .71 .7 .7 .7 .7 .69 .69 .69 .69 .69 .69 .69 .69 .69 .68 .68 .68 .68 .68 .67 .67 .67 .66 .66 .65 .64 .64 .64 .63 .61
b) 6 of the outcomes are less then or equal to .65 and I also believe the customs officers because the average of my probabilities came out to be .7074 which is close to their .70 probability.
ReplyDeletee) i agree with don. When my calculator was finally done with its simulation, the data was a lot more symmetric. The mean was .700656, so i believe the customs agent even more with the bigger sample size. The distribution was from .658 to .7318. The distribution of the larger sample size was smaller and had less variability than the distribution of the smaller sample size, making the data more reliable.
ReplyDeletec)The center for my data was at .69, the spread is from .61 to .85 and my data appeared to be skewed to the right.
ReplyDeletePart (A): randBin(100,0.7,50)/100 store in L1
ReplyDelete.73 .75 .75 .69 .71 .58 .74 .72 .72 .7 .76 .65 .59 .73 .72 .72 .69 .71 .69 .73 .72 .72 .68 .73 .67.66 .61 .69 .72 .73 .67 .71 .79 .74 .76 .7 .71 .67 .78 .75 .74 .7 .82 .73 .72 .83 .65 .69 .69 .65
Part (B): Numbers (-) or (=) to .65: four
Part (C): spread .58 to .83 center:.69 The graph was symmetrical.
Part (c)
Part D: NormalCDF(.58, .65, .69) = .0278 <--probability of getting a sample proportion of .65 or less.
ReplyDeleteD)using the center that i got before, i used normalcdf(.65,.7,.695,.049) and that gave me the answer .36142 or changed to a percent 36.142%. This means that there is about a 36 percent chance of getting lower then a .65.
ReplyDeletePart (E): When using a sample of 1000 the data is more correct because the sample size is increased; the standard deviation is then decreased, and the sampling distribution is closer to normal. My calculator is not responding right now when using a sample of 1000...i guess its too much data for it xD. But according to Amanda's data from part e. the data is more reliable with the larger sample size since the distribution and the variability came out smaller.
ReplyDeletee)When I used 1000 for the sample size, the data came out to be much closer to the customs officers probability. The mean was .70021 and the distribution was from about .66 to .735. This shows how the greater the number of people, the more accurate the custom officers are.
ReplyDeletea)When I put my numbers in descending order I got the values
ReplyDelete.85 .79 .79 .79 .78 .77 .77 .77 .77 .76 .74 .74 .74 .74 .73 .73 .72 .71 .71 .71 .71 .7 .7 .7 .7 .69 .69 .69 .69 .69 .69 .69 .69 .69 .68 .68 .68 .68 .68 .67 .67 .67 .66 .66 .65 .64 .64 .64 .63 .61
b) 6 of the outcomes are less then or equal to .65 and I also believe the customs officers because the average of my probabilities came out to be .7074 which is close to their .70 probability.
c)The center for my data was at .69, the spread is from .61 to .85 and my data appeared to be skewed to the right.
d)Using the center that i got before, i used normalcdf(.65,.7,.695,.049) and that gave me the answer .36142 or changed to a percent 36.142%. This means that there is about a 36 percent chance of getting lower then a .65.
e)When I used 1000 for the sample size, the data came out to be much closer to the customs officers probability. The mean was .70021 and the distribution was from about .66 to .735. This shows how the greater the number of people, the more accurate the custom officers are.
Also, the standard deviation to the sample size 1000 is .01404 and the standard deviation of the sample size of 100 was .04548, about 4 times as large. So the data is again, more reliable with a bigger sample size.
ReplyDeleteA.) In descending order: .83, .78, .77, .76, .76, .76,.76,.74,.73,.73,.73,.73,.72,.72,.72,.72,.72,.72,.72,.72,.72,.72,.71,.71,.71,.70,.70,.69,.69,.69,.69,.69,.69,.69,.68,.68,.67,.67,.67,.66,.65,.65,.65,.65,.65,.64,.63,.62,.60,.57
ReplyDeleteB.) Less than or equal to 0.65 = 10 numbers. When I looked at my average, it came out to be .6996; this value is close to .70, thus I do believe the Customs agent.
C.) The shape is symmetrical. The center is .705. The spread is .57 to .83.
D.) Normalcdf(.57, .65, .71) = .0317
E.) The data was much more reliable. The graph was symmetric. The mean was .7015. With a bigger sample size, the distribution has less variability. The spread was from .6538 to .7317.
d. normalcdf(.57,.65,.71) = .03175 probability of getting a sample propotion less then or equal to .65
ReplyDeletee. my data was more reliable and the graph was symmetrical looking. my mean was .7014. the spread was .671 to .736.
ReplyDeletea) in descending order my data is : .78 .78 .76 .76 .75 .75 .74.74 .74 .73 .73 .73 .72 .71 .7 .7 .7 .69 .69 .69.69 .69 .68 .68 .68 .68 .68 .67 .67 .67 .67 .66 .66 .66 .66 .66 .66 .66 .65 .64 .64 .64 .64 .64 .63 .63 .61 .6 .57 .54
ReplyDeleteb) Less than or equal to 0.65 is 12. When I calculated the mean it came to equal .682. This is extremely close to .7, so I too also believe the customs officer.
c) The center for my data is at .68. The spread is from .54 to .72. The data appears to be partially symmetrical but at the same time skewed more so to the right.
d) I decided to use normalcdf(.65,.7,.68,.049). This came to equal .3882 of 38.82%. There is a 38.82% chance of getting lower than .65
e) working on it!
ok so for part e my calculator would not do it! i didnt know how much longer i should wait! part d i did normalcdf(.65,.7, .7064, .0477) and got .3281
ReplyDeletethe rule of thumb allows you to ensure that normal calculations are accurate enough for statistical purposes. The 50 to 100 does not show an accurate calculation in the first thumb rule. Using a bigger number to work from will give you a more accurate set of data.
ReplyDeletee) after waiting forever for the calculator to finish simulating, you can see from the results that the variability is not as strong. The results from the 1000 are more accurate and precise. The graph has a more symmetrical look to it. The standard deviation is .017029, the mean became .69, and the min and max range became much smaller, The min = .657 and max =.738.