Building Better Batteries
Everyone wants to have the latest technological gadget. That’s why iPods, digital cameras, PDAs, Game Boys, and camera phones have solid millions of units. These devices require lots of power and can drain traditional alkaline batteries quickly. Battery manufacturers are constantly searching for ways to build longer-lasting batteries. In July 2005, Panasonic began marketing its new Oxyride battery in the United States. According to the results of preliminary testing, Oxyride batteries produced more power and lasted up to twice as long as alkaline batteries.
Battery manufacturers must constantly measure battery lifetimes to ensure that their production process is working properly. Because testing a battery’s lifetime requires the battery to be drained completely, the manufacturer wants to test as few batteries as possible. As part of the quality control process, the manufacturer selects a sample of batteries to test at regular intervals throughout production. By looking at the results from the sample, the manufacturer can determine whether the entire batch of batteries produced meets specifications.
At a particular battery production plant, when the process is working properly, AA batteries last an average of 17 hours with a standard deviation of 0.8 hour. Quality control inspectors select a random sample of 30 batteries during each hour of production and then drain them under conditions that mimic normal use. Here are the lifetimes (in hours) of the batteries from one such sample:
16.91 18.83 17.58 15.84 17.42 17.65 16.63 16.84 15.63 16.37
15.80 15.93 15.81 17.45 16.85 16.33 16.22 16.59 17.13 17.10
16.96 16.40 17.35 16.37 15.98 16.52 17.04 17.07 15.73 16.74
Do theses data suggest that the process is working properly?
1. Assuming the process is working properly (mean = 17, standard deviation =0.8), what are the shape, center, and spread of the distribution of sample means for random samples of 30 batteries? Justify each of your answers.
2. For the random sample of 30 batteries, =16.70 hours. Calculate the probability of obtaining a sample with a mean lifetime of 16.70 hours or less if the production process is working properly. Show your work.
3. Based on your answer to Question 2, do you believe that the process is working properly? Justify your answer.
The plant manager also wants to know what proportion of all the batteries produced that day lasted less than 16 hours, which he has declared "unsuitable." From past experience, about 10% of batteries made at the plant are unsuitable. If he can feel relativley certain that the proportion of unsuitable batteries "p" produced that day is less that 0.10, he might take the chance of shipping the whole batch of batteries to customers.
4. Assuming that the actual proportion of unsuitable batteries produced that day is 0.10, what are the shape, center, and spread of the distribution of sample porportions for random samples of 480 batteries? Justify each of your answers.
5. On this particular day, a total of 480 batteries were tested during 16 hours of production. Forty of these batteries were unsuitable. Calculate the probability of obtaining a sample in which such a samll proportion of batteries is unsuitable if p = 0.10. Show your work.
6. Based on your answer to question 5, what advice would you give the plant manager? Why?
Resources:
www.stat.sc.edu/~west/javahtml/CLT.html
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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hello group....How is everyone doing????
ReplyDeleteok since everyone else is doing it in invisible font....I will do it in regular font.....
ReplyDeleteok so i plugged in the data...graphed it and i got that the graph seems to be skewed right and there is seems to be a outlier....I also did vars 1 of the data and got
xbar: 16.7
Sx: .712
tetax:.700
min: 15.63
med::16.685
max: 18.83
1)Assuming the process is working properly, the shape is skewed right, when you plug in the data the center is 16.7, the spread is 3.2 because you subtract 15.63 from 18.83.
ReplyDelete2)to get the standard deviation you divide .8 by the sqaure root of 30 and you get .1461. Then to calculate the probability you tdo normal cdf (10^-99,16.7,17,.1461) and you get .02
3) Based on my answer to question 2 I do not believe that the process is working properly because the probability of getting a sample size so small is unlikely.
ReplyDelete4)When you plug in your numbers to the standard deviation equation, you get .0137. The square root of p(1-p)/n. The rule of thumb justifies this because our population is at least 10 times as large as the sample.
5)For this problem you use normalcdf(10^-99,40/480,.1,.0137) which equals .1119
ReplyDelete6) My advice to the plant manager is don't send out these batteries because they suck.
this isn't my class but i love jacquie <3
ReplyDeleteOk so I finally figured out how to comment on this thing. So to question one I also figured out thAt it is skewed right but not extremly because of the graph being that the skeweness it slanted and that the numbers are close to each other as in 16.8 and 17. For question two , since Jackie has written what I got I will say in words and evalutate it by saying that the probability is not high because it was only 20 percent.
ReplyDeleteFor question three. For the probAbilty of .20 I'd agree with Jackie in saying it is not working because the probabilty of .20 is not very accurate at all.
ReplyDeleteWoooozz.... What is question four saying ? Help please anyone :)
ReplyDeleteSorry guys but you got the speciAl one in ur group. Does number five mean how what is the probabilty of getting a sAmple of forty batteries with a .10 in a population of 480. And do we disregard the 16 hours?
ReplyDeleteSince.... I don't get five.my advice is to make an easier problem .
ReplyDeleteIn five is probabilty the same thing as proportoion?
ReplyDeleteSO i finally figured out how to do this...if you want to learn or know.....well ask me in school
ReplyDelete