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 everyone :)
ReplyDelete-Courtney Acker
Hello llama (:
ReplyDeleteCrap, this problem is not fun.
I made a curve with the standard deviation and the mean. I'll up load it to facebook in a minute.
Now, maybe a smart person can take it from here?
http://i956.photobucket.com/albums/ae42/jcifproductions/Jackie statsblogspot.jpg
ReplyDeleteAs for the data, 20 out of 30 of the samples fall to the left of the mean, so is it right skewed?
ReplyDelete& there isn't much of a range because the data is all fairly close together...
Hii Sniffers !!!! :)
ReplyDeleteyeahh the data is skewed right. The center is about 16.70, i think. The spread is the max-min which would be 18.83-15.63=3.2
I hope thats right lol
-Courtney Acker
For question #2 i did normalcdf(-10^99,16.70,17,.8) and got .3538. I did the lower bound which is -10^99, the upper bound which is 16.70, the mean which is 17, and the standard deviation which is .8.
ReplyDelete-Courtney Acker
For question #3, I don't think the process is working correctly because shouldn't the answer for number two be close to 50% since the mean is 17? What do you guys think??
ReplyDelete-Courtney Acker
look to the central limit theorem to help you justify your answer in part 1. What does the central limit theorem suggest?
ReplyDeletecourtney, your standard deviation is incorrect. How do you calculate the standard deviation of a sample mean from a normal population?
ReplyDeletehello everyone :)
ReplyDeletems. criscuolo I'm confused, how is her standard deviation incorrect?
& I do not understand the central limit theorem.
View the applet on central limit theorem. Review the rules of thumb, what do these concepts allow you to assume about your sample. How are the sample standard deviations related to the population standard deviations?
ReplyDeleteI do not understand any of this /:
ReplyDeleteme 2
DeleteWhat is the central limit theorem and what does it suggest? How do the two rules of thumb apply to this problem? How do you determine the standard deviation of the sampling distribution of p hat.
ReplyDeleteThe central limit theorem suggests that when n is large, the sampling distribution of the sample mean (x bar) is close to the normal distribution N(mu, sigma divided by the square root of n). The stnd. dev. is sigma divided by the square root of n. Rule of thumb 1 says that the population is at least 10 times as large as the sample and rule of thumb 2 is np is greater than or equal to 10 and n(1-p) is greather than or equal to 10.
ReplyDeleteI understand all that i just dont get how to get p or n.
-Courtney Acker
Also, the standard deviation of a sampling distribution is the square root of p(1-p) divided by n.
ReplyDelete-Courtney Acker
1. the shape is normal. using the central limit theorem, we get the distribution of sample means is normal with same mean = 17 which is also the center. variance = (sigma) to the second power over n which is (.8)^2 over 30 which equals .02133. this variance is the measure of spread.
ReplyDelete2. P(x>16.70)
p(z>16.70-17 over .146) stddev = sqrt (.02132)
=p(z>-2.0547)=.9798
shape = skewed right
ReplyDeletefor number two we are finding the probability that the mean lifetime is less than or equal to 16.70.
ReplyDeletenormalcdf(10^-99,16.70,17,.146)
=.019949
the standard deviation is .146 because it is a sample distribution so you take the standard deviation and divide it by the square root of n.. which is .08/square root of 30
ReplyDeletewhy did you use .146 when i did it used .8 so its like
ReplyDeletenormalcdf(-10^99,16.70,17,.8) = .35383
Correct Tina, except it is .8 not .08
ReplyDeleteNUMBER 3 -
ReplyDeleteThe process is not working properly because probably of getting a sample so small is not likely.
NUMBER 4-
ReplyDeletem=.10 n=480
find the std dev by finding the square root of p(1-p) over n
the answer to that is .0137 justify = rule of thumb
#5
ReplyDelete40/480 is the proportion
then do nrmlcdf(-10^99,40/480,.1,.0137) = .1118
#6
the advice i would give to the plant manager is don't send out the batteries because the probability is so low
gabbi, be sure you understand the answer to question 5. what does the probability mean??
ReplyDeletenow that i finally have a computer to use....
ReplyDeletems. criscuolo tina's calculations r still correct for number 2 she just typed .08 instead of .8 right because when i did the calculations i still got the answer that she did
also i think gabi is trying to say for #6 that if the plant manager is looking for such a low probability of unsuitable batteries to chance sending the whole batch out that he shouldn't because the answer for #5 .1118 was higher than what he should chance so he shouldn't send out the entire batch of those batteries because getting such a low probability of unsuitable batteries is really unlikely
im with cathy. i got .8 too. i would say for 6 that the plant manager shouldnt send out the shipment of batteries becasue the probabiity is about .112 and that would be too big of a risk to send out the whole shipment
ReplyDelete.
ReplyDeletemaggie the probability is .10 not m so that is why the equation sq. rt P(1-P)/n works in this case since you are given the p and the n. And for number 5 the question stated that plant manager wouldnt risk sending them out if it was higher than 10% so this is what justifies gabi's answer since on this day it was alittle over 11%.... this is chris young i cant post on my name this is the worst site
ReplyDeleteand cathy no..
ReplyDeleteme finding this 10 years later is insane. Thank you guys.
ReplyDelete