#1
I haven't been paying attention in class recently, and my book's at school, so I have no idea how to do the worksheet. Anyways, I've absorbed some info in my sleep, so I do understand some of the concepts. Anyways. Here's the problem.

An insurance company advertises that 90% of accident claims get settles within 30 days. A group randomly selects 104 of last years claims from the company's files, and finds only 89 of them were settled within 30 days. Are they falsely advertising?

I'm pretty sure a Z test would be appropriate here, but how would I find the standard error without the population deviation?

What I've got so far:

n=104
x=89/104
u=true percentage of accidents that get settled in 30 day

Null hypothesis: u=90%
Alt. hypothesis: u<90%


I've no idea what conditions I need to check....

Help would be appreciated. I've also got another problem, but one thing at a time. Thanks.
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#2
Since you admit to not paying attention, I refuse to help and motion for nobody else to. Enjoy being poor.
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#3
n + x = u<cos^-1(90)

n + x/0=u

u = don't sleep in school
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#4
nah if its just a worksheet.. go in get your book and do it.. pretty straight forward,, you'll learn

i forgot stats, but its pretty easy from what i remember. If you're gonna get us to do your homework for ya, then piss off
#5
I'm just looking for 2 things: How to get the population deviation, and the conditions needed to be checked. I can't get my book because my school is a closed campus.
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#6
It's actually a z test for population proportions.

P = The proportion of all people who get settles within 30 days.

Null: P = .90, Alternative: P < .90

p* = sample proportion = 89/104

Conditions:
SRS (given), normality: 10(p*) > 10; 10(1-p*) > 10, independence: 10(n) < N

z = (p*-P) / sqrt((P(1-P)/n)

...modes and scales are still useless.


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#7
Xiaoxi. Do you regularly play in CSP? I see a guy with your screen name in the server almost all the time.

Anyways, back on topic. For that last bit. So the SD would be calculated with the sample proportion (89/104) or the (excuse the lack of a more technical term) given proportion (.9)?

Thanks.
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#8
Quote by abcdboy
Xiaoxi. Do you regularly play in CSP? I see a guy with your screen name in the server almost all the time.

Anyways, back on topic. For that last bit. So the SD would be calculated with the sample proportion (89/104) or the (excuse the lack of a more technical term) given proportion (.9)?

Thanks.

There is no standard deviation involved in solving for proportion. All the answers you need are in the formula I gave you.

What's CSP? I play CS:S but I don't go by that screen name.

...modes and scales are still useless.


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#9
Huh? Isn't the square root of p(1-p)/n the equation for s?

But yeah, would you use the sample prop. or the given prop.?

CSP is a server. There's a guy that's on pretty often with your login name. Is it a meme or something?

Thanks. Appreciate it.
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#10
Quote by abcdboy
Huh? Isn't the square root of p(1-p)/n the equation for s?

But yeah, would you use the sample prop. or the given prop.?

CSP is a server. There's a guy that's on pretty often with your login name. Is it a meme or something?

Thanks. Appreciate it.

s is the standard deviation yielded from purely sample data (no known theoretical data). In figuring out s, the AP question will provide you a list a data to enter so that you can figure out the s based on the data. s is usually used for population mean based on sample data only and you would use the T test for that test of statistics.

But yes, you can think of root(p(1-p)/n) as the standard deviation in terms of population.

The sample proportion is p* while the given proportion is P. Use the formula to figure out the z score. In the middle of the z distribution graph, P = .90 is z = 0. The z-score from what you will obtain using the previously mentioned formula is the probability of getting your sample proportion results in context of P = .90. If the probability is .05 or lower, it's statistically significant and you would reject the null because there is only 5% or less that you would get the result if the null was true.

And no, Xiaoxi is not a meme. It's my real name. And I don't play on that server.

...modes and scales are still useless.


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#11
i hate stats with a passion. its the most usless class ever. and i cant help you, i dont pay attention either
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#12
Quote by Xiaoxi
s is the standard deviation yielded from purely sample data (no known theoretical data). In figuring out s, the AP question will provide you a list a data to enter so that you can figure out the s based on the data. s is usually used for population mean based on sample data only and you would use the T test for that test of statistics.

But yes, you can think of root(p(1-p)/n) as the standard deviation in terms of population.

The sample proportion is p* while the given proportion is P. Use the formula to figure out the z score. In the middle of the z distribution graph, P = .90 is z = 0. The z-score from what you will obtain using the previously mentioned formula is the probability of getting your sample proportion results in context of P = .90. If the probability is .05 or lower, it's statistically significant and you would reject the null because there is only 5% or less that you would get the result if the null was true.

And no, Xiaoxi is not a meme. It's my real name. And I don't play on that server.


Yeah after reading that, I'm fairly confident I will fail the AP test in may. Like I understand it all completely, but there is no way I could explain it.
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#13
Quote by kyurah
Yeah after reading that, I'm fairly confident I will fail the AP test in may. Like I understand it all completely, but there is no way I could explain it.

Weird. I think it's pretty straightforward for the most part if you just follow the formulaic steps to doing these problems. Buy an AP guidebook or something.

...modes and scales are still useless.


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#14
Ah. Thank you. After I calculate the Z score, I plug it into normalcdf, which yields a P value of .1295, so I can conclude that it is probable that the proportion they came up with occurred because of sampling variability...right?

I've also got another problem, if you're willing to help...

Some cigarettes sold in the US claim to have "low tar". How much less would a smoker inhale by smoking low tar brands instead? Samples of 15 brands from each type (low tar, and non low tar) are randomly chosen from 1206 varieties. Find a 95% c-interval.

I guess values aren't really important here, as you're not the one doing the problem. But yeah, I think this would be a 2 sample z interval. Would I get S for each type the same way, with (p(1-p)/n)^.5?

EDIT: Yeah, the thing I love about this class is that the formulas are the cure-alls. For example. Every Z test will require the same panacea. You just keep applying it over and over again.
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Last edited by abcdboy at Mar 11, 2008,
#15
Quote by abcdboy
Ah. Thank you. After I calculate the Z score, I plug it into normalcdf, which yields a P value of .1295, so I can conclude that it is probable that the proportion they came up with occurred because of sampling variability...right?
Well, I didn't actually do the calculations, so I don't know. I think you're supposed to use normalpdf though...

Also, on your TI 83, go to Stat -> Test -> 1 sample z prop test, then enter the information and select >P, you should get the right answer.

And for understanding's sake, remember that your answer depends on your hypothesis assertion. Shade in the right side and right direction so that you get the right P value. But yes, 12.95% is not significant, therefore you will fail to reject the null.


Some cigarettes sold in the US claim to have "low tar". How much less would a smoker inhale by smoking low tar brands instead? Samples of 15 brands from each type (low tar, and non low tar) are randomly chosen from 1206 varieties. Find a 95% c-interval.

I guess values aren't really important here, as you're not the one doing the problem. But yeah, I think this would be a 2 sample z interval. Would I get S for each type the same way, with (p(1-p)/n)^.5?

Unless we have different terminologies, I think you are misunderstanding. s is for sample-based data only, which you would use T test or T interval for. Z test is for proprtion or for theoretical data (data with a given standard deviation), and the standard deviation is usually notated as σ, which you use the Z test/interval for.

You need to give me the full problem in order for me to determine exactly what type of problem it is.

In general, just remember that if the problem tells you a percentage or proportion (x out of y), the problem is a Z test/interval for proportion. If the problem gives you the theoretical SD, the problem is a Z test/interval for mean. If the problem gives you a table of data or tells you that the SD is yielded from a sample, the problem is a T test/interval for mean.

...modes and scales are still useless.


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#16
Oh. Well...

Type Mg/cig

Reg 18, 10, 14, 15, 15, 12, 17, 11, 14, 17, 12, 14, 15, 15 ,12

Low 9, 5, 10, 4, 8, 9, 9, 3, 7, 12, 6, 10, 8, 11, 8

Referring back to problem one. What would be the difference in a 1 prop z test and a z test? Like, how does the calculator calculate them differently? It'd be pretty much the same thing, right? I'm a little unsure of my terminology. Is P(P=.85)=P(-1.5)=.1295 demonstrating correct usage?

Thanks.
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#17
Quote by abcdboy
Oh. Well...

Type Mg/cig

Reg 18, 10, 14, 15, 15, 12, 17, 11, 14, 17, 12, 14, 15, 15 ,12

Low 9, 5, 10, 4, 8, 9, 9, 3, 7, 12, 6, 10, 8, 11, 8

Referring back to problem one. What would be the difference in a 1 prop z test and a z test? Like, how does the calculator calculate them differently? It'd be pretty much the same thing, right? I'm a little unsure of my terminology. Is P(P=.85)=P(-1.5)=.1295 demonstrating correct usage?

Thanks.

You mean difference between z and t test? They use almost the exact same formula for population mean, but z uses σ while t uses s. s is calculated using a formula that you should have learned earlier during the year but both you and I forgot right now, lol.

And no, that last part is a bit unclear to me. I don't know what you're referring to.

Anyways, back to the problem at hand: because you're given sample data, the problem obviously wants you to find s. You don't really need to do that by hand. Just use your calculator in the Stat section.

Again, because this is from sample only, you're gonna use the T test.

u = mean difference in mg of tar between regular and low tar cigs inhaled by all people.

Although the problem gives you 2 sets of data, this is actually a 1 sample t test because the thing you're really trying to find is the mean difference, which would be 1 set of sample. To do this, enter both sets of data into your calculator, and then for L3 (mean difference), it would be L1(reg)-L2(low), which gives you the differences of mg of tar.

Now the tricky part is stating the null and alt.

Because we don't know that if there really is a difference between reg and low tar cigs, and in the perspective of the company, they're trying to prove AGAINST the assertion that there is no difference, null: u = 0.

The company thinks that there is a difference, more specifically, there would be more than 0 mg of difference between reg and low. Therefore the alt: u > 0.

You're on your own for the rest.

...modes and scales are still useless.


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#18
Are you sure the second one is a T test? Can't σ be figured from the data? Because the means are being compared, wouldn't it be a 2 samp Z int?

Also...the first one is a 1 samp Z prop test...right?

I've got the problems all figured out except for the damned conditions and labeling what kinda test it is.

Thanks again.
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