Sythe Users: What Do They Know? Do They Know Things?? Lets Find Out! #6

[Accounts Shop]

Walterette, Walter's wife, is angry at the overlords for what they did to her husband yesterday.

The Overlords pretend that 90% of the population was happy with what happened to Walter yesterday, but she refuses to believe that this is true.

She plans to show the whole population that all the Overlords are doing is lie to keep control on their dictatorship.

She wants to show that there is no way that 90% of the population was happy with her husband's beheading so she walks down the street and ask 20 citizen of her town how they felt about her husband's faith.

She then found out that only 75% of these 20 citizens were happy about it, which leads her to believe her assumption that it is false that 90% of citizens are happy about it

It is not feasible to ask the whole country one by one so she has to find a way to prove that what the overlords are saying is in fact false, or not.

Question: Is Walterette right in saying that the Overlords are lieing when they say that 90% of the country was happy with Walter's death? Justify

 

[Superfluous]

no. her sample size is too small to make any meaningful conclusions. if she repeats her experiment a whole bunch of times, she might get somewhere, but, as you can imagine, a random sample of 20 people out of a population will vary from the population's mean opinion about walter.

fyi - it would help if we knew the size of the population. if it's 22 people, it's a pretty good sample. if it's 22,000, it's not.

 

make a confidence interval for the proportion and see if .9 is within the limits. I got (.560,.940) so I would say no, but I haven't done statistics in forever so my math is probably wrong

 

[Accounts Shop]

You can assume that the population is big enough to make it infeasible to ask everyone.

Although her sample is small, you can check whether they are lieing or no

 

[Accounts Shop]

make a confidence interval for the proportion and see if .9 is within the limits. I got (.560,.940) so I would say no, but I haven't done statistics in forever so my math is probably wrong

you can't use a confidence interval to solve this problem.

a confidence interval only gives an interval [a,b] inside which you can assert with 100(1 - alpha)% of confidence that the value you want to estimate is inside that interval

This is not what we want to do here.

Walterette want's to verify wheter or not she can assume that the Overlords are lieing based on her sample

 

You can assume that the population is big enough to make it infeasible to ask everyone.

Although her sample is small, you can check whether they are lieing or no

Binomial table lookup for p=0.9, n=20, x=5 is 0.989 according to this table.
On my phone so can't give you better than that.

I'm thinking if you reduce it to a loaded coin toss question: Null hypothesis is that the coin gives heads 90% of the time, alternative is that it does not give heads 90% of the time. A sample of 20 giving 15 heads is 98.9% likely under the null. The evidence is insufficient to reject the null.

The collected data does not disprove the overlords claim.

 

Alternative with a chi squared test:
K = ((15-18)^2)/20 =9/20 = 0.45

Gives a similar result on a chi square table

 

[Accounts Shop]

Alternative with a chi squared test:
K = ((15-18)^2)/20 =9/20 = 0.45

Gives a similar result on a chi square table

You can't use chi-squared for this problem as it isn't an unilateral test, but you are close

 

You can't use chi-squared for this problem as it isn't an unilateral test, but you are close

It is two sided. Your null is =90%, alt is =/= 90%

Also you didn't comment on the binomial stat above

 

[Accounts Shop]

It is two sided. Your null is = 90%, alt is =/= 90%

Also you didn't comment on the binomial stat above

you can't use the chi-squared law when testing proportions hypothesis, you have to use another distribution law.

you could use chi-squared if you had to test a variability hypothesis

As for the binomial stat,

A sample of 20 giving 15 heads is 98.9% likely under the null. The evidence is insufficient to reject the null.

eventhough it 98.9% , there is still 1.1% chance that the null hypothesis will be rejected.

so you can't use the binomial table lookup to prove that walterette can't reject her null hypothesis, meaning that the overlords might be lieing

 

you can't use the chi-squared law when testing proportions hypothesis, you have to use another distribution law.

Pretty sure you can actually. In fact that's basically all you do in chi squared tests: comparing what you got to what you expected in a finite sample.

As for the binomial stat,



eventhough it 98.9% , there is still 1.1% chance that the null hypothesis will be rejected.

so you can't use the binomial table lookup to prove that walterette can't reject her null hypothesis, meaning that the overlords might be lieing

No matter what statistical method is used none of them will give you 100% certainty. You could use a z or t test or anything else you liked and you'd still have at best 99% CI. I.e. 1% chance it's not in the interval

Only way to be certain is to find the population mean by interviewing every member of the population.

Also this reads all wrong. The p value is 0.989. That means highly insignificant. There is no evidence whatsoever to reject the null. There's not "1% chance to reject the null", that's not how p values work.

Also I think I know what you're angling at. There's a proportion test statistic but fuck me if I'm getting my stat text out to look it up. Already answered as far as I'm concerned.

 

[Superfluous]

any question that makes me even think about the horror that is looking at a table of F-statistics is a bad question

 

[Accounts Shop]

Pretty sure you can actually. In fact that's basically all you do in chi squared tests: comparing what you got to what you expected in a finite sample.



No matter what statistical method is used none of them will give you 100% certainty. You could use a z or t test or anything else you liked and you'd still have at best 99% CI. I.e. 1% chance it's not in the interval

Only way to be certain is to find the population mean by interviewing every member of the population.

Also this reads all wrong. The p value is 0.989. That means highly insignificant. There is no evidence whatsoever to reject the null. There's not "1% chance to reject the null", that's not how p values work.

Also I think I know what you're angling at. There's a proportion test statistic but fuck me if I'm getting my stat text out to look it up. Already answered as far as I'm concerned.

you are right about the CI, my bad.

The easiest way to test the validity of a null hypothesis when it comes to proportions is to use the standardized normal law and then you just check whether the absolute value of Z(null hypothesis) > Z of (alpha/2). <alpha/2 since we are dealing with normal table and we want P(z <= -Z) or P(z >= Z)>

formula to use would be pretty straightforward,

o = real proportion
ô = estimated proportion
alpha = 0.05

|Zo| = |(ô - o)/([o*(1-o)]/n)^(1/2)|

then you just verify is |Zo| > |Z alpha/2| which in our case , |Zo|*> 1.96 (read on the normal table)

doing the mats we arrive at |Zo| ~ 2.236 which is > 1.96 thus the null hypothesis can be rejected

so Walterette has all the evidence she needs to show that it is false for the Overlords to state that 90% of the citizens are happy with what hapenned to Walter yesterday

 

Unless I'm mixing this up, this is beginner statistics. Seems like someone is a freshman in college

You need 10 successes and 10 failures to be able to test this, I think. So you can't even go off of just a sample size of 20 in this instance.

 

[Accounts Shop]

Unless I'm mixing this up, this is beginner statistics. Seems like someone is a freshman in college

You need 10 successes and 10 failures to be able to test this, I think. So you can't even go off of just a sample size of 20 in this instance.

why would you need 10 success and 10 failures?

You can test the validity of the null hypothesis with the given sample.

have the null hypothesis H0: p = 0.90 VS H1: p != 0.90 (p < 0.90 or p > 0.90)

from there, you have a bilateral test you need to do to check wheter the null hypothesis can be rejected or no

 

why would you need 10 success and 10 failures?

You can test the validity of the null hypothesis with the given sample.

have the null hypothesis H0: p = 0.90 VS H1: p != 0.90 (p < 0.90 or p > 0.90)

from there, you have a bilateral test you need to do to check wheter the null hypothesis can be rejected or no

It applies to something, not 100% sure if it applies to this instance.

 

[Accounts Shop]

It applies to something, not 100% sure if it applies to this instance.

but in this sample, there is 15 "success" (i.e. people happy about the faith) vs only 5 failures,

thus, you can't approximate the binomial distribution with the normal distribution as 'n' is not big enough.

Also, i'm not sure this would even work for the problem at hand.

for example, if 75% people of the 20 (15) said they were happy, my maths tell that the null hypothesis can be rejected as it is > than |Z alpha/2|. Now, if instead of 75% we input 78% as the happiness rate, then |Zo| instead of being > 1.96, it is < 1.96, because |Zo| with ô:0.78 = 1.78 < 1.96 => Null hypothesis not rejected.

 

you are right about the CI, my bad.

The easiest way to test the validity of a null hypothesis when it comes to proportions is to use the standardized normal law and then you just check whether the absolute value of Z(null hypothesis) > Z of (alpha/2). <alpha/2 since we are dealing with normal table and we want P(z <= -Z) or P(z >= Z)>

formula to use would be pretty straightforward,

o = real proportion
ô = estimated proportion
alpha = 0.05

|Zo| = |(ô - o)/([o*(1-o)]/n)^(1/2)|

then you just verify is |Zo| > |Z alpha/2| which in our case , |Zo|*> 1.96 (read on the normal table)

doing the mats we arrive at |Zo| ~ 2.236 which is > 1.96 thus the null hypothesis can be rejected

so Walterette has all the evidence she needs to show that it is false for the Overlords to state that 90% of the citizens are happy with what hapenned to Walter yesterday

I think you're quite incorrect. The binomial test will give an exact result whereas the proportion test gives a normal approximation to the binomial distribution. I'll give you a better explanation when I get on my PC.

 

So firstly you needed to use a binomial distribution for this problem.
Here's the answer via R:

> binom.test(15, 20, p=0.9)

	Exact binomial test

data:  15 and 20
number of successes = 15, number of trials = 20, p-value = 0.04317
alternative hypothesis: true probability of success is not equal to 0.9
95 percent confidence interval:
 0.5089541 0.9134285
sample estimates:
probability of success 
                  0.75

At 95% CI you have weak evidence that the population proportion is between 0.51 and 0.91. This is not sufficient to reject the null, because while significant, the interval agrees with the null at 0.9.


Further:
You can only construct a confidence interval using se=1.96*(p(1-p)/n)^0.5 if np> 5 and n(1-p) > 5. In this case the latter requirement fails (n(1-p)) = 2 and you cannot use this approximation for your standard error.

Also I fucked up originally with the binomial post, because those tables are P(x = X) not P (x <= X). Fortunately we don't have to use tables in the real world, we have R. (The chi squared also correctly tells you that the evidence is insufficient to reject the null.)