Testing of population variance

[tzx9633]

Homework Statement

with an individual lines at its various windows , a post office finds that the standard deviation for normally distributed waiting times for customers on Friday afternoon is 7.2 minutes . The post office expreiemets with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes.with a significance level of 5% , test the claim that a single line cause lower variation among waiting times for customers

Homework Equations

The Attempt at a Solution

my chi square test value = 5.67 , but teh chi sqaure critical at α = 0.05 and v = 24 is 36.415 , i do n't have enough evidenvce to reject H0 ,
H0 = 7.2
H1= <7.2 ,

But the ans provided is REJECT H0 , why ?

I think the ans provided is incorrect , correct me if i am wrong .

 

[mjc123]

What do you mean by H0 = 7.2? If the question is "Can we say that there is significant improvement...?" then surely H0 = 0 (i.e. difference between means = 0 quintals).
I'm not sure what you've been doing, but I don't think chi-square is the right test for this. Try a t-test.

 

[tzx9633]

What do you mean by H0 = 7.2? If the question is "Can we say that there is significant improvement...?" then surely H0 = 0 (i.e. difference between means = 0 quintals).
I'm not sure what you've been doing, but I don't think chi-square is the right test for this. Try a t-test.

question edited , refer back

 

[mjc123]

And the question is...?

 

[tzx9633]

And the question is...?

with a significance level of 5% , test the claim that a single line cause lower variation among waiting times for customers

 

[mjc123]

Please don't edit your original post like that - it makes my posts meaningless to anyone else reading the thread. To make a significant correction like yours you should put it in a new post.
I still don't understand what you're doing with the chi-squared. This looks like a case for an F-test.

 

[Ray Vickson]

Homework Statement

with an individual lines at its various windows , a post office finds that the standard deviation for normally distributed waiting times for customers on Friday afternoon is 7.2 minutes . The post office expreiemets with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes.with a significance level of 5% , test the claim that a single line cause lower variation among waiting times for customers

Homework Equations

The Attempt at a Solution

my chi square test value = 5.67 , but teh chi sqaure critical at α = 0.05 and v = 24 is 36.415 , i do n't have enough evidenvce to reject H0 ,
H0 = 7.2
H1= <7.2 ,

But the ans provided is REJECT H0 , why ?

I think the ans provided is incorrect , correct me if i am wrong .

You are wrong, but I cannot tell why because you do not give enough details.

(1) Your H0 and H1 are correct; you are correct in trying to use the ch-squared distribution.
(2) Exactly how did you compute your so-called "chi-squared" value? I cannot get a number like your 5.67.
(3) Exactly what test did you perform? Just saying "chi-square" without giving formulas is no help at all.
(4) I get that H0 is strongly rejected---about as strongly rejected as anything I have ever seen in my life, so strongly rejected as to be virtually impossible.

 

[WWGD]

Shouldn't this be an F-test for equality of variances? Where does the $\chi^2$ come from?

 

[Ray Vickson]

Shouldn't this be an F-test for equality of variances? Where does the $\chi^2$ come from?

The F-test is for comparing two sample variances. In the current problem, we are given the value of one variance exactly, as if it had infinitely many degrees of freedom (or that is the implication we are probably supposed to make). So the ratio involves $F(24,\infty) = \chi^2(24).$

 

[WWGD]

Ah, yes, ANOVA, my bad.

 

[tzx9633]

You are wrong, but I cannot tell why because you do not give enough details.

(1) Your H0 and H1 are correct; you are correct in trying to use the ch-squared distribution.
(2) Exactly how did you compute your so-called "chi-squared" value? I cannot get a number like your 5.67.
(3) Exactly what test did you perform? Just saying "chi-square" without giving formulas is no help at all.
(4) I get that H0 is strongly rejected---about as strongly rejected as anything I have ever seen in my life, so strongly rejected as to be virtually impossible.

Sorry , i have uploded an image so that the concept is more clear .. P/s : I have posted my working inside the photo (beside the formula)

 

[Ray Vickson]

Sorry , i have uploded an image so that the concept is more clear .. P/s : I have posted my working inside the photo (beside the formula)

I don't look at posted images; just type it out.

 

[tzx9633]

I don't look at posted images; just type it out.

chi square test = (n-1)(sample standard deviation^2) / (population standard deviation ^2)
chi square test = (25-1)(3.5^2) / (7.2^2)= 5.67

 

[Ray Vickson]

chi square test = (n-1)(sample standard deviation^2) / (population standard deviation ^2)
chi square test = (25-1)(3.5^2) / (7.2^2)= 5.67

OK, so then? What do you do next? What number related to the $\chi^2(24)$ do you use now? In post #1 you said you used 36.415. Why?

 

[tzx9633]

OK, so then? What do you do next? What number related to the $\chi^2(24)$ do you use now?

As pointed out earlier , chi square value when α = 0.05 and v = 24 is 36.415 , but chi square test value = 5.67 , so we need to reject H0 when chi square test > chi square critical ...
So , my ans is do not reject Ho

Anything wrong with my answer ?

 

[WWGD]

Look at the statement of your $H_0, H_A$.

 

[tzx9633]

huh

Look at the statement of your $H_0, H_A$.

nuh ?

 

[WWGD]

huh

nuh ?

Domani, mi amico , dovo sortire de la Starbucks adesso. Chi vediamo ( Chi parlamo domani)

 

[tzx9633]

Domani, mi amico , dovo sortire de la Starbucks adesso. Chi vediamo ( Chi parlamo domani)

?

 

[Ray Vickson]

As pointed out earlier , chi square value when α = 0.05 and v = 24 is 36.415 , but chi square test value = 5.67 , so we need to reject H0 when chi square test > chi square critical ...
So , my ans is do not reject Ho

Anything wrong with my answer ?

Yes, you are badly on the wrong track. However, I must leave it up to you to find your error; if I say any more I will be violating PF rules about giving away answers.

 

[WWGD]

?

So sorry for that, they were closing the coffee shop and it was intended for another forum.

 

[WWGD]

huh

nuh ?

You could expand your statement of your initial-alternative hypotheses to include critical values, so that you have a "roadmap" for what you need to do in your problem. e.g.:

$H_0: \sigma =7.2$
$H_A: \sigma < 7.2$ or $\sigma \neq 7.2$
Test at the $\alpha=0.05$ level.
$\chi^2$ test Critical value =36.451 .
This gives you a guideline. I know it is kind of a hassle and slows you down, but I think it is good that until you're not an expert, you do things this way. But it is ultimately up to you; just a suggestion.

 

[tzx9633]

You could expand your statement of your initial-alternative hypotheses to include critical values, so that you have a "roadmap" for what you need to do in your problem. e.g.:

$H_0: \sigma =7.2$
$H_A: \sigma < 7.2$ or $\sigma \neq 7.2$
Test at the $\alpha=0.05$ level.
$\chi^2$ test Critical value =36.451 .
This gives you a guideline. I know it is kind of a hassle and slows you down, but I think it is good that until you're not an expert, you do things this way. But it is ultimately up to you; just a suggestion.

So , i shouldn't reject H0 , am i right ? if not , can you tell me why i am wrong ?

 

[WWGD]

So , i shouldn't reject H0 , am i right ? if not , can you tell me why i am wrong ?

Sorry, I cannot give you yes/no answers, it is against the Forum's policies, but, accepting/not accepting depends on comparing , as you described in a previous post, the critical value with the statistic you obtained-- also in a way you yourself described. EDIT: Informally, the question is: if the true population variance is 7.2 (and this variance has a specific distribution, described through the $\chi^2$ statistic you used) , is it likely, due to random variation alone, that I will obtain a sample variance will take the value 3.5? I mean, even if the true pop. variance is 7.2., it is likely that you sample variance will not be identical to 7.2., but, is the number you obtained within the "reasonable" values you are likely to obtain if the pop. variance was 7.2? This reasonability depends on the way your data is distributed.

 

[Ray Vickson]

So , i shouldn't reject H0 , am i right ? if not , can you tell me why i am wrong ?

One more hint: when you ask if your $\chi^2$ is greater than 36.451 you are asking if the experimental data point to a significantly larger variance than the original $7.2^2$. Aren't you supposed to be testing if the new variance is smaller than $7.2^2?$ Certainly the experimental value of $3.5^2$ will never, ever, be declared larger than $7.2^2$ at any significance level whatsoever, and you don't even know any statistics tools to tell you that!

You are falling into the bad habit of just using canned formulas without really thinking about what they mean and whether they are appropriate for a particular problem. That can be fatal.

 

[WWGD]

If I may, hope I don't throw of this discussion with this question: how does one determine the distribution of statistics under different conditions, e.g., with this case: how do we know that when the pop. variance is known, the sample distribution satisfies the $\chi^2$ used?

 

[Ray Vickson]

If I may, hope I don't throw of this discussion with this question: how does one determine the distribution of statistics under different conditions, e.g., with this case: how do we know that when the pop. variance is known, the sample distribution satisfies the $\chi^2$ used?

Those are theorems that are proved in more "mathematical" courses. When tailoring material to beginners, it is often necessary to just present statements like that as facts, because their proofs will often involve much more complicated and advanced mathematics than would be appropriate in a first course or an "applied" course. Also, you will typically be given access to tables of values that you can look at when solving specific problems. However, there are also lots of on-line resources that you can use if your tables do not include what you need. Most spreadsheets will have statistical packages that allow by-passing tables altogether and just calling on the spreadsheet to do the computation. Some good hand-held calculators will even have some of the standard statistical distributions included in their capabilities.

All that being said, I am going to engage in a bit of a discussion; it will not be a how-to post for solving a particular problem, but will instead be a discussion on how to structure such problems appropriately, so that you can figure out for yourself what type of analysis to do. It will be a bit long, and I will write it in the context of your current problem, but the same general notions apply to other types of problems as well. I think I can write the following without violating PF policy, because I am not going to tell you exactly how to solve your problem---only how to think about it.

OK, so you have a current system giving a true variance of $v_0 = 7.2^2$. You do some sampling from a new system and get a sample variance of $v = 3.5^2.$ Obviously, $v \neq v_0$, but are the differences "significant"? In particular, we really want to know if the new system has smaller variance than the old, and the single sample point we have suggests that it does. However, in any particular experiment you are bound to get sample variances different from $v_0$, just by chance; in fact, about half the time you will get larger values and about half the time smaller values. Therefore, actually getting a smaller value is really not a surprise. What may be a surprise is getting a value so much smaller than $v_0$. How likely is it that---if the true variance is $v_0$--- you can get a value as small as your observed $v?$ Are the chances 25%? Is the chance 1 in a million?

In a typical hypothesis test we pick a value such as 5% and ask whether our observed $v$ has a less than 5% chance of being seen when $v_0$ is really true. That is, if our observation has a chance of more than 5% we declare that we are not surprised enough to declare that the null hypothesis is false. Of course, it may be true or it may be false---we just do not have strong enough evidence. You should also realize that there is nothing special or magical about the value 5%; other values can be used, and often are. (Alternatively, we can try to give a so-called "p-value", which is the actual probability of an observation as small as the one we observed.)

So, to answer your original question, you need to figure out how likely it would be to see a value as small as $v = 3.5^2$ when the true variance is $v_0 = 7.2^2.$ If $v_0$ were true, the calculated value of $24\: v$ would have probability distribution $v_0 \; \chi^2(24)$, so that $24\: v/v_0$ would have distribution $\chi^2(24).$ Since we have available a table of percentiles for the distribution $\chi^2(24)$ we can see how likely it would be to get a value as small or smaller than the one we actually see.

I have tried to keep this discussion as free as possible of formulas. The aim is to understand intuitively what is happening and how to look at a problem. Once that has been established, then---and only then---can we turn to statistical formulas, because at that point we will have a firm grasp of what formulas to use, and how to use them. Just trying to use some textbook or on-line formulas without such understanding is a really, really bad policy.

 

[Ray Vickson]

If I may, hope I don't throw of this discussion with this question: how does one determine the distribution of statistics under different conditions, e.g., with this case: how do we know that when the pop. variance is known, the sample distribution satisfies the $\chi^2$ used?

Sorry: I did not notice that my response above was to you rather than the OP. Most of it was intended for the OP, not for you as such.

 

[WWGD]

Sorry: I did not notice that my response above was to you rather than the OP. Most of it was intended for the OP, not for you as such.

No problem, Ray.