# Homework Help: Testing of population variance

1. Nov 29, 2017

### tzx9633

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. 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 .

Last edited by a moderator: Nov 29, 2017
2. Nov 29, 2017

### 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.

3. Nov 29, 2017

### tzx9633

question edited , refer back

4. Nov 29, 2017

### mjc123

And the question is...?

5. Nov 29, 2017

### tzx9633

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

6. Nov 29, 2017

### 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.

7. Nov 29, 2017

### Ray Vickson

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.

8. Nov 29, 2017

### WWGD

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

9. Nov 29, 2017

### Ray Vickson

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 $F(24,\infty) = \chi^2(24).$

10. Nov 29, 2017

### WWGD

11. Nov 29, 2017

### tzx9633

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)

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12. Nov 29, 2017

### Ray Vickson

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

13. Nov 29, 2017

### tzx9633

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

14. Nov 29, 2017

### Ray Vickson

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?

15. Nov 29, 2017

### tzx9633

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 ?

16. Nov 29, 2017

### WWGD

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

17. Nov 29, 2017

### tzx9633

huh
nuh ?

18. Nov 29, 2017

### WWGD

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

19. Nov 29, 2017

### tzx9633

????

20. Nov 29, 2017

### Ray Vickson

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.

21. Nov 30, 2017

### WWGD

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

22. Nov 30, 2017

### WWGD

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.

23. Nov 30, 2017

### tzx9633

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

24. Nov 30, 2017

### WWGD

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.

Last edited: Nov 30, 2017
25. Nov 30, 2017

### Ray Vickson

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.