# In a sample of 600 men

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• Dec 4th 2009, 09:32 PM
zorro
In a sample of 600 men
Question :

In a sample of 600 men from a certain large city , 450 are found to be smokers. In another sample 900 from other large city, 650 are smokers. Do the data indicate that the cities are significantly different with respect to the prevalence of smoking among men?
• Dec 5th 2009, 01:15 AM
CaptainBlack
Quote:

Originally Posted by zorro
Question :

In a sample of 600 men from a certain large city , 450 are found to be smokers. In another sample 900 from other large city, 650 are smokers. Do the data indicate that the cities are significantly different with respect to the prevalence of smoking among men?

Well what have you done? What tests do you know for comparing proportions?

CB
• Dec 5th 2009, 09:52 AM
babbagandu
Maybe I am wrong, but shouldn't just get the ratios and compare them?
• Dec 8th 2009, 09:10 PM
zorro
I dont know from where to start
Quote:

Originally Posted by CaptainBlack
Well what have you done? What tests do you know for comparing proportions?

CB

I dont know from where to start....
• Dec 8th 2009, 10:35 PM
CaptainBlack
Quote:

Originally Posted by zorro
I dont know from where to start....

Look at your notes and/or text for testing the difference of two proportions.

Or google finds:

http://stattrek.com/AP-Statistics-4/...px?Tutorial=AP

despite what looks like a type in this it still looks a nice explanation:
http://www.pinkmonkey.com/studyguide...8/s0808f01.asp

CB
• Dec 9th 2009, 03:32 AM
coder
Quote:

Originally Posted by zorro
Question :

In a sample of 600 men from a certain large city , 450 are found to be smokers. In another sample 900 from other large city, 650 are smokers. Do the data indicate that the cities are significantly different with respect to the prevalence of smoking among men?

is the answer as obvious as i think or is it some kind of trick question?

ratio = (650/900)/(450/600)
• Dec 9th 2009, 04:01 AM
CaptainBlack
Quote:

Originally Posted by coder
is the answer as obvious as i think or is it some kind of trick question?

ratio = (650/900)/(450/600)

If it is obvious it is not obvious to you. Why don't you look at the links on the previous post to yours.

CB
• Dec 9th 2009, 11:32 AM
T. Student
The question does not provide you with a level of significance or a level of confidence. My intro textbook always provides this information in such a question. Since "significantly different" is used in the question, you probably should try the significance level approach. There are two significance levels used in my textbook, 1% and 5%. Choose the level which results in the most powerful test. Do you know which percentage gives your test the most power? Read your textbook or go to CaptainBlack's links. Spoonfeeding you the answer is not good for your learning curve.
• Dec 9th 2009, 03:33 PM
zorro
Is this correct?
Quote:

Originally Posted by CaptainBlack
Look at your notes and/or text for testing the difference of two proportions.

Or google finds:

Hypothesis Test for Difference Between Proportions

despite what looks like a type in this it still looks a nice explanation:
PinkMonkey.com Statistics Study Guide 8.15 Test for difference between proportions

CB

Is this correct >

$P_1$ = $\frac{x_1}{n_1}$= $\frac{450}{600}$= $0.75$

$P_2$ = $\frac{x_2}{n_2}$= $\frac{650}{900}$ = $0.72$

CaptainBlack in the link provided there is a mention of the term z
It didnt understand what z stands for ......
• Dec 9th 2009, 04:09 PM
T. Student
You now have both p-hats. All of the information about z and everything else you need to know is clearly explained in any intro to statistics textbook. Which textbook are you using?
• Dec 9th 2009, 08:36 PM
CaptainBlack
Quote:

Originally Posted by zorro
Is this correct >

$P_1$ = $\frac{x_1}{n_1}$= $\frac{450}{600}$= $0.75$

$P_2$ = $\frac{x_2}{n_2}$= $\frac{650}{900}$ = $0.72$

CaptainBlack in the link provided there is a mention of the term z
It didnt understand what z stands for ......

The basic idea is that as the sample sizes are large and the probabilities are in the suitable range we can assume that the sampling distributions of the two proportions are normal with standard deviations $\sqrt{p_i(1-p_i)/n}$. Now the z-score arises when we construct a test under these assumptions as a test statistic with a standard normal distribution that we can use to look up in a table.

CB
• Dec 12th 2009, 12:10 AM
zorro
Is this correct?
Quote:

Originally Posted by CaptainBlack
The basic idea is that as the sample sizes are large and the probabilities are in the suitable range we can assume that the sampling distributions of the two proportions are normal with standard deviations $\sqrt{p_i(1-p_i)/n}$. Now the z-score arises when we construct a test under these assumptions as a test statistic with a standard normal distribution that we can use to look up in a table.

CB

My work Please let me know if it is correct or no

$P_1 : 0.75$

$P_2 : 0.72$

$H_0 : P_1 = P_2$

$H_a : P_1 \neq P_2$

Formulate the Analysis Plan :

-Significant level : 0.05
-Test Method
$. \quad$ Analysis of Sampled data
$. \qquad$Pooled Sampled proportion: $p=\frac{((p_1 * n_1) + (p_2 * n_2))}{n_1 + n_2}$=0.7
$p_1$ : sampled proportion
$p_2$ : sampled proportion
$n_1$ : size of 1st sample
$n_1$ : size of 1st sample

Standard Error (SE) = $\sqrt{p*(1-p) \left[ \frac{1}{n_1} + \frac{1}{n_2} \right]}$ = 0.025

Test Statistic (z) = $\frac{(P_1-P_2)}{SE}$= 1.2

z<-1.2=: 0.11507

Is this correct??????