In a sample of 600 men

**zorro** · December 4th 2009, 09:32 PM

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?

**CaptainBlack** · December 5th 2009, 01:15 AM

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

**babbagandu** · December 5th 2009, 09:52 AM

Maybe I am wrong, but shouldn't just get the ratios and compare them?

**zorro** · December 8th 2009, 09:10 PM

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....

**CaptainBlack** · December 8th 2009, 10:35 PM

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

**coder** · December 9th 2009, 03:32 AM

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)

**CaptainBlack** · December 9th 2009, 04:01 AM

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

**T. Student** · December 9th 2009, 11:32 AM

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.

**zorro** · December 9th 2009, 03:33 PM

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 ......

**T. Student** · December 9th 2009, 04:09 PM

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?

**CaptainBlack** · December 9th 2009, 08:36 PM

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

**zorro** · December 12th 2009, 12:10 AM

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??????

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In a sample of 600 men

I dont know from where to start

Is this correct?

Is this correct?

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