Thread: Quick question on confidence intervals

Ten engineering schools in the US were surveyed. The sample contained 250 electrical engineers, 80 being women, and 175 chemical engineers, 40 being women. Compute a 90% confidence interval for the difference between the porportion of women in these two fields of engineering. Is there a significant difference between the two proportions?

I calculated the confidence interval and got:

$\displaystyle 0.091+/-0.043$

I'm a bit stuck on answering the bolded part though.

I would say that there is a significant difference between the two proportions because the 90% confidence interval doesn't include 0.

Is this correct? Thanks

Originally Posted by downthesun01

Ten engineering schools in the US were surveyed. The sample contained 250 electrical engineers, 80 being women, and 175 chemical engineers, 40 being women. Compute a 90% confidence interval for the difference between the porportion of women in these two fields of engineering. Is there a significant difference between the two proportions?

I calculated the confidence interval and got:

$\displaystyle 0.091+/-0.043$

I'm a bit stuck on answering the bolded part though.

I would say that there is a significant difference between the two proportions because the 90% confidence interval doesn't include 0.

Is this correct? Thanks

What did you get for the SE of the difference of the two proportions?

The 90% confidence interval is the observed difference +/- 1.645*SE

CB

$\displaystyle SE=\frac{0.043}{1.645}\approx 0.0261$

Originally Posted by downthesun01

$\displaystyle SE=\frac{0.043}{1.645}\approx 0.0261$

No, I know what its value is and how to calculate it, what I want to know is how you calculated it, or rather how you calculated the length of the confidence interval (it's a bit like a hint that I think that it is wrong).

CB

$\displaystyle \sqrt{\frac{(0.32)(0.68)}{250}+{\frac{(0.229)(0.77 1)}{175}}}=\sqrt{0.001879308571}\approx 0.043$

Originally Posted by downthesun01

$\displaystyle \sqrt{\frac{(0.32)(0.68)}{250}+{\frac{(0.229)(0.77 1)}{175}}}=\sqrt{0.001879308571}\approx 0.043$

The 90% confidence interval is the observed difference +/- 1.645*SE

So the confidence interval is: .... ?

CB

$\displaystyle pr(0.091-0.043<P_{a}-P_{b}<0.091+0.043)=0.9$

$\displaystyle pr(0.48<P_{a}-P_{b}<0.139)=0.9$

If there's something that you're trying to help me see, I'm totally not getting it sorry.

Originally Posted by downthesun01

$\displaystyle pr(0.091-0.043<P_{a}-P_{b}<0.091+0.043)=0.9$

$\displaystyle pr(0.48<P_{a}-P_{b}<0.139)=0.9$

If there's something that you're trying to help me see, I'm totally not getting it sorry.

Read what has been posted in this thread, we know what the observed difference Ob, is (0.091), we know what the standard error SE of this difference is (0.0433..) and we know the 90% confidence interval is [Ob-1.645.SE, Ob+1.645.SE], so ....

What I want you to see is that the confidence interval is:

$\displaystyle [0.091-1.645\times 0.0433, 0.091+1.645\times 0.0433]\approx [0.091-0.071,0.091+0.071]=[0.020,0.162]$

In other words I want you to calculate the confidence interval correctly.

CB

Oh. You're right. That's what I have written down on paper. Typing all this through my phone is a bit tedious. My mistake.

However, it still leaves me wondering the difference between the two proportions is significant. I'll just say yes. Thanks

Originally Posted by downthesun01

Oh. You're right. That's what I have written down on paper. Typing all this through my phone is a bit tedious. My mistake.

However, it still leaves me wondering the difference between the two proportions is significant. I'll just say yes. Thanks

Since zero is not in the confidence interval, the difference is significant at this level.

CB