Question

Can someone please help me with this question on hypothesis testing?

The Pew Research Group conducted a poll on [capital punishment opinions]. Of the 580 Catholics surveyed, 180 indicated that they favored capital punishment; of the 600 seculars surveyed, 238 favored capital punishment. Is there a significant difference in the proportion of individuals in these groups in favor of capital punishment for persons under the age of 18? Use the α = 0.01 level of significance.

Answer 1

the information given is not specific as to: " in favor of capital punishment for persons under the age of 18". they just favor capital punishment in general is what it reads ... we would have to assume that they survey was in respect to the "under 18" that is being asked about.

Answer 2

so, what is your thoughts on how to process this information?

Answer 3

also, what do you have available to do calculations with? ti83 maybe?

Answer 4

It is supposed to be about people under 18. Sorry I wasn't exactly clear on that, I was trying to avoid typing out the whole problem, since it was pretty long-winded

Answer 5

I do have a ti 84 I can use

Answer 6

you know how to locate your stats menu?

Answer 7

yeah

Answer 8

under tests, find the 2propZtest

Answer 9

ok found it

Answer 10

youve got x1, n1 and x2,n2 given in your information

Answer 11

and we are wanting to determine if: p1 not= p2

Answer 12

Ok. So that is the alternative hypothesis?

Answer 13

yes, Ho is p1 = p2
so, Ha is p1 not= p2

Answer 14

ok got it

Answer 15

the level of significance is compared to your p-value

Answer 16

Ok, so then how exactly do I compare the two?

Answer 17

if your P-value is greater than your significance level, then we have a better chance of failing to reject the null

if it is less then the significance level, we reject the null.

basically, .01 is our threshold, anything less than that is unlikely to be within our null boundaries.

Answer 18

Ok I think I get it

Answer 19

the more complicated method

subtract the means, and divided by the sqrt ofthe sum of the variances

\[z=\frac{u_1-u_2}{\sqrt{var_1+var_2}}\]

(180/580-238/600)/(sqrt((180(580-180))/(580)+((238(600-238))/(600)))) = -.0053

so we have a z value, if i did it right, or -.0053 whose probability in a tail is about 49%

so 2 tails is about .9800

just curious, what was your p-value?

Answer 20

The p-value I got from the calculator was .999, assuming I typed everything I right

Answer 21

seems like we agree then.

what was your zscore?

Answer 22

we would have had to gotten a z score of like 3 or more in order to even approach that .01 sig-level

Answer 23

It came up as -3.099

Answer 24

hmm, wish my ti83 was working right now so that i could dbl chk those results ... but i could have messed up the mathing :)

Answer 25

Its fine, at least I have a pretty good idea of what I'm doing now. Thanks for your help! :)

Answer 26

yep, good luck :) i think my mistake was dividing the top parts ... when i just subtract the numbers and not the proportions i get a -3.5 ish

Answer 27

Ok that makes sense... Thanks again!