Can someone show me how the heck to do this ap stats problem?

Test Preparation AP Test Preparation
1

<p>This isn't a homework problem, just saying. Not cheating at all. Just trying to figure it out.</p>

<p>National polls of about 1000 adults each estimated that 47% of adults logged on to the internet for an hour or more daily and that this increased to 51% by 2005. Using the 4-step process for testing significance in a 2-sample proportion, determine whether or not the change in the internet-use proportion is statistically significant. Follow up your conclusion with a concise conclusion.</p>

<p>If someone could ATLEAST get me started on this I'd appreciate it a lot. Thanks.</p>

2

<p>I can do this. Give me a second to work it out.</p>

3

<ol>
<li><p>conditions
1000(.47)=470>=5
1000(.53)=530>=5
1000(.51)=510>=5
1000(.49)=490>=5
it’s normal.
to calculate SD: all>10,000
SRS - not stated in context. assume it is.</p></li>
<li><p>Hypothesis
null: p1=p2
alternative: p1 not equal p2</p></li>
<li><p>2 Prop Z Test
(used TI-84 to calculate)
p=.0735</p></li>
<li><p>(Significance level is .05) There is enough evidence to reject that there is a change in the number of people using the internet.</p></li>
</ol>

4

<p>^ I think your calculations were wrong. Plus alt. hypothesis is p2>p1 since we assume more people to use the internet in 2005. So you got the wrong test statistic. And you didn’t pool the 2 populations.</p>

<p>You have two different samples from the population, one from 2003 and one from 2005. Therefore…</p>

<p>p^1 = 0.47 and p^2 = 0.51</p>

<p>Step one for finding significance in a two-sample proportion is checking if the conditions satisy the requirements of the test. So…</p>

<p>a. You have two independent and random samples of adults
b. n^p1=470, n^q1=530, n^p2=510, and n^q2 = 490. Because these are all atleast ten, you can attribute your distribution to approximately normal characteristics
c. 1,000 (10) = 10,000, and the population of all adults is > 10,000</p>

<p>Then you have your null and alternate hypotheses:</p>

<p>Ho: p1=p2 (or p1-p2=0) when p1 is the proportion of all adults who used the internet in 2003 and p2 is the proportion of all adults who used the internet in 2005. (since we are generalizing that the time span would not cause a change in the proportion)</p>

<p>Ha: p2>p1, which is what we believe to be true, however there is variation is sampling so we have to check this)</p>

<p>Then calculate the test statistic: (in this case you’ll need to pool the proportions from each year so it will be out of 2000 in the denominator of the test statistic, only because this is a TWO sample proportion… so instead of having two samples of 1000 adults each, pool the two 1000 samples into one 2000 sample</p>

<p>So the test statistic is:</p>

<p>Z = ([470/1000-510/1000]/sqrt([980/2000 X 1020/2000] + [1020/2000 X 980/2000]) = 1.78 (pooled)</p>

<p>(A calculator trick for this is 2nd–>Vars–>6 (2 Prop Z Test) and put in the successed ands n value, it’ll give you the p-value as well)</p>

<p>I won’t lie, I have no idea how to show the work for calculating a P-Value, but from 2nd->Vars you get a P-Value of 0.036.</p>

<p>So then a conclusion would be something like:</p>

<p>If these is no difference between the people who used the interned in 2003 and the people who used the internet in 2005, there is a 3.6% chance of getting a sample difference of 0.04 or greater if a random sample is taken. Since our p-value is smaller than the assumed level of significance of 0.05 (our teacher has us assume .05 if it’s not stated, since we usually use a 95% confidence interval, but idk about yours), our results are not in favor of the null and the null hypothesis is rejected. The change in internet-use proportion, was infact statistically significant.</p>

<p>Yeah, I think that’s right. If you have an answer key let me know hahah</p>

5

<p>According to Jerry’s P-Value ( I have not done the test myself), it is NOT significant at the .05 level meaning that you CANNOT reject the null-hypothesis. The p-value shows how many times by chance alone will the internet usage be over 1 hour. If it was alpha=.10, it would be statistically significant.</p>

6

<p>I’m not sure how you guys are doing it, but if it’s a two sample t test you need the s (standard error). for two sample you would do (47-51-(0, you expect no difference))/sq root s^2/n + s^2/n)</p>

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<p>^ Standard Error is the pooled value in the denominator of the test statistic calculation.</p>

<p>And Jerry forgot to pool so his is wrong. Mine was under .05 so I could reject the null.</p>

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<p>lol… No wonder. We skipped pooling because it won’t be on the AP test.</p>

9

<p>^ You need to pool every time you run a 2-proportion test of significance. When we get one of those on the exam, your whole class will get an automatic 2 on that question because you NEED to pool or else it’ll be mad inaccurate.</p>

10

<p>Well we learned the one sample t test, two sample t test, power, etc. for theese types of problems. Pooling, we either haven’t gotten to yet or is not tested according to my teacher… if it is on the test, then it is in the last 3 chapters of our book.</p>

11

<p>I understand the conditions/hypothesis and the beginning of the test stat. But as soon as you got to the pooling of the 2 proportions, you lost me. Thanks though.</p>

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<p>BreakfastChampion - I assume for the alternative hypothesis that it’s not equal because the context said if it has changed or not. So, in the context, it was not very specific. I would put p2>P1 (changed the condition on the calculator and have the same p-value as yours) if it said that if the statistic has increased significantly or something similar to that.</p>

<p>Plus it’s already pooled when doing the 2 PropZTest on the calculator.</p>

13

<p>Yeah, if you enter it in on the calculator with pooled totals (out of 2000 instead of 1000… in which you’d have to do it like 2 or 4 times that way?) you’ll be all set. But if you selected = instead of <, you’ll obviously get different P-Values like you and I did.</p>

<p>I actually don’t know what’s correct now that I think about it. I’m still thinking p2>p1 is the alternate hypothesis.</p>