for a test of ho: p = 0.5, the z test statistic equals 1.52. find the p-value for ha: p > 0.5.
For a test of ho: p = 0.5, the z test statistic equals 1.52. find the p-value for ha: p > 0.5.
- Fay readan write-up that claimed 26% of Americans deserve to speakmore 보다 one language. She was curious if thisfigure was higher in her city, therefore she tested her null theory that the relationship in her city is the very same as every Americans, 26%. Her alternate hypothesisis it's actually greater than 26%, wherein P to represent theproportion of human being in she city that have the right to speak much more than one language. She uncovered that 40 of 120 people sampled can speak much more than one language. Therefore what's walk on is here'sthe population of her city, she take it a sample, her sample dimension is 120. And then she calculatesher sample proportion which is 40 out of 120 and also this is going to be equal to one-third, which is about equal come 0.33. And also then she calculates the check statistic because that these results was Z isapproximately same to 1.83. We do this in various other videos, however just as a reminderof just how she it s okay this, she's really trying come say fine how plenty of standard deviations abovethe assumed proportion, remember as soon as we're doingthese significance tests we're assuming that thenull hypothesis is true and then we figure outwell what's the probability of gaining something at the very least this extreme or this excessive or more? and then if it's listed below a threshold, then us would disapprove the null theory which would imply the alternative. But that's what this Z statistic is, is how plenty of standard deviations over the suspect proportion is that? for this reason the Z statistic, and we walk this in previous videos, you would uncover thedifference between this, what we gained for our sample, ours sample proportion, and also the assumed true proportion. Therefore 0.33 minus 0.26, all of that over the traditional deviation that the sampling distributionof the sample proportions. And also we've viewed that in vault videos. The is simply going to bethe suspect proportion, so it would be simply this. It would be the assumedpopulation proportion times one, minus the assumed populationproportion end N. In this certain situation, that would be 0.26 times one, minus 0.26, all of that end our N, that's our sample size, 120. And if you calculation this, this should give us approximately 1.83. For this reason they did every one of that because that us. And also they to speak assuming that the necessary conditions are met, they're talk aboutthe necessary problems to assume the the sampling circulation of the sample proportionsis roughly normal and also that's the arbitrarily condition, the regular condition, the independence condition that we have talk around in the past. What is the approximate p value? fine this ns value, this is the P worth would be same to the probability the ina common distribution, we're assuming the thesampling circulation is regular 'cause us met the important conditions, so in a normal distribution, what is the probability of getting a Z greater than or same to 1.83? therefore to aid us visualize this, let's visualize what the sampling circulation would look at like. We're assuming it is about normal. The typical of the samplingdistribution ideal over below would be the assumedpopulation proportion, so that would certainly be ns not. When we put that little zero over there that way the assumed population proportion from the null hypothesis, and that's 0.26, and this an outcome thatwe gained from our sample is 1.83 conventional deviations above the median of the sampling distribution. For this reason 1.83. For this reason that would be 1.83 traditional deviations. And also so what us wanna do, this probability is this area under our normal curve best here. So currently let's get our Z table. So an alert this Z table offers us the area come the left of a certain Z value. We wanted it to the rightof a particular Z value. However a normal distribution is symmetric. So instead of saying anythinggreater 보다 or same to 1.83 typical deviations over the mean, we can say anythingless 보다 or equal to 1.83 conventional deviations listed below the means. For this reason this is an adverse 1.83. And also so we can look at the on this Z table ideal over here, negative 1.8, an unfavorable 1.83 is this right over here. For this reason 0.0336. So there we have it. So this is approximately 0.0336 or a tiny over 3% ora small less than 4%. And so what Fay wouldthen do is compare the to the meaning level that she have to have set before conductingthis meaning test. And also so if she significancelevel to be say 5%, fine then the situationsince this is reduced that that definition level, she would be may be toreject the null hypothesis. She would certainly say hey the probability of acquiring this result assuming that the null hypothesis is true, is below my threshold. It's fairly low.
And so ns will reject it and also it would suggest the alternative. However, if hersignificance level was reduced than this for everything reason, if she has say a 1% significance level, climate she would fail toreject the null hypothesis.