for a test of ho: p = 0.5, the z test statistic equals 1.74. find the p-value for ha: p > 0.5.
For A Test Of Ho: P = 0.5, The Z Test Statistic Equals 1.74. Find The P-Value For Ha: P ≫ 0.5.
- Fay readan write-up that stated 26% that Americans deserve to speakmore than one language. She was curious if thisfigure was higher in she city, therefore she tested she null theory that the ratio in her city is the exact same as all Americans, 26%. Her different hypothesisis it's actually higher than 26%, where P to represent theproportion of people in she city that deserve to speak more than one language. She discovered that 40 that 120 civilization sampled could speak an ext than one language. So what's walking on is here'sthe population of she city, she take it a sample, she sample dimension is 120. And also then she calculatesher sample proportion i beg your pardon is 40 out of 120 and also this is going come be equal to one-third, which is about equal to 0.33. And then she calculates the check statistic for these outcomes was Z isapproximately equal to 1.83. We carry out this in other videos, yet just as a reminderof just how she gets this, she's really trying come say fine how numerous standard deviations abovethe presume proportion, remember when we're doingthese meaning tests we're assuming the thenull theory is true and also then we number outwell what's the probability of acquiring something at the very least this excessive or this too much or more? and then if it's listed below a threshold, then we would disapprove the null hypothesis which would suggest the alternative. Yet that's what this Z statistic is, is how numerous standard deviations above the presume proportion is that? therefore the Z statistic, and we walk this in vault videos, girlfriend would uncover thedifference in between this, what we got for our sample, our sample proportion, and the assumed true proportion. So 0.33 minus 0.26, all of that end the typical deviation that the sampling distributionof the sample proportions. And we've checked out that in vault videos. That is just going to bethe assumed proportion, for this reason it would be just this. It would be the assumedpopulation proportion times one, minus the presume populationproportion over N. In this particular situation, that would certainly be 0.26 times one, minus 0.26, all of that over our N, that's ours sample size, 120. And also if you calculate this, this should give us approximately 1.83. So they did all of that because that us. And also they speak assuming that the necessary conditions are met, they're talk aboutthe necessary conditions to assume the the sampling circulation of the sample proportionsis about normal and also that's the random condition, the regular condition, the independence problem that we have actually talk about in the past. What is the approximate ns value? fine this p value, this is the P value would be equal to the probability of ina common distribution, we're assuming the thesampling distribution is normal 'cause we met the crucial conditions, so in a common distribution, what is the probability of getting a Z higher than or same to 1.83? therefore to assist us visualize this, let's visualize what the sampling distribution would watch like. We're assuming the is approximately normal. The mean of the samplingdistribution right over right here would be the assumedpopulation proportion, therefore that would certainly be ns not. Once we put that small zero over there that means the assumed populace proportion indigenous the null hypothesis, and also that's 0.26, and this an outcome thatwe acquired from ours sample is 1.83 traditional deviations over the typical of the sampling distribution. For this reason 1.83. So that would be 1.83 traditional deviations. And so what we wanna do, this probability is this area under our common curve ideal here. So now let's gain our Z table. So notification this Z table offers us the area come the left that a certain Z value. We want it come the rightof a details Z value. However a normal circulation is symmetric. So rather of saying anythinggreater 보다 or same to 1.83 conventional deviations above the mean, we might say anythingless 보다 or equal to 1.83 typical deviations listed below the means. For this reason this is an adverse 1.83. And so we might look at the on this Z table right over here, an adverse 1.8, an unfavorable 1.83 is this appropriate over here. Therefore 0.0336. So there we have it. For this reason this is roughly 0.0336 or a tiny over 3% ora tiny less 보다 4%. And so what Fay wouldthen perform is compare that to the definition level the she must have collection before conductingthis definition test. And also so if her significancelevel was say 5%, fine then that situationsince this is reduced that that definition level, she would certainly be maybe toreject the null hypothesis. She would say hey the probability of gaining this an outcome assuming that the null theory is true, is listed below my threshold. It's quite low.
And also so i will reject it and it would imply the alternative. However, if hersignificance level was lower than this for every little thing reason, if she has actually say a 1% definition level, climate she would certainly fail toreject the null hypothesis.