And to do that, I'm going to come up here to the top and select Graph –> QQ Plot. Now that my data is open here in StatCrunch, I'm going to construct my normal quantile plot. This is most easily done by constructing a normal quantile plot in StatCrunch, so I'm going to take my data here, click on this icon to the right, and select Open in StatCrunch. And so I will reject it and it would suggest the alternative.OK, the first part of our problem asks us to list the z-scores for the normal quantile plot. She would say hey the probability of getting this result assuming that the null hypothesis is true, is below my threshold. Since this is lower that that significance level, she would be able to Level was say 5%, well then that situation Then do is compare that to the significance level that she should have set before conducting So this is approximately 0.0336 or a little over 3% orĪ little less than 4%. And so we could look at that on this Z table right over here, negative 1.8, negative 1.83 is this right over here. Less than or equal to 1.83 standard deviations below the means. Greater than or equal to 1.83 standard deviations above the mean, we could say anything So notice this Z table gives us the area to the left of a certain Z value. And so what we wanna do, this probability is this area under our normal curve right here. So that would be 1.83 standard deviations. We got from our sample is 1.83 standard deviations above the mean of the sampling distribution. When we put that little zero there that means the assumed population proportion from the null hypothesis, and that's 0.26, and this result that Population proportion, so that would be P not. The mean of the samplingĭistribution right over here would be the assumed Sampling distribution is normal 'cause we met the necessary conditions, so in a normal distribution, what is the probability of getting a Z greater than or equal to 1.83? So to help us visualize this, let's visualize what the sampling distribution would look like. What is the approximate P value? Well this P value, this is the P value would be equal to the probability of inĪ normal distribution, we're assuming that the Is roughly normal and that's the random condition, the normal condition, the independence condition that we have talk about in the past. The necessary conditions to assume that the sampling distribution of the sample proportions And they say assuming that the necessary conditions are met, they're talking about And if you calculate this, this should give us approximately 1.83. In this particular situation, that would be 0.26 times one, minus 0.26, all of that over our N, that's our sample size, 120. Population proportion times one, minus the assumed population The assumed proportion, so it would be just this. So 0.33 minus 0.26, all of that over the standard deviation of the sampling distribution But that's what this Z statistic is, is how many standard deviations above the assumed proportion is that? So the Z statistic, and we did this in previous videos, you would find theĭifference between this, what we got for our sample, our sample proportion, and the assumed true proportion. Well what's the probability of getting something at least this extreme or this extreme or more? And then if it's below a threshold, then we would reject the null hypothesis which would suggest the alternative. Null hypothesis is true and then we figure out These significance tests we're assuming that the The assumed proportion, remember when we're doing Of how she gets this, she's really trying to say well how many standard deviations above We do this in other videos, but just as a reminder And then she calculates the test statistic for these results was Z isĪpproximately equal to 1.83. Her sample proportion which is 40 out of 120 and this is going to be equal to one-third, which is approximately equal to 0.33. The population of her city, she took a sample, her sample size is 120. She found that 40 of 120 people sampled could speak more than one language. Proportion of people in her city that can speak more than one language. Is it's actually greater than 26%, where P represents the She was curious if thisįigure was higher in her city, so she tested her null hypothesis that the proportion in her city is the same as all Americans, 26%. An article that said 26% of Americans can speak
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