standard normal distribution table right of z score

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

Standard normal distribution table right of z score