Calculating a P-value given a z statistic | AP Statistics | Khan Academy

Fay read an article that said 26% of Americans can speak more than one language. She was curious if this figure was higher in her city, so she tested her null hypothesis: that the proportion in her city is the same as all Americans' - 26%. Her alternative hypothesis is it's actually greater than 26%, where P represents the proportion of people in her city that can speak more than one language.

She found that 40 out of 120 people sampled could speak more than one language. So what's going on? Here's the population of her city. She took a sample. Her sample size is 120, and then she calculates her sample proportion, which is 40 out of 120. This is going to be equal to 1/3, which is approximately equal to 0.33.

Then she calculates the test statistic for these results: Z is approximately equal to 1.83. We do this in other videos, but just as a reminder of how she gets this: she's really trying to say, well, how many standard deviations above the assumed proportion? Remember, when we're doing the significance test, we're assuming that the null hypothesis is true.

Then we figure out, well, what's the probability of getting something at least this extreme or more? If it's below a threshold, then we would reject the null hypothesis, which would suggest the alternative. But that's what this Z statistic is: how many standard deviations above the assumed proportion is that?

So the Z statistic, and we did this in previous videos, is found by calculating the difference between what we got for our sample (our sample proportion) and the assumed true proportion. So, 0.33 minus 0.26, all of that over the standard deviation of the sampling distribution of the sample proportions. And we've seen that in previous videos.

That is just going to be the assumed proportion. So it would be the assumed population proportion multiplied by 1 minus the assumed population proportion, all over n. In this particular situation, that would be 0.26 multiplied by 1 - 0.26, all of that over our n, which is our sample size of 120. If you calculate this, it should give us approximately 1.83.

So they did all of that for us. They say, assuming that the necessary conditions are met, they're talking about the necessary conditions to assume that the sampling distribution of the sample proportions is roughly normal. That's the random condition, the normal condition, and the independence condition that we have talked about in the past.

What is the approximate P value? Well, this P value would be equal to the probability in a normal distribution. We're assuming that the sampling distribution is normal because 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?

To help us visualize this, imagine. Let's visualize what the sampling distribution would look like. We're assuming it is roughly normal. The mean of the sampling distribution, right over here, would be the assumed population proportion. So that would be P with a little zero there that indicates the assumed population proportion from the null hypothesis, and that's 0.26.

This result that we got from our sample is 1.83 standard deviations above the mean of the sampling distribution. So, 1.83 standard deviations, and what we want to do is calculate this probability as the area under our normal curve right over here.

Now let's get our Z table. Notice this Z table gives us the area to the left of a certain z value. We wanted it to the right, but a normal distribution is symmetric. Instead of stating anything greater than or equal to 1.83 standard deviations above the mean, we could say anything less than or equal to 1.83 standard deviations below the mean, which is -1.83.

So we could look at that on this Z table right over here. Negative 1.83 gives us the value of 0.336. So there we have it. This is approximately 0.336, or a little over 3%, or a little less than 4%.

Fay would then compare that to the significance level that she should have set before conducting this significance test. If her significance level was, say, 5%, then in that situation, since this is lower than that significance level, she would be able to reject the null hypothesis.

She would say, "Hey, the probability of getting this result, assuming that the null hypothesis is true, is below my threshold. It's quite low, so I will reject it," and it would suggest the alternative. However, if her significance level was lower than this for whatever reason, if she had, say, a 1% significance level, then she would fail to reject the null hypothesis.