faye 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 percent where p represents the proportion of people in her city that can speak more than one language she found that 40 of 120 people sampled could speak more than one language so what's going on is 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 and this is going to be equal to 1 3 which is approximately equal to 0.33 and then she calculates the test statistic for these results was z is approximately equal to 1.83 we count 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 and then we figure out 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 but that's what the z statistic is is well how many standard deviations above the assumed proportion is that so the z statistic and we did this in previous videos you would find the difference 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 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 just this it'd be the assumed population proportion times 1 minus the assumed population proportion over n in this particular situation that would be 0.26 times 1 minus 0.26 all of that over our n that's our sample size 120 and if you calculate this this should give us approximately 1.83 so they did all of that for us and 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 and that's the random condition the normal condition the independence condition that we have talked about in the past what is the approximate p-value well this p-value this is the p-value would be equal to the probability of 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 so 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 not 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 we got from our sample is 1.83 standard deviations above the mean of the sampling distribution so 1.83 so that would be 1.83 standard deviations and so we want to do this probability is this area under our normal curve right over here so now let's get our z table so notice this z table gives us the area to the left of a certain z value we wanted it to the right of a certain z value but a normal distribution is symmetric so instead of saying 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 so this is what negative 1.83 and so we could look at that on this z table right over here negative one let me negative one point eight negative one point eight three is this right over here so 0.0336 so there we have it so this is approximately 0.0 3 or a little over three percent or a little less than four percent and so what faye would then do is compare that to the significance level that she should have set before conducting this significance test and so if her significance level was say five percent well then in that situation since this is lower than that significance level it would 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 and so i'll 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 one percent significance level then she would fail to reject the null hypothesis