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### Course: Statistics and probability > Unit 12

Lesson 3: Tests about a population proportion- Constructing hypotheses for a significance test about a proportion
- Writing hypotheses for a test about a proportion
- Conditions for a z test about a proportion
- Reference: Conditions for inference on a proportion
- Conditions for a z test about a proportion
- Calculating a z statistic in a test about a proportion
- Calculating the test statistic in a z test for a proportion
- Calculating a P-value given a z statistic
- Calculating the P-value in a z test for a proportion
- Making conclusions in a test about a proportion
- Making conclusions in a z test for a proportion

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# Calculating a P-value given a z statistic

In a significance test about a population proportion, we first calculate a test statistic based on our sample results. We then calculate a p-value based on that test statistic using a normal distribution.

## Want to join the conversation?

- If you calculate the z-score, it will actually be 1.75, not 1.83.(26 votes)
- It is not mention in video, but in practice: Calculating the P-value in a z test for a proportion. How do I know it is 2 tail or 1 tail?(7 votes)
- The P-value equation is misleading here. Whether Ha: is p>26 or p<26, P-value = P(z <= -|1.83|). That is the only way you can validly argue that p≠26 is P-value = 2 * P(z <= -|1.83|).(3 votes)
- Realize P(z ≤ -1.83) = P(z ≥ 1.83) since a normal curve is symmetric about the mean. The distribution for z is the standard normal distribution; it has a mean of 0 and a standard deviation of 1. For Ha: p ≠ 26, the P-value would be P(z ≤ -1.83) + P(z ≥ 1.83) = 2 * P(z ≤ -1.83). Regardless of Ha, z = (p̂ - p0) / sqrt(p0 * (1 - p0) / n), where z gives the number of standard deviations p̂ is from p0.(5 votes)

- Could someone please explain how we came to the conclusion that p-value is just p(z>1.83)?(5 votes)
- At2:15why do we not divide by n - 1 to get an unbiased estimate?(4 votes)
- Because the sampling distribution of the sample proportion, whose standard deviation we're calculating, is itself a population and not a sample. We're not trying to estimate anything there, this is a "true" standard deviation.

Think of it this way: while a single sample is part of a population, several samples are collectively a separate thing, a population of samples.

And because of the central limit theorem, the mean of the sampling distribution will be the mean of the parent distribution:

µ[p̂] = p

µ[x̄] = µ(1 vote)

- for this question, am i right in saying that the p value is also known as the probability of getting a sample proportion of 1/3, given that the null hypothesis is true?(2 votes)
- hm. Let me think "loud". For sure the test statistic here is z, and so we run the p-value calculation on our test statistic, namely the probability of z being at least as big as in the sample. Now as we got the reference z value from a sample showing 1/3 sample proportion, yes, I would say this is true what you are saying that

P(z at least as this extreme | H0 is true) = P(sample proportion is at least 1/3 | H0 is true)

or at least I can not imagine a different situation how else we could have an at least this large z value from a population of the same size.

any mistake in my logic?(2 votes)

- Why do we decide what kind of p value we're using based on the alternative hypothesis?

e.g. If our Ha was p > 10, then we would have a one tailed p-value of the probability of getting a sample proportion at least as deviant as our actual sample proportion, given that Ho is true.

What's the logic behind this?(2 votes)- I think that the alternative hypothesis creates its own sampling distribution, which will overlap with the sampling distribution given the Ho is true, from one or both directions (depending on the Ha) like in this video: https://www.khanacademy.org/math/statistics-probability/significance-tests-one-sample/error-probabilities-and-power/v/introduction-to-power-in-significance-tests?modal=1. In this video Sal used an explicit mean in his alternative hypothesis, but presuming the location of the Ha sampling distribution relative to the Ho sampling distribution will allow us to know which tail (if not both) of the Ho sampling distribution will overlap. We can accordingly set a two-tailed or one-tailed significance level. Again, that's just what I think.(2 votes)

- At01:41, Sal says, "...all of that over the standard deviation of the sampling distribution of the sample proportions." Why doesn't he just say "over the standard error"?(1 vote)
- He might be talking about a different equation, I'm taking this course right now and I hate it.(4 votes)

- In Z-Score Table. P value for -1.83 is 0.0336 but for +1.83 is .96638.. Could you please tell me which one to chose.. but Sal told .0336 for both + and - 1.83.(2 votes)
- Sal used a simple shortcut.

A z table indicates the proportion of the area of the distribution TO THE LEFT of a given z score. Given that normal distributions are by definition symmetric around their means, if we're looking for the area of just one tail in the positives, we can either subtract the proportion given by the z table from 1, or simply look at the corresponding negative z-score. To put it more formally:

P(z ≤ -a) = P(z ≥ +a)

Hope that helps!(2 votes)

- Why the population proportion is located in mean point and we count z-statistic from it? I thought proportion is an area in a normal distribution.(2 votes)

## Video transcript

- [Instructor] 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 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 one-third, 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 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
these significance tests 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 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 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 would be the assumed
population proportion times one, minus the assumed population
proportion over N. 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. 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 talk 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 '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. 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 what we wanna do, this probability is this area under our normal curve right 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 means. So this is negative 1.83. 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. So 0.0336. So there we have it. So this is approximately 0.0336 or a little over 3% or
a little less than 4%. And so what Fay 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 5%, well then that situation
since this is lower that 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. And 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 has say a 1% significance level, then she would fail to
reject the null hypothesis.