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AP®︎/College Statistics
Course: AP®︎/College Statistics > Unit 10
Lesson 6: Concluding a test for a population proportionMaking conclusions in a test about a proportion
Example of how to make a conclusion in a test about a proportion.
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- Why can we conclude that more than 50% of adults support the tax increase? I can only conclude that p is not equal to 50%(4 votes)
- The proportion for the sample is greater that p in H0.(1 vote)
- Why do significance levels exist?(1 vote)
- They exist to test the validity of a hypothesis in the set criteria. For example in the scientific world there exists a sig level of .05 to ensure that all scientists make conclusions using the same criteria and the idea of significance is n't something that can be made up of the individual researcher.(2 votes)
- Why isn't answer B correct? With this alpha, we would fail to find evidence suggesting that p > 0.5, so we can conclude that p < 0.5 no?(1 vote)
- Remember that the sample proportion is 113∕200 > 0.5, which in itself makes it more likely than not that the population proportion is also more than 0.5
The significance test doesn't tell us that the population proportion isn't more than 0.5, it only tells us that we don't have enough evidence to safely say that it is.(2 votes)
- I don't understand the two-sided probability test(1 vote)
- Assume a significance level of alpha equals 0.1α=0.1 and use the given information to complete parts (a) and (b) below.
Original claim: Women have heights with a mean equal to 154.8154.8 cm. The hypothesis test results in a P-value of 0.05490.0549.(0 votes)
Video transcript
- [Instructor] A public opinion survey investigated whether a majority, more than 50 % of adults, supported a tax increase to help fund the local school system. A random sample of 200 adults showed that 113 of those sampled supported the tax increase. Researchers used these results to test the null hypothesis is
that the proportion is 0.5. The alternative hypothesis is that it's greater than 0.5 where
P is the true proportion of adults that support the tax increase. They calculated a test statistic of Z is approximately equal to
1.84 and a corresponding P value of approximately 0.033. Assuming the conditions
for inference were met, which of these is an
appropriate conclusion? And we have our four conclusions here. At any point in encourage
you to pause this video and see if you can answer it for yourself, but now we will do it together. And just to make sure we
understand what's going on, before we even cut to the
chase and get to the answer. So what we do is we have this population and we are going to sample it. So N is equal to 200. From that sample we can calculate a sample proportion of adults that support the tax increase. We see 113 out of 200 support it, which is going to be equal to, let's see that is the same thing as 56.5%. So 56.5% and so the key is
to figure out the P value. What is the probability of getting a result this much above the assumed proportion or greater, at least this much above
the assumed proportion? If we assume that the
null hypothesis is true. And if that probability, if that P value is below
a preset threshold, if it's below our significance level, they haven't told it to us yet, it looks like they're gonna
give some in the choices, well then we would reject
the null hypothesis which would suggest the alternative. If the P value is not lower than this, then we will fail to
reject the null hypothesis. Now to calculate that P value, to calculate that probability, what we figure out is how many in our sampling distribution, how many standard deviations above the mean of the sampling distribution and the mean of the sampling distribution would be our assumed
population proportion, how many standard deviations above that mean is this right over here? And that is what this test statistic is. And then we can use this
to look at a Z table and say all right, well in a normal
distribution what percentage or what is the area under the normal curve that is further than
1.84 standard deviations or at least 1.84 or
more standard deviations above the mean? And they did for us as well. So really what we just need to do is compare this P value right over here to the significance level. If the P value is less than
our significance level, then we reject, reject our null hypothesis. And that would suggest the alternative. If this is not true, then we would fail to
reject the null hypothesis. So let's look at these choices and if you didn't answer
it the first time, I encourage you to pause the video again. So at the alpha is equal
to 0.01 significance level, they should conclude that more than 50% of adults support the tax increase. So if the alpha is 1/100, the P value right over here is over 3/100. It's roughly 3.3%. So this is a situation where our P value, our P value is greater
than or equal to alpha. In fact it's definitely
greater than alpha here and so here we would fail to reject, we would fail to reject
our null hypothesis. And so we wouldn't
conclude that more than 50% of adults supported the tax increase. 'cause remember our null
hypothesis is that 50% do and we're failing to reject this. So that's not gonna be true. At that same significance level, they should conclude that less than 50% of adults support the tax increase. No, we can't say that either. We just failed to reject
this null hypothesis that the true proportion is 50%. So at the alpha equals to
5/100 significance level, they should conclude that more than 50% of adults support the tax increase. Well yeah, in this situation we have our P value which is 0.033. It is indeed less than
our significance level. In which case we reject, reject the null hypothesis. And if we reject the null hypothesis, that would suggest the alternative. That the true proportion
is greater than 50%. And so I would pick this
choice right over here. And then choice D at that
same significance level, they should conclude that less than 50% of adults support the tax increase. No, not the situation at all. If we're rejecting our null
hypothesis right over here, then it should suggest this alternative.