we calculated a 95\(\%\) confidence interval for the true population proportion of UCLA students who travelled outside the US. \[(0.26,0.44)\] A 95\(\%\) confidence interval of 26\(\%\) to 44\(\%\) means that

We are 95\(\%\) confident that the true population proportion of UCLA students who travelled outside the US is between 26\(\%\) and 44\(\%\).

95\(\%\) of random samples of size n = 100 will produce confidence intervals that contain the true population proportion.

The true population proportion,p, may be outside the interval,but we would expect it to be somehwat close to \(\hat{p}\)

In our random sample of 100 students we had found that 35 of them have at some point in their lives travelled outside the US,\(\hat{p}\)= 0.35.

It is difficult to decide how close is close enough, or how far is too far, and this decision should not be made subjectively.

In Statistics when testing claims we use an objective method called hypothesis testing

Given a sample proportion, ￼ , and sample size, n, we can test claims about the population proportion, p.

We call these claims hypotheses

Our starting point, the status quo, is called the null hypothesis and the alternative claim is called the alternative hypothesis.

If our null hypothesis was that p = 0.35 and our sample yields ￼ = 0.35, then the data are consistent with the null hypothesis, and we have no reason to not believe this hypothesis.

This doesn’t prove the hypothesis but we can say that the data support it.

If our null hypothesis was different than p=0.35, lets say p=30 and our sample yields \(\hat{p}\), then the data are not consistent with the hypothesis and we need to mke choices as to whethere this inconsistency is large enough to not believe the hypothesis.

If the inconsistency is significant, we reject the null hypothesis

Research conducted a few years ago showed that 35\(\%\) of UCLA students had travelled outside the US. UCLA has recently implemented a new study abroad program and results of a new survey show that out of the 100 randomly sampled students 42 have travelled abroad. Is there significant evidence to suggest that the proportion of students at UCLA who have travelled abroad has increased after the implementation of the study abroad program?

population proportion used to 0.35

new sample proportion is 0.42

Testing the claim that the population proportion is now greater than 0.35 .

But is this difference statistically significant, i.e. are the data inconsistent enough?

We do a formal hypothesis test to answer this question.

Null hypothesis, denoted by Ho, specifies a population model parameter of interest and proposes a value for that parameter (p). \[H_0: p=0.35\]

Alternative hypothesis, denoted by HA, is the claim we are testing for.\[H_A :p > 0.35\]

Even though we are testing for the alternative hypothesis, we check to see whether or not the null hypothesis is plausible.

If the null hypothesis is not plausible, we reject the null hypothesis and conclude that there is sufficient evidence to support the alternative. If the null hypothesis is plausible, we fail to reject the null hypothesis and conclude that there isn’t sufficient evidence to support the alternative.

The same logic used in jury trials is used in statistical tests of hypothesis

We begin by assuming that the null hypothesis is true.

Next we consider whether the data are consistent with this hypothesis.

If they are, all we can do is retain the hypothesis we started with. If they are not, then like a jury, we ask whether they are unlikely beyond a reasonable doubt.

Hypothesis testing is very much like a court trial

this doesnt prove the hypothesis but we can say that the data supports it

H_0:Defendant is innocent

H_A:Defendant is guilityWe then present the evidence-collect data

Then we judge the evidence - “Could these data plausibly have happened by chance if the null hypothesis were true?

If they were very unlikely to have occurred, then the evidence raises more than a reasonable doubt in our minds about the null hypothesis.

Ultimately we must make a decision. How unlikely is unlikely?

If the evidence is not strong enough to reject the presumption of innocent, the jury returns with a verdict of “not guilty”

The jury does not say that the defendant is innocent.

All it says is that there is not enough evidence to convict, to reject innocence.

The defendant may, in fact, be innocent, but the jury has no way to be sure.

Said statistically, we fail to reject the null hypothesis.

We never declare the null hypothesis to be true, because we simply do not know whether it’s true or not.

Therefore we never“accept the null hypothesis”

In Statistics we can quantify our level of doubt.

How unlikely is it to get a random sample of 100 students where 42 have travelled abroad if in fact the true population proportion is 35\(\%\)?

How unlikely is it to get a random sample of 100 students where 42 have travelled abroad if in fact the true population proportion is 35\(\%\)? To answer this question we use the model proposed by the null hypothesis as a given and calculate the probability that the event we have witnessed could happen.

Prob( Observed or more extreme outcome \(\vert\) H_0 true)=Prob(\(\hat{p}\) > 0.42 \(\vert\) p=0.35)

This probability quantifies exactly how suprised we are to see our results and is called the \(\textbf{p-value}\)

First we calculate the test statistic(Z-score)

Test statistic used for hypothesis testing for proportions is a z-score \[\frac{obs-mean}{SD}\]

Remember CLT: in calculating z-score we use the mean and SD by THE CLT, and observed value is \[\hat{p} \approx N\left(mean=p,SD=\sqrt{\frac{p(1-p)}{n}}\right)\]

\(\textbf{P comes from null hypothesis}\)

Random and Independent: The sample is collected randomly and the trials are independent of each ot

Large Sample:

Sample has at least 10 successes, \(np \geq 10\)

at least 10 failures \(n(1-p) \geq 10\)

Large Population: If the sample is collected without replacement, then the population size is at least 10 times the sample size.

We have that \(H_0:p=0.35\) and H_A:\(p>0.35\) \[z-score=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.42-0.35}{\sqrt{\frac{0.35*0.65}{100}}}=1.47\] and p-value\[Prob(\hat{p} > 0.42 \vert p=0.35=prob(z>1.47)=1-0.9292=0.0708\]

When the data are consistent with the model from the null hypothesis, the p-value is high and we are unable to reject the null hypothesis.

In that case, we have to“retain” the null hypothesis.

We can’t claim to have proved it; instead we fail to reject the null hypothesis and conclude that the difference we observed between the null hypothesis (p = 0.35) and the observed outcome (\(\hat{p}\)￼ = 0.42) is due to natural sampling variability (or chance).

If the p-value is low enough, we reject the null hypothesis since what we observed would be very unlikely if in fact the null model was true.

We call such results statistically significant

we compare the p value to a give \(\alpha\):

if pvalue < \(\alpha\) \(\rightarrow\) Reject H_0There is sufficient evidence to suggest to H_A is plausible

if p value > \(\alpha\) <

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