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random variables and distributions
Hi Jordan,
I approached this as a one sample proportion test using a normal distribution to estamate probability( Z tables)we are given a P = 0.20 and a Phat = $(20/60)$ = 0.33
calculate your test statistic using $(P hat  P)/ SE $
SE = $sqrt((P(1P)/n)$
Take this value to the Z table and find the pvalue.
Since the question states "more than 20 of the sixty" , you will take the value of
(1pvalue).
I hope I did this correctly. 
random variables and distributions
@Beau said:
Hi Jordan,
I approached this as a one sample proportion test using a normal distribution to estamate probability( Z tables)Yes, this is exactly right!
The estimate $\hat{p}$ is a random variable whose mean and standard deviation are based on the actual proportion $p=0.2$ and the sample size $n=60$. More precisely, the mean of $\hat{p}$ is $np$ and the standard deviation of $\hat{p}$ is
$\sqrt{p(1p)/n}$. Thus, you use a normal distribution with those values for mean and standard deviation to do your estimate. 
random variables and distributions
Hi Jordan,
I approached this as a one sample proportion test using a normal distribution to estamate probability( Z tables)we are given a P = 0.20 and a Phat = $(20/60)$ = 0.33
calculate your test statistic using $(P hat  P)/ SE $
SE = $sqrt((P(1P)/n)$
Take this value to the Z table and find the pvalue.
Since the question states "more than 20 of the sixty" , you will take the value of
(1pvalue).
I hope I did this correctly. 
random variables and distributions
Final exam review question #4
I think this is drawing from random variables and distributions. I don't remember how to approach this problem. Anyone have any advice? _

X2 Chisquared

X2 Chisquared

comparing data sets
Final exam review question #10
I got a test statistic of 1.88329
For which .9699 is the Pvalue
I don't really know how to evaluate this though... if we're shooting for 95% confidence (which is implicit in the question because its not stated), because the pvalue is so large I think I need to subtract it from 1, which gives .0301, which exceeds .025 so we would fail to reject the null?

hypothesis statement and conclusion
I agree that you would reject the null neither of your reasons are relevant.
Remember that the slope computation of 5.7984 arises from a sample. If you collect another sample, you'll get a different slope. Neither of those are the actual slope but, presumably, they are close to the slope. The question is  "how confident are you that the slope is not zero"? This depends not only on the computed value but also the standard error. Ultimately, we express our level of confidence using the $p$value. That's how hypothesis testing works!!

zscore vs tscore
No, the formula for computing the test statistic for one sample mean or one sample proportion, no matter the sample size. The only difference is which distribution you end up comparing it to  the normal distribution or the $t$distribution.

zscore vs tscore
reference: final exam review problem #5 part (d)
I know a tscore should be solved for when working with <30 data. But is the formula different than the zscore formula?
I know its df:3 and twotailed with 95% confidence, so cut off should be 3.18