# Best Recent Content

• ### 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 P-hat = $(20/60)$ = 0.33

calculate your test statistic using $(P hat - P)/ SE$

SE = $sqrt((P(1-P)/n)$
Take this value to the Z table and find the p-value.
Since the question states "more than 20 of the sixty" , you will take the value of
(1-pvalue).
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
$\sqrt{p(1-p)/n}$. 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 P-hat = $(20/60)$ = 0.33

calculate your test statistic using $(P hat - P)/ SE$

SE = $sqrt((P(1-P)/n)$
Take this value to the Z table and find the p-value.
Since the question states "more than 20 of the sixty" , you will take the value of
(1-pvalue).
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 Chi-squared

I got 2.32 for the $\chi^2$-stat - probably just a rounding difference.

• ### X2 Chi-squared

Problem 12

Did anyone else get 2.25 for a chi-squared statistic, DF=2 with a t*value of 4.30? Thus fail to reject null?

• ### comparing data sets

Final exam review question #10

I got a test statistic of 1.88329

For which .9699 is the P-value

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 p-value 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

@jordan said:
reference final exam review question #7 part (a) and (d):
I feel like we would reject the null because of two reasons:
I. Because the slope is not 0 in the table (its 5.7984)
II. The line is literally slanted on the graph itself

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!!

• ### z-score vs t-score

@jordan said:
reference: final exam review problem #5 part (d)

I know a t-score should be solved for when working with <30 data. But is the formula different than the z-score formula?

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.

• ### z-score vs t-score

reference: final exam review problem #5 part (d)

I know a t-score should be solved for when working with <30 data. But is the formula different than the z-score formula?

I know its df:3 and two-tailed with 95% confidence, so cut off should be 3.18