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Mathematics 9 Online
OpenStudy (anonymous):

In a set of hypotheses, the smaller the p-value, The stronger the evidence against the possibility that the observed results arose simply by chance The larger the level of significance The greater the practical importance of the results The wider the 95% confidence interval will be The larger the z value will be

OpenStudy (anonymous):

@ybarrap

OpenStudy (anonymous):

B?

OpenStudy (ybarrap):

I think of the p-value as "the probability that the difference observed was due to randomness". Do NOT take this literally, but that's how to think about it. So, if the p-value is 0, for example then there is 0% chance that the difference observed was due to random variations and you would accept the null hypothesis. Using this way of thinking, which do you think is the correct answer?

OpenStudy (ybarrap):

Small p-value means accept the null. Large p-value means reject it. Now what do you think?

OpenStudy (anonymous):

A?

OpenStudy (ybarrap):

yes!

OpenStudy (anonymous):

haha!

OpenStudy (ybarrap):

small p-value means little chance of alternative

OpenStudy (anonymous):

can you help me with more questions?

OpenStudy (ybarrap):

k

OpenStudy (anonymous):

Suppose the SAT scores for reading vary normally with a mean µ= 475 and a standard deviation of sigma equals 100 Ten thousand students go through a training program to improve their score. The average SAT scores for the students are found to be larger than 475 and statistically significant with a p-value of 0.0028. Which of the following can be concluded? We cannot say if the score improved at all. The average score is improved by training by at least 100 points The score must have been improved by a large number to produce such a small p-value. We cannot say by exactly how much the SAT reading score is improved, only that the observed improvement is unlikely to have arisen by chance. We cannot say by exactly how much the average score is raised.

OpenStudy (anonymous):

i was going to say c?

OpenStudy (ybarrap):

What would you say is the null hypothesis? Because a small p-value using that as the hypothesis means that you accept it.

OpenStudy (anonymous):

hmmm.

OpenStudy (anonymous):

ohh b!

OpenStudy (anonymous):

because its below .5

OpenStudy (anonymous):

well pvalue exactly

OpenStudy (ybarrap):

p-values don't say anything about how much something is larger, only that it is. So say that something is significant means small p-value p-values only say was difference between observed and expected due to chance or not. If your null hypothesis is that average was higher, and you get a small p-value, then you accept the null hypothesis.

OpenStudy (anonymous):

so not b or c

OpenStudy (anonymous):

are you saying its too small to determine?

OpenStudy (ybarrap):

no, smallness only tells us whether it is significant or not . That's it.

OpenStudy (anonymous):

D!

OpenStudy (ybarrap):

We can't say how much it went up only that any difference could not be because of randomness

OpenStudy (anonymous):

so it is D orE hmm

OpenStudy (anonymous):

E for sure because d is just too in depth

OpenStudy (anonymous):

To access the accuracy of a standard weight scale in a laboratory, a known weight is 20 grams and is repeatedly weighed. The scale readings are normally distributed with a standard deviation of sigma equals 0.01 grams. Suppose a scientist weighs the standard weight 4 times and obtains a mean weight of x-bar is 20.01 grams. Construct a 99% confidence interval for the mean of repeated measurements of the weight. (20.004, 20.016) (20.000, 20.020) (19.997, 20.023) (19.994, 20.006) (19.987, 20.013)

OpenStudy (anonymous):

@ybarrap

OpenStudy (anonymous):

i was saying it was D

OpenStudy (ybarrap):

eh? $$ \large{ \overline{x}\pm\cfrac{z_{((1-c)/2}}{\sqrt{n}} }\\ c=0.99\\ \sigma=.01\\ \overline{x}=20.01\\ n=4 $$ And for z, I'll let you look up in a table. When you do this, you'll get $$ (20,20.02) $$

OpenStudy (ybarrap):

FYI- Two-tailed(99%): z < -2.576 and z > 2.576

OpenStudy (ybarrap):

Left out the sigma \(\sigma\). $$ \large{ \overline{x}\pm\cfrac{z_{((1-c)/2}\sigma}{\sqrt{n}} }\\ c=0.99\\ \sigma=.01\\ \overline{x}=20.01\\ n=4 $$

OpenStudy (anonymous):

Alighty I understand! i Have to note this down! Hold on! One more question!

OpenStudy (ybarrap):

k

OpenStudy (anonymous):

The following hypotheses were tested Ho:u=0 Ha:u>0 h sub 0 mu equals 0 and h sub a mu greater than 0 And a p-value of 0.023 is obtained. What can be concluded? The chance that the alternate hypothesis is true is 0.023 The chance that the null hypothesis is true is 0.023 The chance of obtaining a value of the test statistic as or more extreme than the value actually obtained if the null is true is 0.023 The chance of obtaining a value of the test statistic which is > 0 is only 0.023 if the null is true The chance that the null hypothesis is true is 0.977

OpenStudy (anonymous):

Didn't we go over this? I dont want to make you work all over again

OpenStudy (anonymous):

let me try this on my own

OpenStudy (ybarrap):

of course I'll let you!!

OpenStudy (anonymous):

I know it has something to do with null hypothesis

OpenStudy (ybarrap):

yes, so with a null hypothesis is associated a p-value, if it is small then the null is more likely to be try

OpenStudy (ybarrap):

*true not try

OpenStudy (ybarrap):

remember what I said, the p-value is NOT the probability that the alternative is false or true, it is a measure, but you can THINK of it as a probability -- but it does not mean that it is the probability of true or false, so you can eliminate some of your options

OpenStudy (anonymous):

B.!

OpenStudy (anonymous):

none of these answers were correct....

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