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Mathematics 22 Online
OpenStudy (anonymous):

PLEASE HELP! See attachment.

OpenStudy (anonymous):

OpenStudy (anonymous):

@phi Can you help with this.

OpenStudy (phi):

can you rule out any of the choices for (a) ?

OpenStudy (anonymous):

i think it is supposed to be 63.9

OpenStudy (phi):

what is the null hypothesis H0 ?

OpenStudy (phi):

the null hypothesis is that the height did not change.

OpenStudy (phi):

The hypothesis H1 is "women are taller today" the null hypothesis H0 is "women's height has not changed"

OpenStudy (anonymous):

so 63.9

OpenStudy (phi):

I don't understand. none of the answers are a single number.

OpenStudy (phi):

women's average height was 63.7. if the height does not change, what is their average height?

OpenStudy (anonymous):

i'm so confused.

OpenStudy (phi):

yes, I was thinking that... but it's hard to zero in on what you are missing...

OpenStudy (anonymous):

So the answer is ether A,C,or D?

OpenStudy (phi):

yes. The idea is you have a guess (hypothesis) that height changed (increased) or maybe not... the maybe not is the null hypothesis... in other words , the guess is wrong. if the height did not change, then the height is the same as it used to be mu = 63.7

OpenStudy (phi):

now you need the "guess" written as an equation the height has increased. mu is bigger than it was... mu > 63.7

OpenStudy (anonymous):

so C?

OpenStudy (phi):

choice A mu ≠ 63.7 is the guess that the height is shorter or taller. but we are guessing taller choice D mu < 63.7 means height got shorter. choice C mu > 63.7 means height is taller. so yes, choice C

OpenStudy (anonymous):

OK now Part (b)

OpenStudy (phi):

from wikipedia p-values are the probability of obtaining a test statistic at least as extreme as the one that was actually observed,

OpenStudy (phi):

in other words, they did a test, and measured the average height to be 63.9 However, they did not measure all women, only 45 of them, so there is a chance they got a few tall women (or not enough short women), to skew the average. In other words, if they measured *every* women, they would find the average to be 63.7, but because they only measured 45, they got the wrong answer. What is the chance that the average did not change, and they were "unlucky" 13% (the p-value)

OpenStudy (phi):

can you rule out any of the choices in (b) ?

OpenStudy (anonymous):

is it B?

OpenStudy (phi):

you have to read carefully. they write "a random sample of 45 women results in a mean height of 63.9" re-phrased: "the sample mean height = 63.9"

OpenStudy (phi):

try to get the idea: population mean (millions of women) sample mean ( a handful of women) sample mean should equal population mean, but only if the "sample size" is large (the larger the better)

OpenStudy (phi):

now you need to interpret the definition of the p-value p-values are the probability of obtaining a test statistic at least as extreme as the one that was actually observed, probability of obtaining a test statistic test statistic here is the "sample mean" (the average computed using 45 women) at least as extreme as .... that means the "sample mean" could have been even larger p-value is the probability mu > 63.9 where mu is the "sample mean"

OpenStudy (phi):

or I should say, p-value is the probability mu≥ 63.9 (mu greater than or equal to 63.9)

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