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Statistics right?

Right..

I can help.

So have you covered the t distribution yet?

|dw:1354857899830:dw|

|dw:1354857941165:dw|

Thanks so much for the help. Im guessing thats a sqrt(n) on the bottom?

Does this make sense so far?

|dw:1354858016253:dw|

Now, in class do you use a graphic calculator or table?

Both. :/ But I prefer the calculator.

ok

Following so far?

I think so..

If so, how does that come into play? or am i trialing off??

|dw:1354858281536:dw|

Right, so we need to know the total area under that portion of the curve, right?

yes

And the lines are 2 std deviations from the mean, right?

Almost

I'll return to that later. So, for now, what is the area where the horizontal lines are?

|dw:1354858667169:dw|

|dw:1354858861005:dw|

that means the area under the curve up to that point is 1-0.025 = 0.975

Following so far

That will a be the p value that you can enter into the calculator.
so invnorm(0.975)

it will be close to 2, but not exactly 2.

what result do you get for the z score?

You still there?

You should get something like 1.96 for the z value

.\1.959

Ah. Makes sense now.

The invnorm() command converts a cumulative probability to a zscore

It's important to calculate the z scores right or these type of problems will be difficult.

Do you think you can finish from here?

I can try. :) Thanks so much for the help.

Use this formula and the values that I just solved for:

|dw:1354859367648:dw|

z=1.96, s=5, n=22

Awesome. Only thing really left to do now is plug in and apply. Thanks so much.

Let me know what you get....

so |dw:1354859570561:dw|

88+1.96(5/22)
88-1.96(5/22)
right? so 88.44 and 87.55 ??

2.09 is the margin error

85.9 to 90.1

...woah

it's sqrt 22

ah.... caught me

got it

Is that it?

thats it. :) Thanks

Yep!