At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I can help.
So have you covered the t distribution yet?
I'm willing to work through it to show you how to do this problem. If you haven't covered the t-distribution. Then Just use this formula:
Thanks so much for the help. Im guessing thats a sqrt(n) on the bottom?
Does this make sense so far?
Now, in class do you use a graphic calculator or table?
Both. :/ But I prefer the calculator.
Following so far?
I think so..
Because its a 95% confidence interval, wouldnt it include all values withen 2 standard deviations of the mean?
If so, how does that come into play? or am i trialing off??
Right, so we need to know the total area under that portion of the curve, right?
So basically, you're looking for the area under the curve right up to one of horizontal lined regions.
And the lines are 2 std deviations from the mean, right?
I'll return to that later. So, for now, what is the area where the horizontal lines are?
So the horizontal lines add up to .05. So x+x=.05 That means x=.025. That is the area under one of the horizontal striped .
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
basically ,that's the reason why 95% of the population is around 2 standard deviations from the mean.
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:
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....
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
it's sqrt 22
ah.... caught me
Is that it?
thats it. :) Thanks