Question

I'm not sure if I'm overthinking this question, so could someone please walk me through this? :):
'If the margin of error in an estimate for the mean weight of a shipment is + or -2 pounds at a confidence level of 95 percent, what will be the margin of error at a confidence level of 98 percent? Be sure to show how you arrived at your answer.'
Thanks

Answer 1

http://www.flashandmath.com/mathlets/statistics/meansci/ciformula2.jpg

This formula will be applied for confidence interval range. (or use z* instead of t* if it is a population)

So the margin of error is just z*s/sqrt(n)) Find z* (critical z value based on your tables), for 95% CI and then rearrange the equation to find s/sqrt(n)
Then multiply this s/sqrt(n) by z* for 98% CI and you're done
The point is the standard error, i.e. s/sqrt(n) does not change, but the z* given by confidence level does.

Answer 2

if the involved distribution is the gaussian distribution, then we can write:
\[1.96 \cdot \sigma = 2\]
wherein \(\sigma\) is the standard deviation
please find \(\sigma\)

Answer 3

@mww ahh okay, that makes sense! Thank you so much :) also @Michele_Laino tysm

Answer 4

:)
the requested uncerainty, is:
\[2.33 \cdot \sigma = ...?\]

Answer 5

Margin of error is the z score times the standard error.\[ME = z* SE\]since you know the margin of error, and the z score for 95% confidence (1.96), you can find the Standard Error.

Then just plug in the z score for 98% confidence (get it from a z table) and multiply by the same SE you just worked out.

Answer 6

Though it looks like you might want to use t* since i'm guessing it's a sample.

Answer 7

awesommme thank you all your explanations helped a lot !! @agent0smith we're using z :) ty!!