Answer:
We want to construct a confidence interval at 99% of confidence, so then the significance level would be 
and the value of 
. And for this case since we know the population deviation is not appropiate use the t distribution since we know the population deviation and the best quantile assuming that the population is normally distributed is given by the z distribution.
And if we find the critical value in the normal standard distribution or excel and we got:
And we can use the following excel code:
"=NORM.INV(0.005,0,1)"
Step-by-step explanation:
For this case we have the following info given:
We want to construct a confidence interval at 99% of confidence, so then the significance level would be and the value of . And for this case since we know the population deviation is not appropiate use the t distribution since we know the population deviation and the best quantile assuming that the population is normally distributed is given by the z distribution.
And if we find the critical value in the normal standard distribution or excel and we got:
And we can use the following excel code:
"=NORM.INV(0.005,0,1)"
Answer:
yes
Step-by-step explanation:
they are both triangles
Answer:
Mean track length for this rock specimen is between 10.463 and 13.537
Step-by-step explanation:
99% confidence interval for the mean track length for rock specimen can be calculated using the formula:
M±
where
- M is the average track length (12 μm) in the report
- t is the two tailed t-score in 99% confidence interval (2.977)
- s is the standard deviation of track lengths in the report (2 μm)
- N is the total number of tracks (15)
putting these numbers in the formula, we get confidence interval in 99% confidence as:
12±
=12±1.537
Therefore, mean track length for this rock specimen is between 10.463 and 13.537
I'm gonna say C for this one
Answer:
That's what I think the answer is
CB = CA - AB
= -8 - 29
= -37
