A group of statistics students decided to conduct a survey at their university to estimate the average (mean) amount of time students spent studying per week. They sampled 554 students and found a mean of 22.3 hours per week. Assuming a population standard deviation of six hours, what is the 95% level of confidence? A. [16.3, 28.3] B. [21.64, 22.96] C. [21.80, 22.80] D. [20.22, 22.0]

C. [21.80, 22.80]

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find M as such

In which is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 22.3 - 0.5 = 21.8 hours.

The upper end of the interval is the sample mean added to M. So it is 22.3 + 0.5 = 22.8 hours

So the correct answer is:

C. [21.80, 22.80]

step-by-step explanation:

(s)he  is right it would be 5.5% because it is smaller than 15.5% or anything smaller than 15.5% in general

sum= when you add is how much the whole amount is

distance= when you use angles and it means how far it goes

step-by-step explanation: