In order to estimate the average time spent on the computer terminals per student at a local university, data were collected from a sample of 64 business students over a one-week period. Assume the population standard deviation is 1.2 hours. a) With a 0.98 probability, the margin of error is approximately?b) If the sample mean is 10 hours, then the 98% confidence interval is approximately ?c) The standard error of the mean is?

a) 0.3495

b) (9.65, 10.35)

Step-by-step explanation:

We are given the following in the question:

Sample mean, = 10 hours

Sample size, n = 64

Alpha, α = 0.02

Population standard deviation, σ = 1.2 hours

a) Margin of error

Formula

Putting values, we get,

b)  the sample mean is 10 hours, then the 98% confidence interval

Putting the values, we get,

b) well because you get well

javeir earned 56 dollors more than kyle so kyle and javier had 14w but javier made 56 dollors more so 14w plus 56 equals 70w so kyle has 14w

step-by-step explanation: hope i plz give me brainliest with the points so this is not constant cause it says last week so its clearly not every week