I am having trouble calculating the margin of error and on how to interpret the confidence level.

Let's practice on how to calculate the margin of error (if any) when we are going to find the confidence interval. (if you believe there is no margin of error, please briefly explain it). 1) In a time use study 20 randomly selected managers were found to spend a mean time of 2.4 hours per day on paperwork. The standard deviation of the 20 scores was 1.30 hours. Calculate the margin of error for the process of constructing a 98% confidence interval for the mean time spent on paperwork by all managers. (Do not need to find the confidence interval). 2) A national health organization warns that 30% of the middle school students nationwide have been drunk. Concerned, a local health agency randomly and anonymously surveys 110 of the middle 1212 middle school students in its city. Only 21 of them report having been drunk. Calculate the margin of error for the process of creating a 95% confidence interval for the proportion of the city's middle school students who have been drunk. (Do not need to find the confidence interval). 3) A process has been developed that can transform ordinary iron into a kind of super iron called metallic glass. Metallic glass is three to four times stronger than the toughest steel alloys. To estimate the mean temperature, u, at which a particular type of metallic glass becomes brittle, 25 pieces of this metallic glass were randomly sampled from a recent production run. Each piece was subjected to higher and higher temperatures until it became brittle. The temperature at which brittleness first appeared was recorded for each piece in the sample. The following results were obtained: x = 480 F and s = 11 F. Calculate the margin of error for the process of construct a 95% confidence interval to estimate o.1) The first sample had 30 trees with a mean circumference of 15.71 inches, and standard deviation of 4.63 inches. A 95% confidence interval for the mean circumference of aspen trees from this data is found as (13.98, 17.44). Interpret this confidence interval. 2) Suppose in a state with a large number of voters that 56 out of 100 randomly surveyed voters favored Proposition X. A 95% confidence interval for the proportion of all voters who favored Proposition X can be found as (0.462, 0.653). Interpret this confidence interval. 3) Given that x = 50, s = 5.02, n = 16 Assume that the underlying population is normally distributed. Hence, a 95% confidence interval for the population standard deviation can be found as (3.804, 8.194). Interpret this confidence interval