Kindly allow the step by step process with formulas and how to calculate the Z-table (if required) (no internet and website)

question-1

A survey of an urban university showed that 750 of 1100 students sampled attended a home football game during the season. Using the 99% level of confidence, what is the confidence interval for the proportion of students attending a football game?

Question -2

The 95% Confidence Interval for the weights (in pounds) of a special type of rock has found to be 4.30, 8.46. What will be the margin of error? (Hint: Sample mean +/- margin of error = confidence interval)

Question-3

In a large city, the average number of lawn mowings during summer is normally distributed with mean and standard deviation = 8.7. If I want the margin of error for a 90% confidence interval to be 3, I should select a simple random sample of size (4 decimal points)

Question 4

In a large city, the average number of lawn mowings during summer is normally distributed with mean and standard deviation = 7.3. If I want the margin of error for a 99% confidence interval to be 4, I should select a simple random sample of size (4 decimal points)

Question -5

The standard error of the mean for a sample of size 179 is 25. In order to cut the standard error of the mean in half (to 12.5) we must select a sample size of

Question 6

42 credit card holders are selected at random. For each, their current credit card balance is recorded. The average for these 42 people is= $600. Assume that the current balance of all credit card holders follows a normal distribution with unknown mean , and that a 85.3% confidence interval for is found to be $600 34.6. Find the standard deviation of the population?