1) A confidence interval is to be found using a sample of size 79 and a known population deviation of 2.327. If the critical value should be a z-score, type the number 0 below If the critical value should be a t-score, type the number 1 below *The computer is looking for either the input 0 or the input 1. It will not recognize anything else you type in

2) A confidence interval is to be found using a sample of size 682 and the sample deviation of 4.844. If the critical value should be a z-score, type the number 0 below If the critical value should be a t-score, type the number 1 below *The computer is looking for either the input 0 or the input 1. It will not recognize anything else you type in

3) What would be the degree of freedom for a sample of size 91?

4) A sample of weights of 29 boxes of cereal yield a sample average of 17.5 ounces. What would be the margin of error for a 99% CI of the average weight of all such boxes if the sample deviation is 0.58 ounces? The population of all such weights is normally distributed. Round to the nearest hundredth

5) If a sample average is found to be 98.54, and the margin of error is calculated to be 0.05, then the upper end of the confidence interval for mu would be ______

6) If a sample average is found to be 98.61, and the margin of error is calculated to be 0.10, then the lower end of the confidence interval for mu would be ______

7) Use your TI83 to find the lower end of the interval requested: A 95.0% confidence interval for the average weight of a box of cereal if a sample of 24 such boxes has an average of 17.10 ounces with a sample deviation of 0.224 ounces. The population of all such weights is normally distributed round to the nearest hundreth of an ounce

8) Use your TI83 to find the lower end of the interval requested: A 99.5% confidence interval for the average healthy human body temperature if a sample of 16 such temperatures have an average of 98.67 degrees F with a sample deviation of 0.244 degrees F The population of all such temperatures is normally distributed round to the nearest hundreth of a degree

9) Use your TI83 to find the lower end of the interval requested: A 99.0% confidence interval for the average healthy human body temperature if a sample of 11 such temperatures have an average of 98.58 degrees F with a sample deviation of 0.265 degrees F The population of all such temperatures is normally distributed round to the nearest hundreth of a degree

10) Use your TI83 to find the upper end of the interval requested: A 95.0% confidence interval for the average height of the adult American male if a sample of 23 such males have an average height of 57.5 inches with a sample deviation of 3.5 inches. The population of all such heights is normally distributed round to the nearest hundreth of an inch

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