Question

QUESTION 1

A sample of weights of 50 boxes of cereal yield a sample average of 17 ounces. What would be the margin of error for a 90% CI of the average weight of all such boxes if the population deviation is 0.48 ounces?

Round to the nearest hundredth

QUESTION 2

A sample of heights of 117 American men yields a sample average of 57.04 inches. What would be the margin of error for a 95.44% CI of the average height of all such men if the population deviation is 3.8 inches?

Round to the nearest hundredth

QUESTION 3

A sample of weights of 37 boxes of cereal yield a sample average of 17.8 ounces. What would be the margin of error for a 95% CI of the average weight of all such boxes if the population deviation is 0.35 ounces?

Round to the nearest hundredth

QUESTION 4

A sample of weights of 25 boxes of cereal yield a sample average of 16.5 ounces. What would be the margin of error for a 96% CI of the average weight of all such boxes if the sample deviation is 0.73 ounces?

The population of all such weights is normally distributed.

Round to the nearest hundredth

QUESTION 5

:

A confidence interval is to be found using a sample of size 676 and the sample deviation of 5.89.

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

Answer 1

1)

sample mean, xbar = 17
sample standard deviation, σ = 0.48
sample size, n = 50

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

ME = zc * σ/sqrt(n)
ME = 1.64 * 0.48/sqrt(50)
ME = 0.11

2)

sample mean, xbar = 57.04
sample standard deviation, σ = 3.8
sample size, n = 117

Given CI level is 95.44%, hence α = 1 - 0.9544 = 0.0456
α/2 = 0.0456/2 = 0.0228, Zc = Z(α/2) = 2

ME = zc * σ/sqrt(n)
ME = 2 * 3.8/sqrt(117)
ME = 0.70

3)

sample mean, xbar = 17.8
sample standard deviation, σ = 0.35
sample size, n = 37

Given CI level is 95.44%, hence α = 1 - 0.9544 = 0.0456
α/2 = 0.0456/2 = 0.0228, Zc = Z(α/2) = 2

ME = zc * σ/sqrt(n)
ME = 2 * 0.35/sqrt(37)
ME = 0.12

4)

sample mean, xbar = 16.5
sample standard deviation, s = 0.73
sample size, n = 25
degrees of freedom, df = n - 1 = 24

Given CI level is 96%, hence α = 1 - 0.96 = 0.04
α/2 = 0.04/2 = 0.02, tc = t(α/2, df) = 2.17

ME = tc * s/sqrt(n)
ME = 2.17 * 0.73/sqrt(25)
ME = 0.32

5)

number 1 becaus esample deviation is given