Question

QUESTION 22

Use your TI83 to find the upper end of the interval requested:

A 95% confidence interval for the average weight of a box of cereal if a sample of 26 such boxes has an average of 17.14 ounces with a sample deviation of 0.228 ounces. The population of all such weights is normally distributed

round to the nearest hundreth of an ounce

QUESTION 23

Use your TI83 to find the upper end of the interval requested:

A 90% confidence interval for the average healthy human body temperature if a sample of 14 such temperatures have an average of 98.59 degrees F with a sample deviation of 0.214 degrees F The population of all such temperatures is normally distributed

round to the nearest hundreth of a degree

QUESTION 24

Use your TI83 to find the upper end of the interval requested:

A 95% confidence interval for the average waiting time at the drive-thru of a fast food restaurant if a sample of 169 customers have an average waiting time of 105 seconds with a population deviation of 34 seconds

round to the nearest hundredth of a second

QUESTION 25

Use your TI83 to find the lower end of the interval requested:

A 95% confidence interval for the average weight of a box of cereal if a sample of 28 such boxes has an average of 17.18 ounces with a sample deviation of 0.23 ounces. The population of all such weights is normally distributed

round to the nearest hundreth of an ounce

Answer 1

22)

Given

95% C.I

n = 26

M = 17.14

s = 0.228

M = 17.14
t = 2.06

s_M = standard error = √(s²/n)
s_M = √(0.228²/26) = 0.04

μ = M ± t(s_M)
μ = 17.14 ± 2.06*0.04
μ = 17.14 ± 0.0921

95% CI [17.0479, 17.2321].

95% CI [17.05, 17.23]. (Round to hundredth of ounce)

23)

n = 14

M = 98.59

s = 0.214

M = 98.59
t = 1.77
s_M = √(0.214²/14) = 0.06

μ = M ± t(s_M)
μ = 98.59 ± 1.77*0.06
μ = 98.59 ± 0.1013

90% CI [98.4887, 98.6913].

90% CI [98.49, 98.69].

24)

n = 169

M = 105

s = 34

M = 105
t = 1.97
s_M = √(34²/169) = 2.62

μ = M ± t(s_M)
μ = 105 ± 1.97*2.62
μ = 105 ± 5.16

95% CI [99.84, 110.16].

25)

n = 28

M = 17.18

s = 0.23

M = 17.18
t = 2.05
s_M = √(0.23²/28) = 0.04

μ = M ± t(s_M)
μ = 17.18 ± 2.05*0.04
μ = 17.18 ± 0.0892

95% CI [17.0908, 17.2692].

95% CI [17.09, 17.27].