Question

QUESTION #1

How hot is the air in the top (crown) of a hot air balloon? Information from Ballooning: The Complete Guide to Riding the Winds, by Wirth and Young (Random House), claims that the air in the crown should be an average of 100°C for a balloon to be in a state of equilibrium. However, the temperature does not need to be exactly 100°C. What is a reasonable and safe range of temperatures? This range may vary with the size and (decorative) shape of the balloon. All balloons have a temperature gauge in the crown. Suppose that 53 readings (for a balloon in equilibrium) gave a mean temperature of x = 97°C. For this balloon, σ ≈ 20°C.

(a) Compute a 95% confidence interval for the average temperature at which this balloon will be in a steady-state equilibrium. (Round your answers to one decimal place.)

lower limit	°C
upper limit	°C

(b) If the average temperature in the crown of the balloon goes above the high end of your confidence interval, do you expect that the balloon will go up or down? Explain.

It will go up because hot air will make the balloon rise.

It will go down because hot air will make the balloon fall.    

It will go up because hot air will make the balloon fall

It will go down because hot air will make the balloon rise.

QUESTION # 2

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 14 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with σ = 0.28 gram.

When finding an 80% confidence interval, what is the critical value for confidence level? (Give your answer to two decimal places.)

z_c =

(a)

Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.)

lower limitupper limitmargin of error

Answer 1

#1.
sample mean, xbar = 97
sample standard deviation, σ = 20
sample size, n = 53

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (97 - 1.96 * 20/sqrt(53) , 97 + 1.96 * 20/sqrt(53))
CI = (91.6 , 102.4)

Lower limit = 91.6
upper limit = 102.4

b)
It will go up because hot air will make the balloon rise.

#2.
sample mean, xbar = 3.15
sample standard deviation, σ = 0.28
sample size, n = 14

Given CI level is 80%, hence α = 1 - 0.8 = 0.2
α/2 = 0.2/2 = 0.1, Zc = Z(α/2) = 1.28
zc = 1.28

ME = zc * σ/sqrt(n)
ME = 1.28 * 0.28/sqrt(14)
ME = 0.10

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (3.15 - 1.28 * 0.28/sqrt(14) , 3.15 + 1.28 * 0.28/sqrt(14))
CI = (3.05 , 3.25)