Question

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 ≈ 19°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 fall.It will go down because hot air will make the balloon fall.     It will go down because hot air will make the balloon rise.It will go up because hot air will make the balloon rise.

Answer 1

a.
TRADITIONAL METHOD
given that,
standard deviation, σ =19
sample mean, x =97
population size (n)=53
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 19/ sqrt ( 53) )
= 2.6
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 2.6
= 5.1
III.
CI = x ± margin of error
confidence interval = [ 97 ± 5.1 ]
= [ 91.9,102.1 ]
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DIRECT METHOD
given that,
standard deviation, σ =19
sample mean, x =97
population size (n)=53
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 97 ± Z a/2 ( 19/ Sqrt ( 53) ) ]
= [ 97 - 1.96 * (2.6) , 97 + 1.96 * (2.6) ]
= [ 91.9,102.1 ]
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interpretations:
1. we are 95% sure that the interval [91.9 , 102.1 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean

b.
Given that,
population mean(u)=100
standard deviation, σ =19
sample mean, x =97
number (n)=53
null, Ho: μ=100
alternate, H1: μ<100
level of significance, α = 0.05
from standard normal table,left tailed z α/2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 97-100/(19/sqrt(53)
zo = -1.149
| zo | = 1.149
critical value
the value of |z α| at los 5% is 1.645
we got |zo| =1.149 & | z α | = 1.645
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : left tail - ha : ( p < -1.149 ) = 0.125
hence value of p0.05 < 0.125, here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=100
alternate, H1: μ<100
test statistic: -1.149
critical value: -1.645
decision: do not reject Ho
p-value: 0.125
we do not have enough evidence to support the claim that If the average temperature in the crown of the balloon goes down.