8.48
(a) Construct a 95 percent confidence interval for the true mean.
sample mean = x-bar = 546
sample sd= s =170.378
α=0.05
t(α/2,n-1)=t(0.025,19)=2.093
CI at 95% = x-bar±t(α/2,n-1)s/√n = 546±2.093(170.378)/√20
= (266.762, 426.238)
(b) Why might normality be an issue here?
Normality may be an issue here since there is a lower limit and an upper limit on the area. Lower
limit being zero and upper limit being the full page area.
(c)
(c) What sample size would be needed to obtain an error of ±10 square millimeters
with 99 percent confidence?
Since we don´t know n, we asume that =s=170.38
n = (z(0.005)/E)^2
n = (2.5758*170.38/10)^2
n = 1926.02
Answer: n=1927
(d) If this is not a reasonable requirement, suggest one that is.
This is not areasonable requirement since 1927>1591
We can select a lower level of confidence (95% for example) or a greatest error (20 or 30)
8.64 Biting an unpopped kernel of popcorn hurts! As an experiment, a self-confessed connoisseur
of cheap popcorn carefully counted 773 kernels and put them in a popper. After popping, the
unpopped kernels were counted. There were 86.
(a) Construct a 90 percent confidence interval for the proportion of all kernels that would not
pop.
p-hat = 86/773
=0.1
n=773
CI at 90% = p-hat ± z(0.05)[p-hat(1-p-hat)/n]=
86/773± 0.0113
Answer: (0.0927;0.1299)
(b) Check the normality assumption.
We have to check that n(P-hat)>5 and n(1-p-hat)>5
n(p-hat)=773*0.1113=86.03
n(1-p-hat)=773*(0.8887)=686.97
Answer: Normality assumption is verified
(c) Try the Very Quick Rule. Does it work well here? Why, or why not?
The very quick rule requires a p-hat so close to 0.5, then it does not work here
(p-hat =0.1113)
(d) Why might this sample not be typical
Because the person doing the experiment is a "self-confessed connoisseur of cheap popcorn" and
not a normal popcorn consumer.