Name __________________________________
STAT 100.004, HW #7
Question 1. [Chapter 21, Exercise 3] The Baltimore Sun (Haney, 21 February 1995) reported on a
study by Dr. Sara Harkness in which she compared the sleep patterns of 6-month-old infants in the
United States and the Netherlands. She found that the 36 U.S. infants slept an average of 13 hours out
of every 24, whereas the 66 Dutch infants slept an average of 15 hours. If the standard deviation for
each group is 0.5 hours, compute an approximate 98% confidence interval for the difference in average
sleep time for 6-month-old Dutch and U.S. infants.
Show all your work.
Solution: This is a problem involving the difference of two means.
Sample di↵erence of means: 15
13 = 2
Multiplier for 98% CI: 2.33
s✓
◆2 ✓
◆2
0.5
0.5
p
Standard error of di↵erence:
+ p
= 0.104
36
66
Final confidence interval: 2 ± 2.33 ⇥ 0.104
Optional: Compute an approximate 90% confidence interval for the average sleep time for 6-monthold babies in the United States.
Solution: This is a problem involving a single mean.
Sample mean: 13
Multiplier for 90% CI: 1.64
0.5
Standard error of mean: p = 0.083
36
Final confidence interval: 13 ± 1.64 ⇥ 0.083
Question 2. ArecentGalluppollaskedAmericanadultsiftheywereworriedaboutvarioustypes
ofrobberyandtheft.Galluptookarepresentativesampleof1000adults.Themembersofthe
sampleseemedmostconcernedwithidentitytheft,with690adultssayingtheyareworried
abouttheiridentitybeingstolen.Constructanappropriate84%confidenceintervalforthis
situation,andexplaininwordswhatthisintervalrepresents.
Show all your work.
Solution: This is a problem involving a single proportion.
Sample proportion: 690/1000 = 0.69
Multiplier for 84% CI: 1.41
r
0.69 ⇥ 0.31
Standard error of sample proportion:
= 0.015
1000
Final confidence interval: 0.69 ± 1.41 ⇥ 0.015
Optional: How would you change your answer to make it a 99% confidence interval?
Solution: The multiplier for a 99% CI is 2.576. So everything would be the same except we’d replace
1.41 by 2.576 in the final confidence interval.