Assignment Stats 250 W11 HW 6
Graded Point Total: 25 out of 30
a.
0.5 out of 1
points
Feedback:
dbar is reserved for sample mean
difference shouldn't be used here.
See Textbook for Exercise: 11.a.
Your Answer:
population mean difference,
μ
. This is the population mean difference in height between college
women and their mothers. (daughtermother)
Solution:
The parameter of interest is μ
d
= the population mean difference in heights (in inches) between
all college women and their mothers (where difference = daughter height  mother height).
b.
2.5 out of 3
points
Feedback:
wrong notation.
See Textbook for Exercise: 11.b.
Your Answer:
(0+2+1+3+3)/5 = 1.4 =
assuming difference to be heights of daughter  mother
sqrt(((01.4)^2+(21.4)^2+(11.4)^2+(31.4)^2+(31.4)^2) / (51)) = 1.81659 = s
s/sqrt(n) = 1.81659/sqrt(5) = .8124 = s.e.(
)
t* multiplier for a 95% confidence interval with 51 degrees of freedom =2.78
Confidence Interval =
+/t*s.e.(
)
= 1.4+/2.78(.8124)
=[ 8585 3 658]
Required Assignment Stats 250 W11 HW 6
Question 1
(11.37)
See Textbook for Exercise: 11.37.
4/17/2011
www.lecturebook.com/homeworks/sho…
lecturebook.com/homeworks/show/43
1/8
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
=[.8585,3.658]
Solution:
The sample estimate is
= 1.4. The standard error is s.e.(
) = s
d
/sqrt(n) = 1.817/sqrt(5) = 0.813.
The multiplier is t* = 2.78; coming fromTable A.2 using df =
n
 1 = 5  1 = 4 and the .95 column.
Thus, the 95% confidence interval for the population mean difference in heights (daughter less
mother) is given by: 1.4 ± (2.78 × 0.813), or 1.4 ± 2.26, or 0.86 to 3.66.
c.
1 out of 1 points
Feedback:
Technically, this corresponds to a TWO
sided test, you you should say different from, rather
than "taller".
See Textbook for Exercise: 11.c.
Your Answer:
Because 0 in included in our interval above, we do not have sufficient evidence to conclude that
the all college women are taller on average than their mothers.
Solution:
With 95% confidence we would estimate that, on average, college women are somewhere
between 0.86 inch shorter and 3.66 inches taller than their mothers.
These results apply for the
population of motherdaughter pairs represented by this sample.
d.
2 out of 2 points
See Textbook for Exercise: 11.d.
Your Answer:
The researchers could have gotten a larger sample size, or they could have lowered the
confidence level.
Solution:
They could have used a larger sample size, which would have reduce both the standard error and
the multiplier (with the same confidence level). They could also use a lower confidence level; for
instance they could find a 90% confidence interval instead of a 95% confidence interval.
1 out of 2 points
Feedback:
Need to be clear that the normality assumption is
regarding the population of differences.
N
d t
b
l
th t th
QQ
l t i
f
l
Question 2
(Chapter 11)
Recall the paired data from Exercise 11.37 on mother and daughter heights. We are told the sample is
random. Clearly state what is the other condition required for the confidence interval to be valid and how
you would assess that condition.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Winter '10
 Gunderson
 Statistics, Normal Distribution, Statistical hypothesis testing, researcher

Click to edit the document details