Introduction to Statistical Inference and Regression
ST 372

Fall 2013
ST372 HW8
934
b. The sample means and standard deviations of the two samples are
x = 13.90 , s1 = 1.225 , y = 12.20 ,
s 2 = 1.010 . The pooled variance estimate is s =
2
p
m 1 2 n 1 2 4 1
4 1
2
2
s1 +
s2 =
(1.225) +
(1.010 ) = 1.260 , so
m + n 2
Introduction to Statistical Inference and Regression
ST 372

Spring 2008
ST 372 Spring 2007 Introduction to Statistical Inference and Regression
Instructor: Email: Office: Phone: Office hour: Dr. Judy Huixia Wang [email protected] Patterson Hall Rm 209 F (919) 5131661 Wednesdays, 3pm5pm (or by appointment)
Lectur
Introduction to Statistical Inference and Regression
ST 372

Fall 2013
ST 372
Midterm Exam 1
Sept 23, 2010
Instructor: Yichao Wu
Your Name (Print):
Your ID:
Notes:
1. There are totally 4 problems.
2. Be sure to show all your work; your partial credit might depend on it.
3. NO CREDIT will be given without supporting work.
4.
Introduction to Statistical Inference and Regression
ST 372

Fall 2013
ST372 HW 5
8.11
a.
b.
H o : 10 v. H a : 10
P ( rejecting H0 when H0 is true) = P ( x 10.1032 or 9.8968 when 10) . Since x is
.2
normally distributed with standard deviation
.04,
n5
P ( z 2.58or 2.58) .005 .005 .01
8.15
a.
P ( z 1.88 when z has a stand
Introduction to Statistical Inference and Regression
ST 372

Fall 2013
ST 372
Midterm Exam 2
Nov 2, 1010
Instructor: Yichao Wu
Your Name (Print):
Your ID:
Notes:
1. There are totally 5 problems plus one bonus question.
2. Be sure to show all your work; your partial credit might depend on it.
3. NO CREDIT will be given withou
Introduction to Statistical Inference and Regression
ST 372

Fall 2013
ST372 HW9
10.1
a. Ho will be rejected if f F.05, 4 ,15 3.06 (since I 1 = 4, and I ( J 1 ) = (5)(3) = 15 . The computed value of
F is f
MSTr 2673.3
2.44 . Since 2.44 is not 3.06 , Ho is not rejected. The data does not indicate a
MSE 1094.2
difference in
Introduction to Statistical Inference and Regression
ST 372

Fall 2013
ST372 HW7
95
a.
Ha says that the average calorie output for sufferers is more than 1 cal/cm2/min below that for nonsufferers.
12
2
2
.22 .42
.1414 , so z
m
n
10
10
z 2.33 ; since 2.90 < 2.33, reject H0.
b.
1 2.33
d.
mn
.1414
At level .01, H0 is re
Introduction to Statistical Inference and Regression
ST 372

Fall 2013
ST372 HW4
7.4
a.
58.3
b.
58.3
c.
58.3
1.963
25
58.3 1.18 57.1,59.5
1.963
100
2.583
100
58.3 .59 57.7,58.9
58.3 .77 57.5,59.1
d. 82% confidence 1 . 82 .18
58.3
1.343
100
2
.09 , so
z 2 z.09 1.34 and the interval is
57.9,58.7 .
e.
22.58 3
n
239
Introduction to Statistical Inference and Regression
ST 372

Fall 2013
ST372 HW1
3.23
a.
b.
c.
d.
P (X = 2) = 0.39 0.19 = 0.20.
P (X > 3) = 1 0.67 = 0.33.
P (2 X 5) = 0.97 0.19 = 0.78.
P (2 < X < 5) = 0.92 0.39 = 0.53.
3.37
1
1
1
E (x) = x( n ) = n x = n ( n(n2+1) ) = n+1 .
2
1
1 n(n+1)(2n+1)
2
21
2
E (x ) = x n = n x = n (
Introduction to Statistical Inference and Regression
ST 372

Fall 2013
ST372 HW2
6.5
Let = the total audited value.Three potential estimators of are 1 = N X ,2 =
y
x
T N D,and 3 = T X . From the data, = 374.6, = 340.6,and d = 34.0.
Y
Knowing N = 5, 000 and T = 1, 761, 300, the three corresponding estimates are
1 = (5, 000)(3
Introduction to Statistical Inference and Regression
ST 372

Fall 2013
ST372 HW 3
6.20
a.
n
n x
ln p x (1 p ) , set it equal to zero and solve for p.
x
x nx
d n
; setting this equal to zero and solving for
ln + x ln ( p ) + (n x ) ln (1 p ) =
dp x
p 1 p
x
3
p yields p = . For n = 20 and x = 3, p =
= .15
20
n
We wis
Introduction to Statistical Inference and Regression
ST 372

Fall 2013
ST372 HW6
846
H0: p = .035 v. Ha: p < .035.
p
We use z
p .035
.035.965 / n
, with the rejection region z z.01 = 2.33. With
.005
15
.61 . Because .61 isnt 2.33, H0 is not rejected. Robots have not
.03 , z
500
.0082
demonstrated their superiority.