Want P(Xbar > 83) = P(Z > (8083)/(8/square root of 49))
= P(Z > ____)
= 1  _______ from Z table
 No CLT needed since X has a normal distribution. Do not check n > 30
For #34: Quiz scores (10 points each) for two quizzes are analyzed for 6 students. The
regression analysis is below. What is the residual for the point where X = 4? Use proper
units and show all work for credit.
DATA:
ANALYSIS:
Quiz 1
Quiz 2
Predictor
Coef
SE Coef
T
P
5
5
Constant
1.533
2.431
0.63
0.562
2
3
Quiz 1
0.9429
0.6242
1.51
0.205
3
3
S = 2.61133
RSq = 36.3%
RSq(adj) = 20.4%
4
10
6
6
1
2
Equation of the line is 1.533 + .9429X.
Predicted Y when X = 4 is
Y =
1.533+.9429(4)=
(plug X = 4 into the equation)
Observed Y when X = 4 is 10 (given)
Residual = Observed Y – Predicted Y = 10  ___________
6
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentFor #3536: Bob heard on the news that 30% of all employees would prefer to work from
home one day a week. SHOW WORK!
35. Bob surveys 8 of his employees at random and counts the number that would
prefer to work from home one day a week. What’s the probability that 2 of Bob’s
employees would prefer to work from home one day per week?
X = number preferring to work from home one day per week
X has a binomial distribution with n = 8 and p = .30
We want P(X = 2). Use the binomial table to get ______________
36. Bill also hears the same newscast as Bob and he conducts the same survey on a
random sample of 80 of his employees. What’s the chance that at least 25 of Bill’s
employees would prefer to work from home?
X = number preferring to work from home one day per week
X has a binomial distribution with n = 80 and p = .30
(n is too large for the table – need normal approximation)
Check conditions for normal approximation: 1) np = 80x.3 = 24 is at least 10;
and 2) n(1p) = 80x.7 = 56 is at least 10.
Use Z = (X – np)/(square root of np(1p))
You want P(X at least 25)
=P(Z at least (2524)/square root (80x.3x.7) =
=P(Z at least _________)
= 1 – P(Z < ______) = ____________
This is an approximate answer.
For #37: Bob measured the length (in cm) and weight (in grams) of a random sample of
mice of a certain species. The results of his analysis are shown below.
Variable
Count
Mean
StDev
Minimum
Median
Maximum
Length
150
168.67
43.15
63.00
163.00
280.00
Weight
150
150.77
19.76
101.00
150.50
208.00
Predictor
Coef
SE Coef
T
P
Constant
39.05
25.09
1.56
0.122
Weight
0.8597
0.1650
5.21
0.000
Pearson correlation of Length and Weight = 0.394
37. ESTIMATE the average weight of all mice of this species in the population. Show
all work and use proper units WITH 95% CONFIDENCE. Best possible answer.
Need a 95% confidence interval for average weight.
Xbar plus/minus 1.96(standard deviation)/square root of n
7
X = weight, n = 150, mean = 150.77, stdev = 19.76
Estimate is: 150.77 plus/minus 19.76/square root of 150
= 150.77 plus/minus ______________
GRAMS = (
,
)
GRAMS
8
This is the end of the preview.
Sign up
to
access the rest of the document.
 Winter '12
 Johnson
 Statistics, Normal Distribution, Standard Deviation, A. Bob

Click to edit the document details