Download Confidence Intervals- take-home exam Part I. Sanchez 98-2

Self assessment exam B - Confidence Intervals - answers
Name: __________________________
1. True-false test.
_True_____ For 10 degrees of freedom, P(t >-1.5) = 0.9177463368
_False_____ s is an unbiased estimator of  .
_True_____ For 10 degrees of freedom, t 0.05  1.812461102
__False____ z 0.05  2.576
__True____ pˆ is an unbiased estimator of p
__True____ For 10 degrees of freedom the ordinate of the t-distribution for t=1.35 is
0.1549.
2. Complete:
a) According to the _Central Limit Theorem__________ as n increases the sampling
distribution of the mean approaches a normal distribution.
b) In a sample of size 26 if 5 conditions must be satisfied, then we have
___21____ degrees of freedom.
c) If samples of size n=9 are taken from a population with mean µ =12 and standard
deviation then the sampling distribution of the mean has a mean  x = 12_____
and a standard deviation  x = 4____________.
d) As n increases the student’s t distribution approaches the standard normal
___distribution (the z-distribution).
e) As n increases the standard error of the mean _decreases___________.


3. The times taken to complete a statistics test by all students are normally distributed
with a mean of 55 minutes and a standard deviation of 9 minutes. Find the probability
that the mean time taken to complete this test by a class of 36 students would be
between 50 and 58 minutes.
Solution:   55,   9
A sample of n = 36 students is observed.
Find P(50  x  58). The distributi on of x has a mean  x    55 and a

9
3
s tan dard deviation  x 

  0.5
n
36 6
P(50  x  58)  Normcdf (50, 58, 55, 0.5)  0.99999999
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4. According to the Centers for Disease Control and Prevention, the average length of stay that
patients to nonfederal short-stay hospitals In the United States was 5.2 days. Assume that this
average was based on a random sample of 36 such hospitals stays and that the sample
standard deviation was 1.8 days. Construct a 98% confidence Interval for the mean length of
all such hospital stays. Show the work.
a )   0.02 ________
b ) Formula : x  z 
2
s
n
___________________
n  36, x  5.2, s  1.8
z 0.01  invNormal 0.01  2.326347877
c) Work :
x  z
2
s
n
 5.2  ( 2.326)
1.8
36
 5.2  0.6978  5.2  0.7
Therefore , 4.5    5.9 is a 98% confidence int erval for 
d )Answer (give answers rounded to one decimal place )
We are 98% confident that the mean length of all such hospital stays is between
___4.5 and 5.9__ days.
5. In problem 4 how large a sample will be needed to find the confidence interval for the mean within 1/2
day, that is, with a maximum error of 0.5 days?
E = 0.5 days. Use   1.8
2
 z 
2


 2.236  1.8 
 2 
n
 E 






.5
  65

Answer : 65 hospitals
6. A company wants to estimate the mean net weight of its Top Taste cereal boxes. A sample of 16 such
boxes produced the mean net weight of 31.98 ounces with a standard deviation of 0.26 ounces.
Construct a 95% confidence interval for the mean net weight of all Top Taste cereal boxes. Assume that
the net weights of all such cereal boxes have a normal distribution.
a )   0.05 _________
b ) Formula : x  t 
2
s
n
_______
n  16, deg rees of freedom  15, x  31.98, s  0.26
t 0.025  invT (0.025, 15)  2.131449536)
c) Work :
x  t
2
s
n
 31.98  2.131
0.26
16
 31.98  0.1386  31.98  0.14
d )Answer (give answers rounded to two decimal place )
We are 95% confident that the mean weight of Top Taste cereal boxes is between
__31.84 and 32.12___ ounces.
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7. It is said that happy and healthy workers are efficient and productive. A company that
manufactures exercising machines wanted to know the percentage of large companies
that provide on-site health club facilities. A sample of 240 such companies showed that
96 of them provide such facilities on site. Construct a 90% confidence interval for the
percentage (p) of all such companies that provide such site.
p̂q̂
a )   0.10 _________
b ) Formula : p̂  z 
_____________________
n
2
96
 0.40, q̂  0.60
240
z 0.05  invNorm (0.05)  1.645
n  240, x  96, p̂ 
c) Work :
p̂  z 
2
p̂q̂
0.40(0.60)
 0.40  1.645
 0.40  0.0520
n
240
0.35  p  0.45
d )Answer (give answers rounded to two decimal place )
We are 90% confident that the percentage (p) of all such companies that provide such site is between
____0.35 and 0.45_________________________________
9. A sample of 50 men studying statistics shows an average grade of 80 points with a
standard deviation of 5 points. A sample of 40 women studying statistics shows an
average grade of 85 points with a standard deviation of 6 points. Use the given sample
to construct a 98% confidence interval for the difference in average grade between all
men and women who study statistics.
Men : n1  50, x1  80, s1  5; Women : n 2  40, x 2  85, s1  6,   0.02
Z 0.01  invNorm (0.01)  2.326
s12 s 22
25 36
Formula : x1  x 2  z 

80  85  2.326

 5  2.75
n1 n 2
50 40
2
 7.75  1   2  2.25
Answer: We have a 98% confidence that the average grade of mean is below the
average of women by somewhere between 7.75 points and 2.25 points.
.
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10. If measurements of the specific gravity of a metal can be looked upon as a random
sample from a normal population with the standard deviation of 0.025 ounces, what Is
the probability that the mean of a random sample of size n=16 will be off by at most
0.01 ounces?

This means that we want the probability that x    0.01
 0.01 x  
0.01

x    0.01   0.01  x    0.01 


0.025

0.025
16
n
16
 0.01
0.01
z
  1.6  z  1.6
0.00625
0.00625
P( 1.6  z  1.6)  normalcdf ( 1.6, 1.6)  0.8904

answer: 0.8904__________
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