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Stats 242.3(02) Review Assignment
Solutions
1.
Notice that from the table we see that in the sample of n1 = 1708 Catholics, 541 of their father's
reached the High school graduate level of education, while for the sample of n 2 = 1358 Protestants, 406
of their father's reached the High school graduate level of education
a)
Determine 95% confidence limits for the proportion of Catholics whose fathers reach the
High School graduate level of education.
ˆ 1  ˆ 
0.3167 1  0.3167 
541
 0.3167,  ˆ 

 .0113
1708
n
1708
ˆ  z0.025 ˆ or 0.3167  1.960 .0113
Thus 95% confidence limits are: 0.2947 to 0.3388
ˆ 
b) Determine 99% confidence limits for the proportion of Protestants whose fathers reach the
High School graduate level of education.
ˆ 1  ˆ 
0.2990 1  0.2990 
406
 0.2990,  ˆ 

 .0124
1358
n
1358
ˆ  z0.025 ˆ or 0.2990   2.576 .0124
Thus 95% confidence limits are: 0.2670 to 0.3310
ˆ 
c)
Test to see if the proportion of Catholics whose fathers reach the High School graduate level
of education is the same as the proportion of Protestants. (use  = 0.05 and  = 0.01)
The Test statistic
ˆ1  ˆ2
x x
541  406
947
z
where ˆ  1 2 

 0.3089
n1  n2 1708  1358 3066
1 1
ˆ 1  ˆ    
 n1 n2 
Thus z 
0.3167  0.2990

0.0178
 1.058
0.0168
1 
 1
0.3089 1  0.3089  


 1708 1358 
Using = 0.05 , z0.025 = 1.960, Since -1.960 < z < 1.960, H0 is accepted.
Using = 0.01 , z0.005 = 2.576, Since -2.576 < z < 2.576, H0 is accepted.
d) Estimate the difference between the proportion of Catholics and the proportion of Protestants
whose fathers reach the High School graduate level of education with a 95% confidence
interval.
95% confidence interval for 1 - 2
ˆ 1  ˆ1  ˆ 2 1  ˆ 2 
ˆ1  ˆ 2  z0.025 1

,
n1
n2
0.0178  1.960
0.3167 1  0.3167 
1708

0.2990 1  0.2990 
1358
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or -0.0151 to 0.0506
2.
A civil engineer wishes to measure the compressive strength of two different types of concrete. A
random sample of 10 specimens of the first type yielded the following data (in psi)
Type I:
3250, 3268, 4302, 3184, 3266,
3297, 3332, 3502, 3064, 3116
10
x
i 1
i
 33581,
10
x
i 1
2
i
This value was in error
 1241571690
whereas a sample of 10 specimens of the second type of concrete yielded the data
Type II:
3094, 3106, 3004, 3066, 2984,
3124, 3316, 3212, 3380, 3018
10
x
i 1
i
 31304,
10
x
i 1
2
i
This value was in error
 1078094016
A. Compute a 95 % confidence interval for the mean strength of the Type I concrete.
10
x   xi n 
i 1
33581
 3358.1
10
 10 2  10 2 
  xi    xi  n 
 i 1

 i 1 

s 
n 1
1241571690  33581 10   352.73
2
9
95% confidence intervals
x  t0.025
s
352.73
or 3358.1  2.262
or 3105.8 to 3610.4
10
n
Compute a 95 % confidence interval for the mean strength of the Type II concrete.
10
x   xi n 
i 1
31304
 3130.4
10
 10 2  10 2 
  xi    xi  n 
 i 1

 i 1 

s 
n 1
1078094016  31304  10   133.15
2
9
95% confidence intervals
x  t0.025
s
133.15
or 3130.4  2.262
or 3035.2 to 3225.6
10
n
B. Compute 95 percent confidence limits for the variance of strength for both types of
concrete.
 n  1 s 2
2
 0.025
to
 n  1 s 2
2
 0.975
95% CI for variance
Type I
Type II
58864.9 to 414670.9
8387.8 to 59087.2
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4.
An embryologist wished to estimate the mean difference in time at which eye pigmentation is first
evidenced in an embryo for 2 varieties of birds. A sample of n = 33 embryos from species A revealed
that it took an average of 74.32 hours for eye pigmentation to appear with a standard deviation of 2.51
hours. A second sample of m =33 revealed that it took an average of 100.21 hours for eye pigmentation
to appear with a standard deviation of 19.44 hours.
a) Estimate the difference in the mean tome at which eye pigmentation appears for the two varieties
of birds, with a 95% confidence interval.
Solution:
95% Confidence limits  x1  x2   z0.025
100.21  74.32   1.960
s12 s22
or

n1 n2
2.512 19.442

or 19.20 to 32.58
33
33
b) Using a 5% significance level test
the Null Hypothesis H0: Variety A’s pigmentaion requires the same period of time to appear as the
time required by variety B
against
the alternative Hypothesis HA: Variety A’s pigmentaion requires a shorter period of time to appear
than the time required by variety B
Solution: The test statistic z 
x1  x2  0
s12 s22

n1 n2

100.21  74.32
2.512 19.442

33
33
= 7.59
We will reject H0 if z > z0.05 = 1.645. Since this is true H0 is rejected for  = 0.05.
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