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Homework 6 Solutions
1. a)
b)
c)
X ~ Binomial( n = 15, p = .25)
µ = np = 15*.25 = 3.75 and σ = [n*p*(1-p)]1/2 = (15*.25*.75) ½ = 1.677
P( X >= 10) =
P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) =
(15 nCr 10)*.2510*.755 + (15 nCr 11)*.2511*.754 + (15 nCr 12)*.2512*.753 +
(15 nCr 13)*.2513*.752 + (15 nCr 14)*.2514*.751 + (15 nCr 15)*.2515*.750 = .0008
d)
X is approximately N(µ =np = 250, σ = [n*p*(1-p)]1/2 = 13.69)
P( X > 275 )
1- P( Z < (275 – 250)/13.69 ) = 1- P( Z < 1.83) = .0336
2. a) Xbar19 ~ Approximately Normal(µ = 190, σ / n) = 35/ 19 = 8.03 )
P( Xbar19 > 210 ) 1- P(Z < (210-190)/8.03) = 1- P(Z < 2.49) = .0064
b)
The key is to note that our question about total weight can be written as a question
about the sample mean. We know how to answer questions about sample means!
P( Total > 4000) = P( Xbar19 > 4000/19 ) 1- P(Z < (210.5263-190)/8.03) =
1 – P(Z < 2.56) = .0052
3.
The 99% confidence interval for µ is:
= ( 105.84 – 2.576*15/sqrt(31) , 105.84 + 2.576*15/sqrt(31) )
= ( 98.90, 112.78 ).
We are 99% confident that true mean IQ score for all seventh-grade girls in the school
district is between 98.90 and 112.78. (Picture shows how to find critical value)
4.
a)
The 95% confidence interval for µ is:
= ( 22 – 1.96*50/sqrt(1000) , 22 – 1.96*50/sqrt(1000))
= (18.90 , 25.10)
We are 95% confident that true mean score gain for all students is between 98.90
and 112.78.
b)
The 95% confidence interval for µ is:
= ( 22 – 1.96*50/sqrt(250) , 22 – 1.96*50/sqrt(250))
= (15.80 , 28.20)
The interval is wider since our sample size is smaller.
c)
The 95% confidence interval for µ is:
= ( 22 – 1.96*50/sqrt(4000) , 22 – 1.96*50/sqrt(4000))
= (20.45 , 23.55)
The interval is narrower since our sample size is larger.
d)
As sample size increases, the margin of error decreases. In fact we can say more.
Notice that when the sample size increases by a factor of four, the margin of error
is cut in half. This is because of the square root in the denominator.
Sample Size = n
250
1000
4000
5.
Margin of Error =
6.20
3.10
1.55
For these data Xbar is 22 and sigma is 50. The standard deviation of the sample
mean is sigma/sqrt(1000) = 1.581. Then the table below compares the different
confidence intervals. As confidence increases the width of the interval increases.
Confidence
Level
90%
95%
99%
Z*
(critical
value)
1.645
1.960
2.576
Margin of Error =
Lower Bound
Upper Bound
2.60
3.10
4.44
19.40
18.90
17.56
24.60
25.10
26.44
6. a)
b)
We are interested in estimating the population parameter, µ, which is the true
average reading of the “standard weight” on this scale. (By true average reading
we mean the average reading that would be obtained in many, many, many trials.)
The 95% confidence interval for µ is:
=
( 10.0022 – 1.96*.0021/sqrt(10) , 10.0022 + 1.96*.0021/sqrt(10))
= (10.0009 , 10.0035)
c)
The claimed true weight of 10 is not in the interval and thus statistically we
would declare that the scale was biased. The difference from 10 is quite small
– we would want to judge whether the bias is practically significant (i.e.
whether it would make a difference in practice). The small bias might matter
if we need very precise measurements, as in a science lab, but probably would
not matter if this was an ordinary home bathroom scale.
d)
Xbar 10 ~ Approximately Normal(µ = 10, σ / n) = .0021/ 10 = .0007 )
P( Xbar10 > 10.0022 ) 1- P(Z < (10.0022-10)/.0007) = 1- P(Z < 3.14) = .0008
There is only a .08% chance of observing an average greater than or equal to
10.0022 assuming that the true weight is 10 grams. This probability calculation
once again suggests that the scale is not free of bias at the 1% level.