Solution to ST361 HW7
1) 7.2 (p.293)
Solution: (a) The sample mean X is an unbiased estimator of  . So a point estimate is x  (103
+156 + 118 + 89 + 125 + 147 + 122 + 109 + 138 + 99)/10 = 120.6
(b) The sample proportion p of homes that use over 100 therms is an unbiased estimator of  .
Since there are 8 home out of 10 that use more than 100 therms, p=8/10=0.8
2) 7.3 (p.293)
Solution: (a) Area =
P[   1  X    1]  P[
1
5 / 10

X 
1

]  P[0.63  Z  0.63]  0.47.
5 / n 5 / 10
(b) The areas are 0.84, 0.95 and 1, respectively for n = 50, 100, 1000.
0.8
0.7
0.5
0.6
probability
0.9
1.0
(c) Here is the graph. As sample size n increases, the probability P[   1  X    1] increases.
0
200
400
600
800
1000
Sample size n
3) 7.7(p 301)
Solution: Since n is large, we use normal approximation to all the problems. (a) The
confidence level = 1 - 2*P[Z  -3.09] = 0.9980. (b) The confidence level = 1 - 2*P[Z  -2.81] = 0.9950
= 99.5%. (c) The confidence level = 1 - 2*P[Z  -1.44] = 0.8501=85%. (d) The confidence level = 1 2*P[Z  -1]= 0.6827=68.3%.
4) 7.12 (p.302)
Solution: Here the sample sizes are large enough to apply the CLT.
(a) A 95% CI for the mean concentration for Mugil liza is
[ X  z0.025
s
s
s
s
, X  z0.025
]  [ X  1.96
, X  1.96
]  [8.82,9.48] .
56
56
56
56
1
(b) A 99% CI for the mean concentration for Pogonias cromis is
X  z0.005
s
s
s
s
, X  z0.005
]  [ X  2.57
, X  2.57
]  [2.52,3.64] .
61
61
61
61
The width of the CI in (b) = 3.64-2.52 =1.12 is greater than the width of the CI in (a) = 9.48 – 8.82 =
0.66 because the confidence level in (b) is higher than that in (a) so that the critical value is larger.
Another reason is that the sample standard deviation s is also larger. The impact of these two
factors is greater than the sample size here (the sample in (b) is only 5 larger than that in (a)).
5) 7.24 (p.311)
Solution: Here n = 487 and the sample proportion p = 0.072. We first check the conditions for the
CI formula for  = the proportion of all births with low birth weight given in class. (1) n = 487  30.
(2) np = 487*0.072 = 35  10, (3) n(1 – p) = 451  10. All conditions are met so we can use the
fomula
[ p  z0.005
p (1  p )
p (1  p )
, p  z0.005
]
n
n
 [0.072  2.57
0.072(1  0.072)
0.072(1  0.072)
, 0.072  2.57
]  [0.042, 0.102]
487
487
6) 7.39 (for part (c), need to calculate sample standard deviation)
420
430
440
450
460
Solution: (a) The box plot is in the following. It is a little skewed to right. But overall, it looks
symmetric.
(b) It is plausible to assume that the data are from a normal distribution. (it is difficult to
assess with this sample size actually)
(c) Assuming normality, then a 95% CI for  (the average degree of polymerization) can be
calculated as
[ X  tn 1,0.025
s
s
15.14
15.14
, X  tn 1,0.025
]  [438.3  2.12 
, 438.3  2.12 
]  [430.51, 446.08]
n
n
17
17
Since 440 is in this interval, so it is a plausible value for  . However, 450 is not in the
interval. So it is not a plausible value for  .
2