Answers Exercises L4 Basic Stats –Applied statistics
1. The MSE’s are σ2, ½ σ2, 10 σ2 + 81μ2 and σ2/10, so T4 is the best estimator of µ.
2. a.
b.
:
c.
, because
are independent.
So:
3. a,
≤
c. var(
= 1/3200
4. a. 121 ml b. The quantities X1, ..., X20 is a random sample of from a population having a N(µ, σ2)distribution. 95%-CI(µ) = (120.0, 122.0)
c. 90%-CI(µ) = (120.2, 121.8) d. 2.3, 1.6 en 1.3 ml.
5. a. 20.0 and s ≈ 0.498 b. The temperatures X1, ..., X6 are independent and all N(µ, σ2)-distributed
(unknown µ and σ2)
c. We can use the formula
, where the
mean and s are computed in a. and c = 2.02 from the t(6-1)-table such that P( T ≤ c) = 0.95
d. We use the formulas
(0.112, 1.08) and
(0.335, 1.04), where s2 =0.4982, n = 6, and c1 =1.15 and c2
= 11.07 from the
-table such that P(
6. a. 95%
c1 ) = P(
c2 ) = 0.05
in which n = 800,
N(0, 1)-table : c = 1.96 such that P( Z ≤ c) = 0.975 : 95%
1.645 such that P( Z ≤ c) = 0.95: 90%
= 329/800 = 0.41 and, from the
= (0.376, 0.444 )
b. Now c =
= (0.381, 0.439 ): less wide than in a. c. For larger n the
width decreases. d. The width of the interval in the formula is
≤ 0.02: using c = 1.96 and
, we find n ≥ 9604
7. a. The salaries X1, …,X15 are independent and N( , σ2)-distributed, unknown  and σ2
s
s
95%-CI()= ( x  c
, xc
) ≈(2168, 2832), where n  15 , x  2500 , s = 600
n
n
and c = 2.14 from the t15-1 -table, such that P(T14 < c)= 0.975
b. We are 95% confident that the mean salary of all starting IT-specialists is between 2168 and 2832
Euro: 95% of intervals computed in this way will contain the real expected salary of starting ITspecialists.
c. 95%-BI()= (
(n  1) s 2
c2
,
(n  1) s 2
c1
)  (439,946) . In which n  15 , s2 = 6002 and c1  5.629 and
2
2
c 2  26.119 such that P (  14  c1)  0.975 and P(  14  c 2)  0.025
8. a. Rejection Region = [11.645,∞)
b. The power for μ = 11, 12, 13, 10, + ∞ is 26% , 64% , 91%, 5% and 100% resp.
c. p-value =
μ = 10) = P(Z ≥ 2.5) = 1- P(Z ≤ 2.5) = 0.62% < α = 5%
(12.5 is also in the rejection region determined in a)
9. a. 1. Is the claim of higher life expectancy of Michelin tyres justified?
4. Test statistic
~ t(20-1) if H0: µ = 40000 is true.
5. Observed value: t =
≈ 2.98
6. The TEST: If T ≥ c => reject H0. From the t(19)-table: Rejection region = [1.73,∞)
7. t = 2.98 is in the rejection region => reject H0
8. The test shows at significance level 5% that Michelin tyres live longer
10. a. = 200.1 and s2 = 7251 b. Summary of the test: statistic
≈ 3.19 > 1.76 =>
reject H0 at 5% level c. p-value = P(T ≥ 3.19| H0 ) ≈ 0.35% < α = 5% => reject H0
11. 8a. Rejection Region = (-∞, 8.04] [11.96,∞)
8c. p-value = 2
μ = 10) = 2P(Z ≥ 2.5) =2[ 1- P(Z ≤ 2.5) ] = 1.24% < α = 5%
(12.5 is also in the rejection region determined in a)
12 b. This is a so called paired sample: the weights within every pair are dependent,
but the differences in weight before and after are independent variables and all N(μ, σ2)distributed, unknown mean difference μ and variance σ2. H0 : μ = 0 and H1 : μ > 0.
is t(14)-distributed: If T ≥ 1.76, then reject H0 .t ≈ 1.57 < 1.76 => accept H0
c. (-0.72,12.72).
13. a. 1. Will National pride have less than 50% of all votes.
2. X = number of national Pride voters among the 1250 voters.
X ~ B(1250, p) , where p is the unknown population proportion.
Test H0: p ≥ ½ and H1: p < ½ , α = 0.01
X ~ B(625, ½ ) if p = ½ : X is approximately N(625, 312.5)
Observed value of X : 575
Left sided test: If X ≤ c, then reject H0
P(X ≤ c |H0 ) ≤ α or P( [X - 625]/√312.5 ≥ [c- ½ - 625]/√312.5 | H0 ) ≤ 0.01
Using the N(0, 1)-table we find: [c- ½ - 625]/√312.5 ≤ -2.33 or c ≤ 584.3. So: c = 584
7. X = 575 < 584 = c => reject H0.
8. The poll shows that the new party leader will have to resign on election day (sign. level 1%)
3.
4.
5.
6.
b. 90%  BI ( p )  ( p̂ - c

p (1 pˆ )
, p̂ + c
n
p (1 pˆ )
n
) = ( 0.443, 0.477)
(c = 1.645 and p̂ = 575/1250 =0.46)
14. 1. Is the new medicine as effective as the standard medicine? 2. X = “the number of cured patients in
the sample” X ~ B (200, p ) the success rate p of the new medicine is unknown.
3. H 0 : p  0.70 and H1 : p  0.70 , α = 0.10
4. Test statistic X if H0 is true: X ~ B(200,0.70) :
normal approximation: X ~ N (np, np(1  p))  N (140,42)
5. Observed X : 148
6. Reject H 0 if X ≥ c2 or X ≤ c1 such that P(X ≥ c2 | H 0 : p  0.70 ) = P( X ≤ c1 | H 0 : p  0.70 ) =
½ α = 0.05
c  0.5  140 

: P( X  c2 )  P( X  c2  0.5)  P  Z  2
  5%
42


c  0.5  140
N (0,1)  table  2
 1.645  c2  140.5  1.645 42  151.2
42
c2 = 152 and c1 = 128 (symmetrical) [ Decision with the p-value ≈ 24% > α]
7. 148 is not in the rejection region => accept H0 .
8. A difference in effectiveness of the new and the standard medicine can not be proven, on 10%-l2v2l
15. Summary: Test H0: σ2 = 42 and H1: σ2 >16 for α =5% . Test Statistic S2:
Observed value s2 = 25
=>
TEST: s2 ≥ c => reject H0. P(S2 ≥ c | σ2 =16) =
if H0 is true.
0.05
=> c = 30.0 conclusion: s2 = 25 < 30.0 = c => do not reject H0
16. Similar to exercise 15, but now we conduct a two-sided chi-square test: H0: σ2 = 2500 and
H1: σ2  2500, α = 10% . TEST: S2 ≤ c1 or S2 ≥ c2 => reject H0
Using the chi-square table with 14 degrees of freedom we find:
=>
2
or c1 = 1173 and
=>
23. 685 or c1 = 4229. s = 7251 is in the rejection region => reject H0 (the variance changed).