MTH/STA 562
Exercise 7.22. This is a random sample of size n = 100 from a population with mean = 14 and standard deviation = 2. By the Central Limit Theorem, (a) P Y > 14:5 (b) P which implies that P
n n o
0:5 P Z> p 2= 100
(
(
)
= P fZ > 2:5g =
MTH/STA 562
POWER OF TESTS AND NEYMAN-PEARSON LEMMA
Denition 1. Suppose that W is the test statistic and RR is the rejection
region for a test of a hypothesis concerning the value of an unknown parameter
6. Then the power of the test, denoted by power (
MTH/STA 562
ELEMENTS OF A STATISTICAL TEST
The main objective of statistics is to make inferences about unknown population parameters on the basis of sample information. That is, how can we reach some conclusion about the population based upon the s
MTH/STA 562 COMMON LARGE-SAMPLE STATISTICAL TESTS
In the preceding section, we have laid out a ground work for hypothesis testing which speci.cally includes four basic components null hypothesis, alternative hypothesis, test statistic, and rejectio
MTH/STA 562 CALCULATING TYPE II ERROR PROBABILITIES AND FINDING THE SAMPLE SIZE FOR THE Z TEST
Consider testing the null hypothesis H0 : against the alternative hypothesis Ha : =
a
=
0
>
0
where 0 and a are speci ed values of , with the rejecti
MTH/STA 562
Exercise 8.68. It is given that Y = 26:6, S = 7:4, and n = 21. For a 95% con dence level with n 1 = 20 degrees of freedom, t0:025 = 2:086. Then a 95% con dence interval for is Y or (23:33; 29:97).
n X i=1 n X i=1
S t0:025 p = 26:6 n
7:
MTH/STA 562
Exercise 7.44. (a) pq < 1 n r pq if and only if 3 < q n 9p < q if and only if n p+3 if and only if (b) 0 < p if and only if 9pq < p2 n n 1 if and only if > 9q p
! ! r
if and only if if and only if if and only if
!
3
pq <1 n 9pq < q2 n
MTH/STA 562
Exercise 8.1. Write
b b
Then Thus,
= b
h
2
E b = b
2
i
h
+ E b
i2
h
i h
= b
h
E b
i
i
+ B:
E b
h
+ 2B b E b
i2
E b
+ B2:
i
Note that
M SE b = E
h
b
=E b and
b
+ 2BE b
i
h
E b
+ B2
Hence,
V ar b = E b
E b
i2
MTH/STA 562
Exercise 8.17. The point estimate of is b = Y = 11:5. With n = 50 and S = 3:5, a bound on the error of estimation is 2
Y
= 2p
n
2 (3:5) p = 0:99: 50
Exercise 8.18. The point estimate of is b = Y = 7:2. With n = 200 and S = 5:6, a bou
MTH/STA 562
Exercise 8.35. Note that Y has a gamma distribution with parameters = 2 and an unknown . Then mY (t) = (1 t) 2 . Now m2Y = (t) = E e = 1
t( 2Y
) = E e( 2t )Y = m Y
!
2
2t
!
2t
= (1
2t)
2
:
This implies that 2Y = has a chi-square
MTH/STA 562
Exercise 8.42. (a) First calculate
b p=
Y 268 = = 0:536: n 500
s
Then an approximate 98% con dence interval for p is
b p
z0:01
s
b b p (1 p) = 0:536 n
2:33
(0:536) (0:464) = 0:536 500
0:052
or (0:484; 0:588). (b) Since the inter
MTH/STA 562
q
p(1 p) n
Exercise 8.58. We need to solve for n in the equation 1:96 (a) If p = 0:9 and B = 0:05, then 1:96
s
= B.
(0:9) (0:1) = 0:05 n
or
n = 139:
(b) If p is unknown, we use p = 0:5 and 1:96
s
(0:5) (0:5) = 0:05 n
or
n = 385:
MTH/STA 562
n X i=1 n X i=1
Exercise 8.81. Calculate Then Y = and S =
2
Yi = 514:4 and
Yi2 = 44; 103:74 with n = 6.
n 1X 514:4 Yi = = 85:73 n i=1 6
1 n 1
n X i=1
Yi2
nY
2
!
=
44; 103:74 5
6 (85:73)2
= 0:502667:
With con dence coe cient
MTH/STA 562
Exercise 10.76. Refer to Exercise 10.2. Since the rejection region is Y the power is 1 = P fY 12j pg. To calculate power, use Table 1. (a) 1 = P fY 12j p = 0:4g = 0:979. (b) 1 = P fY 12j p = 0:5g = 0:868. (c) 1 = P fY 12j p = 0:6g = 0:58
MTH/STA 562 RELATIONSHIP BETWEEN HYPOTHESIS TESTING PROCEDURES AND CONFIDENCE INTERVALS
As usual, let b denote an unbiased estimator of the unknown parameter standard error b. (1) (Two-tailed Test) The rejection region for testing H0 : = Ha : 6= 0 (