Section 8.7
Exercise 8.51. given by A 90% con dence interval for p1
b (p1 b p2 )
p2 is approximately
1:645
s
(a) If n1 = 800, n2 = 640, x1 = 337, x2 = 374, then
b p1 =
b b b b p 1 q1 p 2 q2 + . n
MTH/STA 562
Exercise 7.1. Since the amount of .ll dispensed by a bottling machine is normally distributed with = 1, the sample mean will also be normally distributed, from Theorem 7.1, with mean and
CHAPTER 10
INFERENCE FROM SMALL SAMPLES
10.2 Students t Distribution
From the discussion of the sampling distribution of E in the preceding chapters, a few
points had been discovered:
* When the or
CHAPTER 9
LARGESAMPLE TESTS
OF HYPOTHESES
9.1 Testing Hypotheses about Population Parameters
Often statisticians are not interested solely in estimating unknown population parameters,
but in derivi
CHAPTER 7
SAlVIPLING DISTRIBUTION
7.2 Sampling Plans and Experimental Designs
The way a sample is selected is called the sampling plan or experimental design. Simple
random sampling is commonly used
Section 10.7
Exercise 10.58. It is given that n1 = 16, n2 = 20, s2 = 55:7, and s2 = 31:4. 1 2
2 1 2 2 2 1 2 2:
(a) The null and alternative hypotheses for the test are H0 : The test statistic is = ve
Section 10.6
Exercise 10.49. interval for 2 is It is given that n = 15 and s2 = 0:3214. A 90% con dence (n
2 =2
1) s2
<
2
<
(n
2 1
1) s2
=2
:
With
= 0:10 and df = n
2 =2
1 = 14, we read from
Section 10.5
Exercise 10.35. It is given that n = 10, d = 0:3, s2 = 0:16 (sd = 0:4). d (a) The null and alternative hypotheses for the test are H0 : versus Ha : The test statistic is t= d 0:3 0 p = 2:
Section 10.3
Exercise 10.1. (a) 2:015 (b) 2:306 (c) 1:330 (d) 1:960 Exercise 10.2. (a) A two-tailed test with = 0:01 and 12 df gives t =2 = t0:005 = 3:055. The rejection region is jtj > 3:055; that is
Section 9.6
Exercise 9.43. (a) The hypotheses are H0 : p 1 p2 = 0 versus Ha : p 1 p2 < 0:
(b) This is a left-tailed test. (c) With n1 = 140, n2 = 140, x1 = 74, x2 = 81, the two individual sample b b
Section 9.5
Exercise 9.31. (a) The hypotheses are H0 : p = 0:4 This is a two-tailed test.
b (b) With n = 1400 and x = 529, the sample proportion is p = 529=1400 = 0:378 and the test statistic is
vers
Section 8.9
Exercise 8.68. The parameter on interest is . Then n z 2 =2 2 : B2 = 0:95 or = 0:05, we have
It is given that = 12:7 and B = 1:6. With 1 z =2 = z0:025 = 1:96. Hence, n or n = 243:
(1:96)
Section 8.5
Exercise 8.23. The 90% con dence interval for x 1:645 p . n is given by
(a) If n = 125, x = 0:84, and s2 = 0:086, then p 0:086 = 0:084 0:043 or 0:084 1:645 p 125 (b) If n = 50, x = 21:9,
Section 8.4
Exercise 8.3. The margin of error in estimating Margin of error = 1:96 (a) If n = 30 and
2
is given by
p . n
= 0:2, then p = 1:96 n p 0:2 p = 0:160. 30
1:96 (b) If n = 30 and
2
= 0:9,
Section 7.6
Exercise 7.37. The mean and standard deviation of the sampling distribb ution of the sample proportion p are, respectively, given by (a) If n = 100 and p = 0:3, then
b E (p) = 0:3 b E (p)
Section 7.5
Exercise 7.19. The mean and standard deviation of the sampling distribution of the sample mean x are, respectively, given by E (x) = (a) If n = 36, = 10, and
2
and = 9, then
SE (x) = p :
Section 7.2
Exercise 7.1. You can select a simple random of size n = 20 using Table 10 in Appendix I. First choose a starting point and consider the rst three digits in each number. Since the experime