プリンキピア

付録

11.4: Test of Homogeneity11.4：均質性の検定

83. You conduct a chi-square test of homogeneity for data in a five by two table. What is the degrees of freedom for this test?
83。5行2列の表のデータにカイ二乗均質性検定を行う。この検定の自由度はいくつか。

11.5: Comparison Summary of the Chi-Square Tests: Goodness-of-Fit, Independence and Homogeneity11.5：カイ二乗検定の比較：適合度・独立性・均質性

84. A 2013 poll in the State of California surveyed people about taxing sugar-sweetened beverages. The results are presented in the following table, and are classified by ethnic group and response type. Are the poll responses independent of the participants’ ethnic group? Conduct a hypothesis test at the 5% significance level.
84。2013年にカリフォルニア州で、砂糖入り飲料への課税について意見を調査した。結果を、民族集団と回答の種類で分類して次の表に示す。回答は参加者の民族集団と独立か。有意水準5%で仮説検定を行え。
Ethnic Group \ Response Type
民族集団／回答の種類
Favor
賛成
Oppose
反対
No Opinion
意見なし
Row Total
行合計
White / Non-Hispanic
白人／非ヒスパニック系
234
433
43
710
Latino
ラテンアメリカ系
147
106
19
272
African American
アフリカ系米国人
24
41
6
71
Asian American
アジア系米国人
54
48
16
118
Column Total
列合計
459
628
84
1171
Table B.12表 B.12
85. In a test of homogeneity, what must be true about the expected value of each cell?
85。均質性検定では、各セルの期待度数はどの条件を満たさなければならないか。
86. Stated in general terms, what are the null and alternative hypotheses for the chi-square test of independence?
86。カイ二乗独立性検定の帰無仮説と対立仮説を、一般的な形で述べよ。
87. Stated in general terms, what are the null and alternative hypotheses for the chi-square test of homogeneity?
87。カイ二乗均質性検定の帰無仮説と対立仮説を、一般的な形で述べよ。

11.6: Test of a Single Variance11.6：一つの分散の検定

88. A lab test claims to have a variance of no more than five. You believe the variance is greater. What are the null and alternative hypothesis to test this?
88。ある臨床検査は分散が5以下だとされている。あなたは分散がもっと大きいと考えている。検定する帰無仮説と対立仮説は何か。

Practice Test 3 Solutions模擬試験3の解答

8.1: Confidence Interval, Single Population Mean, Population Standard Deviation Known, Normal8.1：母標準偏差が既知の場合の一つの母平均の信頼区間：正規分布

1. σ n = 4 30 = 0.73
2. normal
2. 正規分布
3. 0.025 or 2.5%; A 95% confidence interval contains 95% of the probability, and excludes five percent, and the five percent excluded is split evenly between the upper and lower tails of the distribution.
3。0.025、すなわち2.5%。95%信頼区間は確率の95%を含み、5%を除く。その5%が分布の上下の裾に等分される。
4. z-score = 1.96; E B M = z α 2 ( σ n ) = ( 1.96 ) ( 0.73 ) = 1.4308
4。z得点 = 1.96。 E B M = z α 2 ( σ n ) = ( 1.96 ) ( 0.73 ) = 1.4308
5. 41 ± 1.43 = (39.57, 42.43); Using the calculator function Zinterval, answer is (40.74, 41.26. Answers differ due to rounding.
5。41 ± 1.43 = (39.57, 42.43)。電卓の Zinterval 機能による答えは (40.74, 41.26)。丸めのため答えが異なる。
6. The z-value for a 90% confidence interval is 1.645, so EBM = 1.645(0.73) = 1.20085.
The 90% confidence interval is 41 ± 1.20 = (39.80, 42.20).
The calculator function Zinterval answer is (40.78, 41.23). Answers differ due to rounding.
6。90%信頼区間の z値は1.645なので、EBM = 1.645(0.73) = 1.20085。90%信頼区間は 41 ± 1.20 = (39.80, 42.20)。電卓の Zinterval 機能による答えは (40.78, 41.23)。丸めのため答えが異なる。
7. The standard error of measurement is: σ n = 4 50 = 0.57
E B M = z α 2 ( σ n ) = ( 1.96 ) ( 0.57 ) = 1.12
The 95% confidence interval is 41 ± 1.12 = (39.88, 42.12).
The calculator function Zinterval answer is (40.84, 41.16). Answers differ due to rounding.
7。測定の標準誤差は、 σ n = 4 50 = 0.57 。 E B M = z α 2 ( σ n ) = ( 1.96 ) ( 0.57 ) = 1.12 。95%信頼区間は 41 ± 1.12 = (39.88, 42.12)。電卓の Zinterval 機能による答えは (40.84, 41.16)。丸めのため答えが異なる。
8. The 99% confidence interval, because it includes all but one percent of the distribution. The 95% confidence interval will be narrower, because it excludes five percent of the distribution.
8。99%信頼区間である。分布の1%を除いたすべてを含むため。95%信頼区間は分布の5%を除くので、より狭い。

8.2: Confidence Interval, Single Population Mean, Standard Deviation Unknown, Student’s t8.2：標準偏差が未知の場合の一つの母平均の信頼区間：スチューデントの t分布

9. The t-distribution will have more probability in its tails (“thicker tails”) and less probability near the mean of the distribution (“shorter in the center”).
9。t分布は裾に含まれる確率が大きく（「裾が厚い」）、平均付近の確率が小さい（「中央の山が低い」）。
10. Both distributions are symmetrical and centered at zero.
10。両方とも対称で、0を中心とする。
11. df = n – 1 = 20 – 1 = 19
11。df = n − 1 = 20 − 1 = 19
12. You can get the t-value from a probability table or a calculator. In this case, for a t-distribution with 19 degrees of freedom, and a 95% two-sided confidence interval, the value is 2.093, i.e.,
t α 2 = 2.093 . The calculator function is invT(0.975, 19).
12。t値は確率表か電卓で求められる。自由度19の t分布で、両側95%信頼区間に使う値は2.093、すなわち t α 2 = 2.093 。電卓では invT(0.975, 19) を使う。
13. E B M = t α 2 ( s n ) = ( 2.093 ) ( 0.3 20 ) = 0.140
98.4 ± 0.14 = (98.26, 98.54).
The calculator function Tinterval answer is (98.26, 98.54).
13。 E B M = t α 2 ( s n ) = ( 2.093 ) ( 0.3 20 ) = 0.140 。98.4 ± 0.14 = (98.26, 98.54)。電卓の Tinterval 機能による答えは (98.26, 98.54)。
14. t α 2 = 2.861. The calculator function is invT(0.995, 19).
E B M = t α 2 ( s n ) = ( 2.861 ) ( 0.3 20 ) = 0.192
98.4 ± 0.19 = (98.21, 98.59). The calculator function Tinterval answer is (98.21, 98.59).
14。 t α 2 = 2.861 。電卓では invT(0.995, 19) を使う。 E B M = t α 2 ( s n ) = ( 2.861 ) ( 0.3 20 ) = 0.192 。98.4 ± 0.19 = (98.21, 98.59)。Tinterval 機能による答えは (98.21, 98.59)。
15. df = n – 1 = 30 – 1 = 29. t α 2 = 2.045
E B M = z t ( s n ) = ( 2.045 ) ( 0.3 30 ) = 0.112
98.4 ± 0.11 = (98.29, 98.51). The calculator function Tinterval answer is (98.29, 98.51).
15。df = n − 1 = 30 − 1 = 29。 t α 2 = 2.045 。 E B M = z t ( s n ) = ( 2.045 ) ( 0.3 30 ) = 0.112 。98.4 ± 0.11 = (98.29, 98.51)。電卓の Tinterval 機能による答えは (98.29, 98.51)。

8.3: Confidence Interval for a Population Proportion8.3：母比率の信頼区間

16. p ′ = 280 500 = 0.56
q ′ = 1 − p ′ = 1 − 0.56 = 0.44
s = p q n = 0.56 ( 0.44 ) 500 = 0.0222
17. Because you are using the normal approximation to the binomial, z α 2 = 1.96 .
Calculate the error bound for the population (EBP):
E B P = z a 2 p q n = 1.96 ( 0.222 ) = 0.0435
Calculate the 95% confidence interval:
0.56 ± 0.0435 = (0.5165, 0.6035).
The calculator function 1-PropZint answer is (0.5165, 0.6035).
17。二項分布の正規近似を使うので、 z α 2 = 1.96 。母比率の誤差限界（EBP）を計算すると、 E B P = z a 2 p q n = 1.96 ( 0.222 ) = 0.0435 。95%信頼区間は 0.56 ± 0.0435 = (0.5165, 0.6035)。電卓の 1-PropZint 機能による答えは (0.5165, 0.6035)。
18. z α 2 = 1.64
E B P = z a 2 p q n = 1.64 ( 0.0222 ) = 0.0364
0.56 ± 0.03 = (0.5236, 0.5964). The calculator function 1-PropZint answer is (0.5235, 0.5965)
18。 z α 2 = 1.64 。 E B P = z a 2 p q n = 1.64 ( 0.0222 ) = 0.0364 。0.56 ± 0.03 = (0.5236, 0.5964)。電卓の 1-PropZint 機能による答えは (0.5235, 0.5965)。
19. z α 2 = 2.58
E B P = z a 2 p q n = 2.58 ( 0.0222 ) = 0.0573
0.56 ± 0.05 = (0.5127, 0.6173).
The calculator function 1-PropZint answer is (0.5028, 0.6172).
19。 z α 2 = 2.58 。 E B P = z a 2 p q n = 2.58 ( 0.0222 ) = 0.0573 。0.56 ± 0.05 = (0.5127, 0.6173)。電卓の 1-PropZint 機能による答えは (0.5028, 0.6172)。
20. EBP = 0.04 (because 4% = 0.04)
z α 2 = 1.96 for a 95% confidence interval
n = z 2 p q E B P 2 = 1.96 2 ( 0.5 ) ( 0.5 ) 0.04 2 = 0.9604 0.0016 = 600.25
You need 601 subjects (rounding upward from 600.25).
20。4% = 0.04 なので EBP = 0.04。95%信頼区間では z α 2 = 1.96 。 n = z 2 p q E B P 2 = 1.96 2 ( 0.5 ) ( 0.5 ) 0.04 2 = 0.9604 0.0016 = 600.25 。600.25を切り上げ、601人が必要である。
21. n = n 2 p q E B P 2 = 1.96 2 ( 0.6 ) ( 0.4 ) 0.04 2 = 0.9220 0.0016 = 576.24
You need 577 subjects (rounding upward from 576.24).
21。 n = n 2 p q E B P 2 = 1.96 2 ( 0.6 ) ( 0.4 ) 0.04 2 = 0.9220 0.0016 = 576.24 。576.24を切り上げ、577人が必要である。
22. n = n 2 p q E B P 2 = 1.96 2 ( 0.5 ) ( 0.5 ) 0.03 2 = 0.9604 0.0009 = 1067.11
You need 1,068 subjects (rounding upward from 1,067.11).
22。 n = n 2 p q E B P 2 = 1.96 2 ( 0.5 ) ( 0.5 ) 0.03 2 = 0.9604 0.0009 = 1067.11 。1067.11を切り上げ、1068人が必要である。