0.2 Practice tests (1-4) and final exams  (Page 21/36)

64 . $p_c=\frac{x_A+x_A}{n_A+n_A}=\frac{65+78}{100+100}=0.715$

65 . Using the calculator function 2-PropZTest, the p-value = 0.0417. Reject the null hypothesis. At the 3% significance level, here is sufficient evidence to conclude that there is a difference between the proportions of households in the two communities that have cable service.

66 . Using the calculator function 2-PropZTest, the p -value = 0.0417. Do not reject the null hypothesis. At the 1% significance level, there is insufficient evidence to conclude that there is a difference between the proportions of households in the two communities that have cable service.

10.4: matched or paired samples

67 . H ₀ : ${\overline{x}}_d\ge 0$
H _a : ${\overline{x}}_d<0$

68 . t = – 4.5644

69 . df = 30 – 1 = 29.

70 . Using the calculator function TTEST, the p -value = 0.00004 so reject the null hypothesis. At the 5% level, there is sufficient evidence to conclude that the participants lost weight, on average.

71 . A positive t -statistic would mean that participants, on average, gained weight over the six months.

11.1: facts about the chi-square distribution

72 . μ = df = 20
$\sigma =\sqrt{2(df)}=\sqrt{40}=6.32$

11.2: goodness-of-fit test

73 . Enrolled = 200(0.66) = 132. Not enrolled = 200(0.34) = 68

74 .

75 . df = n – 1 = 2 – 1 = 1.

76 . Using the calculator function Chi-square GOF – Test (in STAT TESTS), the test statistic is 3.7656 and the p-value is 0.0523. Do not reject the null hypothesis. At the 5% significance level, there is insufficient evidence to conclude that high school most recent graduating class distribution of enrolled and not enrolled does not fit that of the national distribution.

77 . approximates the normal

78 . skewed right

11.3: test of independence

79 .

80 . $\frac{{\left(100-150\right)}^2}{150}=16.67$
$\frac{{\left(150-100\right)}^2}{100}=25$
$\frac{{\left(200-100\right)}^2}{150}=16.67$
$\frac{{\left(50-100\right)}^2}{100}=25$

81 . Chi-square = 16.67 + 25 + 16.67 + 25 = 83.34.
df = ( r – 1)( c – 1) = 1

82 . p -value = P (Chi-square, 83.34) = 0
Reject the null hypothesis.
You could also use the calculator function STAT TESTS Chi-Square – Test.

11.4: test of homogeneity

83 . The table has five rows and two columns. df = ( r – 1)( c – 1) = (4)(1) = 4.

11.5: comparison summary of the chi-square tests: goodness-of-fit, independence and homogeneity

84 . Using the calculator function (STAT TESTS) Chi-square Test, the p -value = 0. Reject the null hypothesis. At the 5% significance level, there is sufficient evidence to conclude that the poll responses independent of the participants’ ethnic group.

85 . The expected value of each cell must be at least five.

86 . H ₀ : The variables are independent.
H _a : The variables are not independent.

87 . H ₀ : The populations have the same distribution.
H _a : The populations do not have the same distribution.

11.6: test of a single variance

88 . H ₀ : σ ² ≤ 5
H _a : σ ² >5

Practice test 4

12.1 linear equations

1 . Which of the following equations is/are linear?

1. y = –3 x
2. y = 0.2 + 0.74 x
3. y = –9.4 – 2 x
4. A and B
5. A, B, and C