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64 . p c = x A + x A n A + n A = 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 : x ¯ d 0
H a : 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
σ = 2 ( d f ) = 40 = 6.32

11.2: goodness-of-fit test

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

74 .

Observed (O) Expected (E) O – E (O – E)2 ( O E ) 2 z
Enrolled 145 132 145 – 132 = 13 169 169 132 = 1.280
Not enrolled 55 68 55 – 68 = –13 169 169 68 = 2.485

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 .

Cell = Yes Cell = No Total
Freshman 250 ( 300 ) 500 = 150 250 ( 200 ) 500 = 100 250
Senior 250 ( 300 ) 500 = 150 250 ( 200 ) 500 = 100 250
Total 300 200 500

80 . ( 100 150 ) 2 150 = 16.67
( 150 100 ) 2 100 = 25
( 200 100 ) 2 150 = 16.67
( 50 100 ) 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

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Source:  OpenStax, Introductory statistics. OpenStax CNX. May 06, 2016 Download for free at http://legacy.cnx.org/content/col11562/1.18
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