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

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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
$\sigma =\sqrt{2\left(df\right)}=\sqrt{40}=6.32$

## 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 $\frac{{\left(O-E\right)}^{2}}{z}$
Enrolled 145 132 145 – 132 = 13 169 $\frac{169}{132}=1.280$
Not enrolled 55 68 55 – 68 = –13 169 $\frac{169}{68}=2.485$

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 $\frac{250\left(300\right)}{500}=150$ $\frac{250\left(200\right)}{500}=100$ 250
Senior $\frac{250\left(300\right)}{500}=150$ $\frac{250\left(200\right)}{500}=100$ 250
Total 300 200 500

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

## 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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that's true
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Probability tells you the likelihood of an event happening. ... The higher the probability, the more likely it is to happen. Probability is a number or fraction between 0 and 1. A probability of 1 means something will always happen, and a probability of 0 means something will never happen...
Saying it's a number between zero and one means it is a fraction so you could remove "or fraction" from you definition.
Carlos
wouldn't be correct to remove fractions, saying a number is justified as probabilities can also be decimals between 0 and 1.
Denzel
Saying "a number" will include it being a decimal which are themselves fractions in another form.
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I will simply say a probability is a number in the range zero to one, inclusive.
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f#\$
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Statistic is the mean of the sample.
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