# 0.2 Practice tests (1-4) and final exams  (Page 24/36)

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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

how can l calculate G. M from the following size 125 133 141 173 182 frequency 7 5 4 1 3
how they find mean population
parts of statistics
what is a mean?
given the sequence 128,64,32 find the 12th term of the sequence
12th number is 0.0625
Thangarajan
why do we use summation notation to represent set of observations
what is the potential outlier ?
A pharmaceutical company claims that their pain reliever capsule is 70% effective. But a clinical test on this capsule showed 65 out of 100 effectiveness
Part of statistics
how to find mean population
what is data value
what is relative frequency
liner regression analysis
Proper definition of outlier?
Extraordinary observation (too distant, high, low etc)