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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.
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.
72 .
μ =
df = 20
$\sigma =\sqrt{2(df)}=\sqrt{40}=6.32$
73 . Enrolled = 200(0.66) = 132. Not enrolled = 200(0.34) = 68
74 .
Observed (O) | Expected (E) | O – E | (O – E)2 | $\frac{{(O-E)}^{2}}{z}$ | |
---|---|---|---|---|---|
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
79 .
Cell = Yes | Cell = No | Total | |
---|---|---|---|
Freshman | $\frac{250(300)}{500}=150$ | $\frac{250(200)}{500}=100$ | 250 |
Senior | $\frac{250(300)}{500}=150$ | $\frac{250(200)}{500}=100$ | 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.
83 . The table has five rows and two columns. df = ( r – 1)( c – 1) = (4)(1) = 4.
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.
88 .
H
_{0} :
σ
^{2} ≤ 5
H
_{a} :
σ
^{2} >5
1 . Which of the following equations is/are linear?
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