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lana [24]
3 years ago
11

in 2004, 11.65 million people between the ages of 18 and 24 voted ( compared to 47.3 million voters between the ages of 45 and 6

4). if the united states has a population of 24.9 million people between the ages of 18 and 24, what percentage of those people voted?
Mathematics
1 answer:
QveST [7]3 years ago
4 0
Namely, what is 11.65 in percentage from 24.9

if we take 24.9 to be the 100%, let's see

\bf \begin{array}{ccllll}
amount&\%\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
24.9&100\\
11.65&x
\end{array}\implies \cfrac{24.9}{11.65}=\cfrac{100}{x}

solve for "x".
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4 years ago
To test the belief that sons are taller than their​ fathers, a student randomly selects 13 fathers who have adult male children.
KATRIN_1 [288]

Answer:

1) B. The differences are normally distributed or the sample size is large

C. The  sample size mus be large

E. The sampling method results in an independent sample

2) The null hypothesis H₀:  \bar x_1 =  \bar x_2

The alternative hypothesis Hₐ: \bar x_1 <  \bar x_2

Test statistic, t = -0.00693

p- value = 0.498

Do not reject Upper H₀ because, the P-value is greater than the level of significance. There is sufficient evidence to conclude that sons are the same height as their fathers  at 0.10 level of significance

Step-by-step explanation:

1) B. The differences are normally distributed or the sample size is large

C. The  sample size mus be large

E. The sampling method results in an independent sample

2) The null hypothesis H₀:  \bar x_1 =  \bar x_2

The alternative hypothesis Hₐ: \bar x_1 <  \bar x_2

The test statistic for t test is;

t=\dfrac{(\bar{x}_1-\bar{x}_2)}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}-\dfrac{s _{2}^{2}}{n_{2}}}}

The mean

Height of Father, h₁,  Height of Son h₂

72.4,      77.5

70.6,      74.1

73.1,       75.6

69.9,      71.7

69.4,      70.5

69.4,      69.9

68.1,       68.2

68.9,      68.2

70.5,       69.3

69.4,       67.7

69.5,       67

67.2,       63.7

70.4,       65.5

\bar x_1  = 69.6      

s₁ = 1.58

\bar x_2 = 69.9

s₂ = 3.97

n₁ = 13

n₂ = 13

t=\dfrac{(69.908-69.915)}{\sqrt{\dfrac{3.97^{2}}{13}-\dfrac{1.58^{2} }{13}}}

(We reversed the values in the square root of the denominator therefore, the sign reversal)

t = -0.00693

p- value = 0.498 by graphing calculator function

P-value > α Therefore, we do not reject the null hypothesis

Do not reject Upper H₀ because, the P-value is greater than the level of significance. There is sufficient evidence to conclude that sons are the same height as their fathers  at 0.10 lvel of significance

8 0
3 years ago
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