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4 years ago

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The upper quartile of a set of data is represented by which of the following? The median of the upper quarter of the data. The m

ode of the upper half of the data. The median of the upper half of the data. The mode of the upper quarter of the data

2 answers:

mrs_skeptik [129]4 years ago

8 0

The upper quartile (Q3) is represented by the median of the upper half of the data

Ganezh [65]4 years ago

8 0

The median of the upper half of the data

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Answer:

A)

<u><em>Null hypothesis:H₀:-</em></u><em> There is no significant difference between in drug resistance between the two states</em>

<u><em>Alternative Hypothesis :H₁:</em></u>

<em>There is significant difference between in drug resistance between the two states</em>

<em>B)</em>

<em>The calculated value Z = 2.7261 > 2.054 at 0.02 level of significance</em>

<em> Rejected H₀</em>

<em>There is a significant difference in drug resistance between the two states.</em>

C)

P - value = 0.0066

<em>P - value = 0.0066 < 0.02</em>

<em>D) </em>

<em>1) Reject H₀ </em>

<em>There is a significant difference in drug resistance between the two states.</em>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given first sample size n₁ = 174</em>

Suppose that, of 174 cases tested in a certain state, 11 were found to be drug-resistant.

<em>First sample proportion </em>

$p_{1} = \frac{x_{1} }{n_{1} } = \frac{11}{174} = 0.0632$

<em>Given second sample size n₂ = 375</em>

Given data Suppose also that, of 375 cases tested in another state, 7 were found to be drug-resistant

<em>Second sample proportion</em>

<em> </em> $p_{2} = \frac{x_{2} }{n_{2} } = \frac{7}{375} = 0.0186$ <em></em>

<u><em>Step(ii):-</em></u>

<u><em>Null hypothesis:H₀:-</em></u><em> There is no significant difference between in drug resistance between the two states</em>

<u><em>Alternative Hypothesis :H₁:</em></u>

<em>There is significant difference between in drug resistance between the two states</em>

<em>Test statistic</em>

<em> </em> $Z = \frac{p_{1}-p_{2} }{\sqrt{PQ(\frac{1}{n_{1} }+\frac{1}{n_{2} } ) } }$ <em></em>

<em> Where </em>

<em> </em> $P = \frac{n_{1}p_{1} +n_{2} p_{2} }{n_{1} +n_{2} }$ <em></em>

<em> </em> $P = \frac{174 (0.0632) + 375 (0.0186) }{174+375 } = \frac{17.9718}{549} = 0.0327$ <em></em>

<em> Q = 1 - P = 1 - 0.0327 = 0.9673</em>

<u><em>Step(iii):-</em></u>

<em></em>

<em> Test statistic</em>

<em> </em> $Z = \frac{p_{1}-p_{2} }{\sqrt{PQ(\frac{1}{n_{1} }+\frac{1}{n_{2} } ) } }$ <em></em>

<em> </em> $Z = \frac{0.0632-0.0186 }{\sqrt{0.0327 X 0.9673(\frac{1}{174 }+\frac{1}{375 } ) } }$ <em></em>

<em> Z = 2.7261</em>

<em> </em>

<em>Level of significance = 0.02 or 0.98</em>

<em>The z-value = 2.054</em>

<em>The calculated value Z = 2.7261 > 2.054 at 0.02 level of significance</em>

<em> Reject H₀ </em>

<em>There is a significant difference in drug resistance between the two states.</em>

<u><em>P- value </em></u>

<em>P( Z > 2.7261) = 1 - P( Z < 2.726)</em>

<em> = 1 - ( 0.5 + A (2.72))</em>

<em> = 0.5 - 0.4967</em>

<em> = 0.0033</em>

we will use two tailed test

<em>2 P( Z > 2.7261) = 2 × 0.0033</em>

<em> = 0.0066</em>

<em>P - value = 0.0066 < 0.02</em>

<em> Reject H₀ </em>

<em>There is a significant difference in drug resistance between the two states.</em>

8 0

3 years ago