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Nastasia [14]
3 years ago
12

California has 143deaths from heart disease and 137deaths from cancer per100,000 residents. Which rate is more extreme compared

to other states? A.Perform any needed computations here: b. Is California's rate for heart disease or for cancer more extreme compared to the other states? Explain:
Mathematics
1 answer:
belka [17]3 years ago
3 0

Answer:

California's rate for heart disease is more extreme than Cancer

Step-by-step explanation:

Given

Represent Cancer with C and Heart Disease with H

H = 143

C = 137

Population = per 100,000

First we need to determine the probability of both.

For H

P(H) = H/100000

P(H) = 143/100000

P(H) = 0.00143

For C

P(C) = C/100000

P(C) = 137/100000

P(C) = 0.00137

By comparison,

0.00143 > 0.00137

So, California's rate for heart disease is more extreme than Cancer.

Reason: P(H) > P(C)

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In a random sample of 300 attendees of a minor league baseball game, 182 said that they bought food from the concession stand. C
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<em>90% confidence interval for the proportion of fans who bought food from the concession stand</em>

(0.5603,0.6529)

Step-by-step explanation:

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

<em>Given sample size 'n' =300</em>

Given data  random sample of 300 attendees of a minor league baseball game, 182 said that they bought food from the concession stand.

<em>Given sample proportion </em>

<em>                        </em>p^{-}  = \frac{x}{n} = \frac{182}{300} =0.606

level of significance = 90% or 0.10

Z₀.₁₀ = 1.645

<em>90% confidence interval for the proportion is determined by</em>

(p^{-} - Z_{0.10}\sqrt{\frac{p(1-p)}{n} }  , p^{-} +Z_{0.10}\sqrt{\frac{p(1-p)}{n} })

(0.6066 - 1.645\sqrt{\frac{0.6066(1-0.6066)}{300} }  ,0.6066+1.645\sqrt{\frac{0.6066(1-0.6066)}{300} })

(0.6066 -  0.0463  ,0.6066 +  0.0463)

(0.5603,0.6529)

<u>final answer</u>:-

<em>90% confidence interval for the proportion of fans who bought food from the concession stand</em>

(0.5603,0.6529)

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Attached file

Step-by-step explanation:

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