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Tanzania [10]
4 years ago
5

A paper reported that in a representative sample of 282 American teens age 16 to 17, there were 71 who indicated that they had s

ent a text message while driving. For purposes of this exercise, assume that this sample is a random sample of 16- to 17-year-old Americans. Do these data provide convincing evidence that more than a quarter of Americans age 16 to 17 have sent a text message while driving? Test the appropriate hypotheses using a significance level of 0.01. (Round your test statistic to two decimal places and your P-value to four decimal places.)
Mathematics
1 answer:
elena-s [515]4 years ago
8 0

Answer:

z=\frac{0.252 -0.25}{\sqrt{\frac{0.25(1-0.25)}{282}}}=0.0776  

p_v =P(z>0.0776)=0.4691  

Since the p value is higher than the significance level so then we can conclude that we have enough evidence to FAIL to reject the null hypothesis and there is no evidence in order to say that the true proportion of teens more than a quarter of Americans age 16 to 17 have sent a text message while driving is higher than 0.25

Step-by-step explanation:

Information provided

n=282 represent the sample selected

X=71 represent the teens indicated that they had sent a text message while driving

\hat p=\frac{71}{282}=0.252 estimated proportion of teens indicated that they had sent a text message while driving

p_o=0.25 is the value that we want to test

\alpha=0.01 represent the significance level

z would represent the statistic

p_v represent the p value

System of hypothesis

We want to check if more than a quarter of Americans age 16 to 17 have sent a text message while driving, and the hypothesis are:

Null hypothesis:p\leq 0.7  

Alternative hypothesis:p > 0.25  

The statistic is given by:

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

Replacing we got:

z=\frac{0.252 -0.25}{\sqrt{\frac{0.25(1-0.25)}{282}}}=0.0776  

Decision

The p value for this case is:

p_v =P(z>0.0776)=0.4691  

Since the p value is higher than the significance level so then we can conclude that we have enough evidence to FAIL to reject the null hypothesis and there is no evidence in order to say that the true proportion of teens more than a quarter of Americans age 16 to 17 have sent a text message while driving is higher than 0.25

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Step-by-step explanation:

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also let B denote the event that an item has a flaw.

a)

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P(B) = 0.1

so

P(A₁ / B ) = 0.9,  P(A₂ / B ) = 0.7

using Baye's rule, the probability that an item has a flaw if it was passed by the first inspector is;

P(B / A₁ⁿ ) =  [P(A₁ⁿ / B ) P(B)] / [ P(A₁ⁿ / B ) P(B) + P(A₁ⁿ / Bⁿ ) P(Bⁿ) ]

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P(A₁ⁿ / B ) = 1 - P(A₁ / B )

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Since there are no false detections, the probability that the first inspector will not detect the flaw, given that item does not have flaws is;

P(A₁ⁿ / Bⁿ ) = 1.0

so the probability that the item has flaw if it was passed by the first inspector is;

P(B / A₁ⁿ ) =  [(0.1) (0.1)] / [ (0.1 ) (0.1) + (1.0) (1 - 0.1) ]

= 0.01 / ( 0.01 + 0.9)

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b)

Also by Bayes rule, the probability that the item has a flaw if it was passed by both inspectors is;

P(B / A₁ⁿ ∩ A₂ⁿ) =  [ P(A₁ⁿ ∩ A₂ⁿ / B) P(B)] / [ P(A₁ⁿ ∩ A₂ⁿ / B) P(B) + P(A₁ⁿ ∩ A₂ⁿ / Bⁿ) P(Bⁿ)]

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P(A₁ⁿ ∩ A₂ⁿ / B) = P(A₁ⁿ / B) P(A₂ⁿ / B)

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since there are no false detections, the probability that both inspectors will not detect the flaw, given that item does not have a flaw is;

P(A₁ⁿ ∩ A₂ⁿ / Bⁿ) = 1.0

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