g A recent newspaper article claims that 68% of car accidents involve some evidence of distracted driving. A police officer questions this claim, and pulls a random sample of 85 accident reports. If 63 of them involve distracted driving, is this enough evidence to show that the proportion of car accidents involve some evidence of distracted driving is significantly greater than than reported by the newspaper? (Use α α=0.10)

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jtellezd jtellezd · Answer 1 · 2020-07-26T03:58:41+02:00

Answer:

We accept the null hypothesis, we don´t have enough evidence to reject it

Step-by-step explanation:

We need to execute a proportion test

Accidents involving evidence of distracted driving p₀ = 68 %

Sample size 85 and p = 63/85 p = 74,11 %

CI ( confidence interval ) = 90 then α = 10 %

α = 0,1

Test of proportion:

Null Hypothesis H₀ p = p₀ p = 68

Aternate Hypothesis Hₐ p > p₀

Critical value z(c) for α = 0,1 , we proceed as follows

We find for α = 0,109 z (c) = 1,23 ( enough approximation)

Computing z(s)

z(s) = ( p - p₀) / √ p₀q₀/n

z(s) = ( 0,7411 - 0,68 ) / √0,68*0,32/ 85

z(s) = 0,0611/ √0,00256

z(s) = 1,21

Now we compare z(s) and z(c)

1,21 < 1,23

z(s) < z(c)

z(s) s in the acceptance region we keep null hypothesis