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)

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

ian gonna lie that is way to much