Answer:
Step-by-step explanation:
Given that:
There are 96 automobile accident fatalities in Connecticut and 44 were alcohol related.
The null hypothesis and the alternative hypothesis for this study can be computed as:
[tex]H_o: P =0.39 \\ \\ H_1:P> 0.39[/tex]
The sample proportion [tex]\hat p[/tex]= 44/96 = 0.45833
Using the Z formula for a single proportion test, we get;
[tex]Z = \dfrac{\hat p - P }{\sqrt\dfrac{{P(1-P)}}{{n}}}[/tex]
[tex]Z = \dfrac{0.45833 - 0.39 }{\sqrt\dfrac{{0.39(1-0.39)}}{{96}}}[/tex]
[tex]Z = \dfrac{0.06833 }{\sqrt\dfrac{{0.39(0.61)}}{{96}}}[/tex]
[tex]Z =1.37[/tex]
The p-value = 1 -P(Z< 1.37)
The p-value = 1 - 0.9147
The p-value = 0.0853
At the level of significance (∝) 0.05 ;
Since p-value is greater than level of significance (∝); we accept the null hypothesis and conclude that there is no sufficient evidence to say that higher percentage of alcohol is related to automobile fatalities.
