The national percentage of automobile accident fatalities that are alcohol related is 39%. In a random sample of 96 automobile accident fatalities in the state of Connecticut, 44 were alcohol related. Is this sufficient evidence to say that Connecticut has a higher percentage of alcohol related automobile fatalities?

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:

The sample proportion = 44/96 = 0.45833

Using the Z formula for a single proportion test, we get;

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.

75 tickets have been sold so far.

step-by-step explanation:

in this question it is clearly mentioned that 45 children's tickets have been sold and this number is 60% of the total tickets sold.

now let's assume the total number of tickets sold be x.

then 60% of x = 45

⇒

⇒ x(.6) = 45

by dividing with .6 on both the sides of the equation

⇒ x = 75

therefore 75 tickets have been sold so far.