Answer:
Yes, it is convincing evidence to conclude that the proportion of red cars that drive too fast on this
highway is greater than the proportion of non-red cars that drive too fast.
Explanation:
From the, we wish to first test;
H0; P_r - P_o = 0
And; H0; P_r - P_o > 0
Where; P_r and P_o are the proportion of red cars and other cars, respectively, who are driving too fast.
We will use a significance level of a = 0.05.
Thus;
The procedure is a two-sample z-test for the difference of proportions.
For, Random Conditions: The policemen chose cars randomly by rolling a die.
10%: We can safely assume that the number of cars driving past the rest area is essentially infinite, so the 10% restriction does not apply.
Large counts: The number of successes and failures in the two groups are 18, 10, 75, and 130—all of which are at least 10.
So, P_r = 18/28 = 0.64
P_o = 75/205 = 0.37
P_c = (18 + 75)/(28 + 205) = 0.4
Thus:
z = [(0.64 - 0.37) - 0]/√[[(0.4 x 0.6)/28] + [(0.4 x 0.6)/205]]
z = 2.73
From the one tailed z-score calculator online, I got P value = 0.003167
Thus, the P-value of 0.0032 is less than a = 0.05, so we reject H0. We
have sufficient evidence to conclude that the proportion of red cars that drive too fast on this
highway is greater than the proportion of non-red cars that drive too fast.