Question

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?

c. Estimate using a 99% confidence interval the proportion of automobile fatalities in

Connecticut that are alcohol related.

Answer 1

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

x = 44

n = 96

P = x/n = 44/96 = 0.458333333

Confidence level = 99%

Critical Z value = 2.5758

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.458333333 ± 2.5758* sqrt(0.458333333*(1 – 0.458333333)/96)

Confidence Interval = 0.458333333 ± 2.5758*0.0509

Confidence Interval = 0.458333333 ± 0.1310

Lower limit = 0.458333333 - 0.1310 = 0.3273

Upper limit = 0.458333333 + 0.1310 = 0.5893

Confidence interval = (0.3273, 0.5893)