The national percentage of automobile accident fatalities that are alcohol related is 39%. In a random...

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.

Solutions

Expert Solution

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)

orchestra answered 1 year ago

Next > < Previous

Related Solutions

Suppose that the dollar amount of damage involved in an automobile accident is an exponential random...

Suppose that the dollar amount of damage involved in an automobile accident is an exponential random variable X with mean $1000. (a) What is the probability that the amount of an accident exceeds $400? (b) If the deductible is $400, that means, the insurance company only pays the amount exceeding $400. What is the expected value of the amount that the insurance company pays per accident?

The data in the table below are the result of a random survey of 39 national...

The data in the table below are the result of a random survey of 39 national flags (with replacement between picks) from various countries. We are interested in finding a confidence interval for the true mean number of colors on a national flag. Let X = the number of colors on a national flag. X Freq. 1 1 2 7 3 18 4 7 5 6 Construct a 95% confidence interval for the true mean number of colors on national...

The data in the table below are the result of a random survey of 39 national...

A study investigated maintenance and repair costs for automobiles. A random sample of 39 automobile owners...

A study investigated maintenance and repair costs for automobiles. A random sample of 39 automobile owners were asked how much they spent on maintenance and repair costs in the last year. The average amount spent was 1387.782 dollars with a standard deviation of 170 dollars. Is there evidence at the 0.05 level that maintenance and repair costs exceeds 1320 dollars annually? HoHo: Select an answer μ p̂ p ȳ ? ≠ = > < HaHa:Select an answer p̂ p ȳ μ ?...

Assume that an uncle of yours was injured on 1/1/17 in an automobile accident. The accident...

Assume that an uncle of yours was injured on 1/1/17 in an automobile accident. The accident was serious, and it was the other driver's fault. As the result of the accident, your uncle is not able to return to his medical practice where he earned $280,000 during calendar year 2016. Your uncle's accident occurred on his 60th birthday (some birthday) and he planned (prior to the accident) to work another eight years (until the end of 2024). Your uncle has...

Suppose that every driver faces a 4% probability of an automobile accident every year. An accident...

Suppose that every driver faces a 4% probability of an automobile accident every year. An accident will, on average, cost each driver $13,000. Suppose there are two types of individuals: those with $104,000.00 in the bank and those with $6,500.00 in the bank. Assume that individuals with $6,500.00 in the bank declare bankruptcy if they get in an accident. In bankruptcy, creditors receive only what individuals have in the bank. Assume that both types of individuals are only slightly risk...

Suppose that every driver faces a 4% probability of an automobile accident every year. An accident...

Suppose that every driver faces a 4% probability of an automobile accident every year. An accident will, on average, cost each driver $7,000. Suppose there are two types of individuals: those with $56,000.00 in the bank and those with $3,500.00 in the bank. Assume that individuals with $3,500.00 in the bank declare bankruptcy if they get in an accident. In bankruptcy, creditors receive only what individuals have in the bank. Assume that both types of individuals are only slightly risk...

Suppose every driver faces a 1% probability of an automobile accident every year. An accident will,...

Suppose every driver faces a 1% probability of an automobile accident every year. An accident will, on an average cost each driver $10,000. Suppose there are two types of individuals: those with $60,000 and those with $5,000 in the bank. Assume that individuals with $5,000 in the bank declare bankruptcy if they get in an accident. In the bankruptcy, creditors receive only what individuals have in the bank. What price are individuals with $5,000 in the bank willing to pay...

The Fitzgeralds (who have a PAP) are involved in an automobile accident that was not their...

The Fitzgeralds (who have a PAP) are involved in an automobile accident that was not their fault on their way home from a dinner party. Although they are unharmed, their car is disabled. The police recommend that they leave the car by the side of the road and take a taxi home. When the Fitzgeralds return to the scene of the accident, they find that their car has been stripped down to the chassis. The Fitzgeralds submit a claim to...

An automobile insurance company has determined the accident rate (probability of having at least one accident...

An automobile insurance company has determined the accident rate (probability of having at least one accident during a year) for various age groups (see Table). Suppose that a policyholder calls in to report an accident. What is the probability that he or she is over 60? over 60 - proportion of total insured .10 & accident rate =0.06 (erased on table and could not add it back on accident ) What is the the probability that he or she is...

Subjects