Confidence Intervals & Stats Workshop

Confidence by flickr user gerriet

A ferry boat captain knows from past experience that 35% of the passengers will get sick in the rough water ahead. The ferry has 126 passengers. What is the probability that at least 50 passengers will get sick?

(b) Solve this as a normal distribution problem. i.e. a normal approximation of the binomial distribution.

(c) Compare your results in (a) and (b). How do they compare?

(a) Solve this as a binomial distribution problem.HOMEWORK

A handy little applet ...

http://onlinestatbook.com/stat_sim/index.html

Solve a Binomial Problem as an Approximation to a Normal Distribution Problem

According to a group promoting safer driving habits, 63% of all drivers in Manitoba wear seatbelts when driving. During a Safety Week road check, 85 cars were stopped. What is the probability that from 50 to 60 (inclusive) drivers were wearing seatbelts.

HOMEWORK

The manager of the Jean Shop knows that 3 percent of all jeans sold will be defective, and the money paid for these pairs of jeans will be refunded. The manager went on holidays for a period of time, and an employee sold 247 pairs of jeans. The employee reported that refunds were given for 14 pairs of jeans.

(c) Does the employer have proof that the employee did something wrong?

(b) Does the employer have reason to be suspicious of the employee?

(a) What is the probability that 14 pairs of jeans were defective?

Calculate a 95 percent confidence interval and the percent margin of error, given the mean, the standard deviation, and the number of trials.

(b) μ = 44.5, σ = 3.82, n = 430(a) μ = 44.5, σ = 3.82, n = 56

(a) 95% confidence interval = µ ± 1.96σ = 44.5 ± 1.96(3.82) = 44.5 ± 7.4995% confidence interval = (37.01, 51.99)Margin of error = ± 1.96σ = 7.49% margin of error = ±13.38%

(b) 95% confidence interval = µ ± 1.96σ = 44.5 ± 1.96(3.82) = 44.5 ± 7.4995% confidence interval = (37.01, 51.99)Margin of error = ± 1.96σ = 7.49% margin of error = ±1.74%

(c) 95% confidence interval = 540 ± 16.66 = (523.34, 556.66)Margin of error = ± 1.96σ = ± 16.66% margin of error = ± 8.33%

HOMEWORK

The student council is planning a spring dance. They need to sell 85 tickets in order to cover their costs. Records show that, on average, 10% of the students enrolled at the school attend dances. This year there are 1160 students enrolled at the school.

Construct a 95% confidence interval for the number of students who will attend the spring dance.

HOMEWORK

Some Senior 4 students in a large high school want to change a tradition at graduation. Instead of wearing the usual cap and gown, they want to wear formal clothes. A quick survey of 96 randomly selected students shows that 41 prefer formal wear.

• How certain can we be that the results would be approximately the same if another survey of 250 students were done? • Is it possible that a majority of students prefer formal wear? • Based on the results of this survey, what are the smallest and largest numbers of students that would likely be in favour of dressing formally?

These questions can be answered by constructing a 95% confidence interval and the margin of error.

Dave is writing a true-false test consisting of 60 questions. Since he has not studied for the test, he decides to guess all the answers. Find a 95 percent confidence interval for his expected mark.

19 times out of 20, his test score would be between 22 and 38 inclusive (out of 60).

A toy manufacturer produces balloons that have a 3 percent defective rate. In a shipment of 4000 balloons, what is the probability that:

(b) between 100 and 130 balloons inclusive will be defective?

(a) fewer than 100 balloons will be defective?

Use a normal approximation to solve this problem.

A producer of hatching eggs acknowledges that 4 percent of all the eggs produced will not hatch. In a shipment of 600 eggs, what is the probability that:

(c) between 20 and 24 inclusive will not hatch?

(b) fewer than 20 will not hatch?

(a) at least 25 will not hatch?

Use this slide and those that follow to prepare for the test tomorrow.

It is known from past experience that, on average, 4.5 percent of all mouse traps produced by a company are defective. If 50 traps produced by the company are selected at random, find the probability that from 3 to 6 inclusive of them are defective.

In a small community, 65 percent of the people speak at least two languages. What is the probability that, if a group of 40 people is randomly selected, at most seven of them speak only one language?

It is known from past experience that, on average, 4.5 percent of all mouse traps produced by a company are defective. If 50 traps produced by the company are selected at random, find the probability that from three to six inclusive of them are defective.

Use a normal approximation to solve this problem.

A survey was conducted in a shopping mall to determine the number of people wearing jeans. A total of 340 people were observed, and 238, which is 70 percent, were wearing jeans.

(d) Calculate the percent margin of error if 950 people had been observed, and 70 percent of them were wearing jeans. How does this answer compare to (c)?

(c) Calculate the percent margin of error.

(b) Calculate a 95 percent confidence interval for the percent of people wearing jeans.

(a) Calculate a 95 percent confidence interval for the number of people wearing jeans.

(a) We are 95 percent confident that the number of people who wear jeans is between 221 and 255 inclusive.(b) We are 95 percent confident that from 65.1 percent to 74.9 percent of the people wear jeans.(c) % margin of error = ± 4.87%(d) % margin of error = ± 2.91%. The margin of error decreases as the sample size increases.

ANSWERS ON NEXT SLIDE

A survey was conducted in a shopping mall to determine the number of people wearing jeans. A total of 340 people were observed, and 238, which is 70 percent, were wearing jeans.

(a) We are 95 percent confident that the number of people who wear jeans is between 221 and 255 inclusive.

(b) We are 95 percent confident that from 65.1 percent to 74.9 percent of the people wear jeans.

(c) % margin of error = ± 4.87%

(d) % margin of error = ± 2.91%. The margin of error decreases as the sample size increases.

For a given set of data, μ = 23 and σ = 4.5.(a) What are μ and σ if 10 is added to each number in the set of data?

(b) What are μ and σ if each number in the set of data is multiplied by 2?

Mary Reed recently entered college and has not decided on a major. She decides to take an aptitude test in order to help her select a major. Her results, as well as the results of the other candidates, are as follows:

(a) Rewrite each of Mary Reed's scores as z-scores.

(c) In which area does she have the least talent?

(b) In which area does she have the most talent?

Talent Mean Standard Deviation Mary Reed's ScoreWriting 58 3.7 46Acting 84 7.9 85Medicine 37 2.8 44Law 49 4.7 41

Tammy and Jamey both applied for the same job. Tammy scored 80 on a provincial aptitude test where the mean was 70 and the standard deviation was 4.2. Jamey scored 510 on the company exam where the mean was 490 and the standard deviation was 10.3. Assuming the company uses these test results as the only criteria for hiring new employees and that both tests are considered to be equal by company officials, who might get the job? Explain your answer.

The following information concerning grades is posted on the bulletin board.

Test Grade Z-Score 55 -2 65 -1 75 0 85 1 95 2

(a) What is the average test grade?

(b) What test score has a z-score of -2.76?