Education.com
Try
Brainzy
Try
Plus

# Inference for Means and Proportions Review Problems for AP Statistics

(not rated)
By McGraw-Hill Professional
Updated on Feb 4, 2011

Review the following concepts if necessary:

### Problems

1. How large a sample is needed to estimate a population proportion within 2.5% at the 99% level of confidence if
1. you have no reasonable estimate of the population proportion?
2. you have data that show the population proportion should be about 0.7?
2. Let X be a binomial random variable with n = 250 and p = 0.6. Use a normal approximation to the binomial to approximate P(X > 160).
3. Write the mathematical expression you would use to evaluate P(X > 2) for a binomial random variable X that has B(5, 0.3) (that is, X is a binomial random variable equal to the number of successes out of 5 trials of an event that occurs with probability of success p = 0.3). Do not evaluate.
4. An individual is picked at random from a group of 55 office workers. Thirty of the workers are female, and 25 are male. Six of the women are administrators. Given that the individual picked is female, what is the probability she is an administrator?
5. A random sample of 25 cigarettes of a certain brand were tested for nicotine content, and the mean was found to be 1.85 mg with a standard deviation of 0.75 mg. Find a 90% confidence interval for the mean number of mg of nicotine in this type of cigarette. Assume that the amount of nicotine in cigarettes is approximately normally distributed. Interpret the interval in the context of the problem.

### Solutions

1.
1. .
2. .
2. .
3. .

(The exact binomial given by the TI-83/84 is 1-binomcdf(250,0.6,160) = 0.087. If you happen to be familiar with using a continuity correction (and you don't really need to be) for the normal approximation, normalcdf (160.5,1000,150,7.75) = 0.088, which is closer to the exact binomial.)

4. . (On the TI-83/84, this is equivalent to 1-binomcdf(5,0.3,2).)
5. P (the worker is an administrator | the worker is female)
6. .

7. A confidence interval is justified because we are dealing with a random sample from an approximately normally distributed population. df = 25 – 1 = 24 t* = 1.711.
8. .

We are 90% confident that the true mean number of mg per cigarette for this type of cigarette is between 1.59 mg and 2.11 mg.

150 Characters allowed

### Related Questions

#### Q:

See More Questions

### Today on Education.com

Top Worksheet Slideshows