Exam 8: Estimating Single Population Parameters

When calculating a confidence interval, the reason for using the t-distribution rather than the normal distribution for the critical value is that the population standard deviation is unknown.

What sample size is needed to estimate a population mean within ±50 of the true mean value using a confidence level of 95%, if the true population variance is known to be 122,500?

The fact that a point estimate will likely be different from the corresponding population value is due to the fact that point estimates are subject to sampling error.

One way to reduce the margin of error in a confidence interval estimate is to lower the level of confidence.

In a sample size determination situation, reducing the margin of error by half will double the required sample size.

A human resources manager wishes to estimate the proportion of employees in her large company who have supplemental health insurance. What is the largest size sample she should select if she wants 95 percent confidence and a margin of error of ± 0.01?

The margin of error is:

A local pizza company is interested in estimating the percentage of customers who would take advantage of a coupon offer. To do this, they give the coupon out to a random sample of 100 customers. Of these, 45 actually use the coupon. Based on a 95 percent confidence level, the upper and lower confidence interval limits are approximately 0.3525 to 0.5475.

A public policy research group is conducting a study of health care plans and would like to estimate the average dollars contributed annually to health savings accounts by participating employees. A pilot study conducted a few months earlier indicated that the standard deviation of annual contributions to such plans was $1,225. The research group wants the study's findings to be within $100 of the true mean with a confidence level of 90%. What sample size is required?

To construct a 99 percent confidence interval where σ is known, the correct critical value is 1.96.

When σ is unknown, the margin of error is computed by using:

The margin of error is one-half the width of the confidence interval.

A point estimate is an unbiased estimator of the true population value. However, error is associated with this estimate.

According to USA Today, customers are not settling for automobiles straight off the production lines. As an example, those who purchase a $355,000 Rolls-Royce typically add $25,000 in accessories. One of the affordable automobiles to receive additions is BMW's Mini Cooper. A sample of 179 recent Mini purchasers yielded a sample mean of $5,000 above the $20,200 base sticker price. Suppose the cost of accessories purchased for all Mini Coopers has a standard deviation of $1,500. Calculate a 95% confidence interval for the average cost of accessories on Mini Coopers.

The manager of the local county fair believes that no more than 30 percent of the adults in the county would object to a fee increase to attend the fair if it meant that better entertainment could be secured. To estimate the true proportion, he has selected a random sample of 200 adults. The manager will use a 90 percent confidence level. Assuming his assumption about the 30 percent holds, the margin of error for the estimate will be approximately ±.169.

The sampling distribution for a proportion has a formula for that standard error that involves using p. Yet when a confidence interval is calculated for a proportion, the standard error formula uses the sample proportion. Why do they differ?

The t-distribution is used for the critical value when estimating a population proportion when the standard deviation of the population is not known.

Confidence intervals constructed with small samples tend to have greater margins of error than those constructed from larger samples, all else being constant.

Chicago Connection, a local pizza company, delivers pizzas for free within the market area. The delivery drivers are paid $2.00 per delivery plus they get to keep any tips. To estimate the proportion of deliveries that result in a tip to the driver, a random sample of 64 deliveries was selected. Of these, 48 times a tip was received. Based on this information, and using a 95 percent confidence level, the margin of error for the estimate is approximately ±.1061.

When finding sample size, cutting the margin of error in half requires that the sample size be four times larger.