Exam 11: Estimation: Describing a Single Population

What is the width of a 99% confidence interval for the population proportion of people that own more than one mobile phone, if a random sample of 50 people was taken, and 10 of these had more than one mobile phone?

A sample of 50 students was asked how much time they spend on average a week in front of a computer. The sample mean and sample standard deviation were 15.8 and 2.7 hours, respectively. Estimate with 95% confidence interval the mean number of hours students spend in front of a computer a week.

How large a sample must be drawn to estimate the proportion of students who prefer statistics over mathematics, to within 0.02 with 95% confidence?

How large a sample of state employees should be taken if we want to estimate with 98% confidence the mean salary to within $2000. The population standard deviation is assumed to be $10 500.

In developing an interval estimate for a population mean, the interval estimate was 62.84 to 69.46. The population standard deviation was assumed to be 6.50, and a sample of 100 observations was used. The mean of the sample was:

Find and interpret a 98% confidence interval for the mean number of animals visited by a veterinarian per day. A random sample of 35 veterinarians, found that they had a sample mean of 25.3 animals and a sample variance of 2.8 animals.

It is possible to construct a confidence interval estimate of the population mean if the population variance is unknown.

A sample of size 200 is to be taken at random from an infinite population. Given that the population proportion is 0.60, the probability that the sample proportion will be greater than 0.58 is:

A random sample of 10 university students was surveyed to determine the amount of time they spent weekly using a personal computer. The times (in hours) were: 13 14 5 6 8 10 7 12 15 3 If the times are normally distributed with a standard deviation of 5.2 hours, estimate with 90% confidence the mean weekly time spent using a personal computer for all university students.

In the formula $\bar { x } \pm z _ { \mathfrak { w } / 2 } \sigma / \sqrt { n }$ , the $\alpha / 2$ refers to:

An interval estimate is an estimate of the range for a population parameter.

The probability of a success on any trial of a binomial experiment is 0.15. Find the probability that the proportion of success in a sample of 300 is more than 12%.

In general, decreasing the confidence level ( $1 - \alpha$ ) will narrow the interval.

Which of the following statements is false?

The problem with relying on a point estimate of a population parameter is that:

For statistical inference about the mean of a single population when the population standard deviation is unknown, the number of degrees for freedom for the t-distribution is equal to n - 1 because we lose one degree of freedom by using the:

A normal population has a standard deviation of 15. How large a sample should be drawn to estimate the population mean to within 1.5 with 95% confidence?

The objective of estimation is to determine the approximate value of:

A survey of 100 retailers revealed that the mean after-tax profit was $80 000. Assuming that the population standard deviation is $15 000, determine the 95% confidence interval estimate of the mean after-tax profit for all retailers.

Suppose that a 95% confidence interval for $\mu$ is given by $\bar { x } \pm 3.25$ . This notation means that if we repeatedly draw samples of the same size from the same population, 95% of the values of $\bar { x }$ will be such that $\mu$ would lie somewhere between $\bar { x } - 3.25$ and $\bar { x } + 3.25$ .