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Exam 10: Estimation: Describing a Single Population

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Question 51

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In developing an interval estimate for a population mean, the population standard deviation  was assumed to be 10. The interval estimate was 50.92 ± 2.14. Had the population standard deviation equaled 20, the interval estimate would have been:

A) 60.92 ± 2.14.
B) 50.92 ± 12.14.
C) 101.84 ± 4.28.
D) 50.92 ± 4.28.

E) None of the above
F) B) and C)

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Question 52

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Suppose that the amount of time teenagers spend on the Internet is normally distributed, with a standard deviation of 1.5 hours. A sample of 100 teenagers is selected at random, and the sample mean is computed as 6.5 hours. Determine the 95% confidence interval estimate of the population mean, changing the population standard deviation to 1.2.

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Question 53

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Find and interpret a 95% confidence interval for the population mean number of laptops owned by an average Australian household, if a random sample of 40 Australian households had a sample mean of 1.3 laptops. The population variance is known to be 4 laptops2.

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%.

A point estimate is defined as:

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 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?

As a manufacturer of golf clubs, a major corporation wants to estimate the proportion of golfers who are right-handed. How many golfers must be surveyed if they want to be within 0.02 with a 95% confidence level: a. assuming that there is no information available that could be used as an estimate of p? b. assuming that the manufacturer has an estimate of p obtained from a previous study that suggests that 75% of golfers are right-handed?

If there are two unbiased estimators of a parameter, the one whose variance is smaller is said to be relatively efficient.

Suppose that the amount of time teenagers spend on the Internet is normally distributed, with a standard deviation of 1.5 hours. A sample of 100 teenagers is selected at random, and the sample mean is computed as 6.5 hours. Determine the 95% confidence interval estimate of the population mean, changing the sample mean to 5.0 hours.

Which of the following statements is (are) correct?

The probability that a confidence interval includes the parameter of interest is either 1 or 0.

The sample mean $\bar { X }$ is a consistent estimator of the population mean µ.

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:

Which of the following is not a part of the formula for constructing a confidence interval estimate of the population mean?

When constructing confidence interval for a parameter, we generally set the confidence level $1 - \alpha$ close to 1 (usually between 0.90 and 0.99) because we would like to be reasonably confident that the interval includes the actual value of the population parameter.

A marketing researcher wishes to determine the sample size needed to estimate the proportion of wine drinkers who prefer a certain brand of wine. How many wine drinkers should be surveyed if the researcher wants to be within 0.025 with 95% confidence?

Recall the rule of thumb used to indicate when the normal distribution is a good approximation to the sampling distribution for the sample proportion $\hat { p }$ . For the combination n = 50; p = 0.05, the rule is satisfied.

An unbiased estimator of a population parameter is an estimator whose expected value is equal to the population parameter to be estimated.

The sample standard deviation is an unbiased estimator of the population standard deviation.