Question 1

A researcher wants to make a 99% confidence interval for a population proportion.The most conservative estimate of the sample size that would limit the margin of error to be within 0.033 of the population proportion is:

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Question 2

A sample of 20 elements produced a mean of 91.4 and a standard deviation of 11.16.Assuming that the population has a normal distribution,the 90% confidence interval for the population mean is:

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Question 3

A random sample of 765 persons showed that 13.5% do not have any health insurance.Based on this sample,the 95% confidence interval for the proportion of all persons who do not have any health insurance,rounded to four decimal places,is:

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Question 4

A researcher wants to make a 99% confidence interval for a population mean.She wants the margin of error to be within 4.6 of the population mean.The population standard deviation is 18.22.The sample size that will yield a margin of error within 4.6 of the population mean is:

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Question 5

A researcher wants to make a 95% confidence interval for a population mean.She wants the margin of error to be within 1.9 of the population mean.The population standard deviation is 11.07.The sample size that will yield a margin of error within 1.9 of the population mean is:

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Question 6

A random sample of 983 families selected from a large city showed that 17.5% of them make $100,000 or more per year.Based on this sample,the 99% confidence interval for the proportion of all families living in this city who make $100,000 or more per year,rounded to four decimal places,is:

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Question 7

A researcher wants to make a 99% confidence interval for a population proportion.A preliminary sample produced the sample proportion of 0.680.The sample size that would limit the margin of error to be within 0.024 of the population proportion is:

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Question 8

A random sample of 188 produced a sample proportion of 0.40.The 98% confidence interval for the population proportion,rounded to four decimal places,is:

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Question 9

The value of t for 19 degrees of freedom and a 90% confidence interval is:

Accepted Answer

For a 90% confidence interval with 19 degrees of freedom, the t-value is 1.729. This value is typically positive as it represents the critical value for the upper tail of the t-distribution.

Question 10

A random sample of 12 life insurance policy holders showed that the mean value of their life insurance policies is $210,000 with a standard deviation of $44,600.Assuming that the values of life insurance policies for all such policy holders are approximately normally distributed,the 99% confidence interval for the mean value of all life insurance policies,rounded to two decimal places,is:

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Question 11

The value of t for 19 degrees of freedom and a 98% confidence interval is:

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Question 12

A random sample of 1,100 adults showed that 32% of them are smokers.Based on this sample,the 90% confidence interval for the proportion of all adults who are smokers,rounded to four decimal places,is:

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Question 13

A random sample of 450 produced a sample proportion of 0.71.The 95% confidence interval for the population proportion,rounded to four decimal places,is:

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Question 14

A sample of 25 elements produced a mean of 123.4 and a standard deviation of 18.32.Assuming that the population has a normal distribution,the 90% confidence interval for the population mean,rounded to two decimal places,is:

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Question 15

A company wants to estimate,at a 95% confidence level,the proportion of all families who own its product.A preliminary sample showed that 30.0% of the families in this sample own this company's product.The sample size that would limit the margin of error to be within 0.045 of the population proportion is:

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Question 16

In a random sample of 562 items produced by a machine,the quality control staff found 6.6% to be defective.Based on this sample,the 95% confidence interval for the proportion of defective items in all items produced by this machine,rounded to four decimal places,is:

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Question 17

A random sample of 23 tourists who visited Hawaii this summer spent an average of $1,456.0 on this trip with a standard deviation of $263.00.Assuming that the money spent by all tourists who visit Hawaii has an approximate normal distribution,the 95% confidence interval for the average amount of money spent by all tourists who visit Hawaii,rounded to two decimal places,is:

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Question 18

A researcher wants to estimate the mean age of all Business Week readers at a 99% confidence level.She wants the margin of error to be within 3.1 years of the population mean.The standard deviation of ages of all Business Week readers is 10.04 years.The sample size that will yield a margin of error within 3.1 of the population mean is:

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Question 19

A random sample of 8 houses selected from a city showed that the mean size of these houses is 1,881.0 square feet with a standard deviation of 328.00 square feet.Assuming that the sizes of all houses in this city have an approximate normal distribution,the 90% confidence interval for the mean size of all houses in this city,rounded to two decimal places,is:

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Question 20

A company wants to estimate the mean net weight of all 32-ounce packages of its Yummy Taste cookies at a 95% confidence level.The margin of error is to be within 0.025 ounces of the population mean.The population standard deviation is 0.096 ounces.The sample size that will yield a margin of error within 0.025 ounces of the population mean is:

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Quiz 8: Estimation of the Mean and Proportion

A researcher wants to make a 99% confidence interval for a population proportion.The most conservative estimate of the sample size that would limit the margin of error to be within 0.033 of the population proportion is:

A sample of 20 elements produced a mean of 91.4 and a standard deviation of 11.16.Assuming that the population has a normal distribution,the 90% confidence interval for the population mean is:

A random sample of 765 persons showed that 13.5% do not have any health insurance.Based on this sample,the 95% confidence interval for the proportion of all persons who do not have any health insurance,rounded to four decimal places,is:

A researcher wants to make a 99% confidence interval for a population mean.She wants the margin of error to be within 4.6 of the population mean.The population standard deviation is 18.22.The sample size that will yield a margin of error within 4.6 of the population mean is:

A researcher wants to make a 95% confidence interval for a population mean.She wants the margin of error to be within 1.9 of the population mean.The population standard deviation is 11.07.The sample size that will yield a margin of error within 1.9 of the population mean is:

A random sample of 983 families selected from a large city showed that 17.5% of them make $100,000 or more per year.Based on this sample,the 99% confidence interval for the proportion of all families living in this city who make $100,000 or more per year,rounded to four decimal places,is:

A researcher wants to make a 99% confidence interval for a population proportion.A preliminary sample produced the sample proportion of 0.680.The sample size that would limit the margin of error to be within 0.024 of the population proportion is:

A random sample of 188 produced a sample proportion of 0.40.The 98% confidence interval for the population proportion,rounded to four decimal places,is:

The value of t for 19 degrees of freedom and a 90% confidence interval is:

The value of t for 19 degrees of freedom and a 98% confidence interval is:

A random sample of 1,100 adults showed that 32% of them are smokers.Based on this sample,the 90% confidence interval for the proportion of all adults who are smokers,rounded to four decimal places,is:

A random sample of 450 produced a sample proportion of 0.71.The 95% confidence interval for the population proportion,rounded to four decimal places,is:

A sample of 25 elements produced a mean of 123.4 and a standard deviation of 18.32.Assuming that the population has a normal distribution,the 90% confidence interval for the population mean,rounded to two decimal places,is:

In a random sample of 562 items produced by a machine,the quality control staff found 6.6% to be defective.Based on this sample,the 95% confidence interval for the proportion of defective items in all items produced by this machine,rounded to four decimal places,is: