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We can reduce the margin of error in an interval estimate of p by doing any of the following except​


A) ​increasing the sample size.
B) ​increasing the planning value p* to .5.
C) ​increasing α.
D) ​reducing the confidence coefficient.

E) All of the above
F) B) and D)

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From a population that is normally distributed, a sample of 25 elements is selected and the standard deviation of the sample is computed.For the interval estimation of μ, the proper distribution to use is the


A) normal distribution.
B) t distribution with 25 degrees of freedom.
C) t distribution with 26 degrees of freedom.
D) t distribution with 24 degrees of freedom.

E) A) and C)
F) C) and D)

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D

It is known that the population variance equals 484.With a .95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is


A) 190.
B) 74.
C) 189.
D) 75.

E) All of the above
F) A) and B)

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A sample of 75 information systems managers had an average hourly income of $40.75 with a standard deviation of $7.00.The value of the margin of error at 95% confidence is


A) 80.83.
B) 7.00.
C) .81.
D) 1.61.

E) C) and D)
F) B) and C)

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The mean of the t distribution is​


A) ​0.
B) ​.5.
C) ​1.
D) ​problem specific.

E) A) and C)
F) A) and B)

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A random sample of 25 employees of a local company has been taken.A 95% confidence interval estimate for the mean systolic blood pressure for all employees of the company is 123 to 139.Which of the following statements is valid?


A) ​95% of the sample of employees has a systolic blood pressure between 123 and 139.
B) ​If the sampling procedure were repeated many times, 95% of the resulting confidence intervals would contain the population mean systolic blood pressure.
C) ​95% of the population of employees has a systolic blood pressure between 123 and 139.
D) ​If the sampling procedure were repeated many times, 95% of the sample means would be between 123 and 139.

E) B) and C)
F) A) and C)

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The manager of a grocery store has taken a random sample of 100 customers.The average length of time it took these 100 customers to check out was 3 minutes.It is known that the standard deviation of the population of checkout times is 1 minute.The standard error of the mean equals


A) .001.
B) .010.
C) .100.
D) 1.000.

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

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To compute the necessary sample size for an interval estimate of a population proportion, all of the following procedures are recommended when p is unknown except​


A) ​use the sample proportion from a previous study.
B) ​use the sample proportion from a preliminary sample.
C) ​use 1.0 as an estimate.
D) ​use judgment or a best guess.

E) A) and B)
F) B) and D)

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From a population that is not normally distributed and whose standard deviation is not known, a sample of 6 items is selected to develop an interval estimate for the mean of the population (μ) .


A) The normal distribution can be used.
B) The t distribution with 5 degrees of freedom must be used.
C) The t distribution with 6 degrees of freedom must be used.
D) The sample size must be increased.

E) A) and B)
F) B) and D)

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A random sample of 144 observations has a mean of 20, a median of 21, and a mode of 22.The population standard deviation is known to equal 4.8.The 95.44% confidence interval for the population mean is


A) 15.2 to 24.8.
B) 19.20 to 20.80.
C) 19.216 to 20.784.
D) 21.2 to 22.8.

E) A) and D)
F) A) and C)

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The t value for a 95% confidence interval estimation with 24 degrees of freedom is


A) 1.711.
B) 2.064.
C) 2.492.
D) 2.069.

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

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The probability that the interval estimation procedure will generate an interval that does not contain the actual value of the population parameter being estimated is the​


A) proportion estimate.
B) ​margin of error.
C) ​confidence coefficient.
D) same as α.

E) B) and C)
F) A) and D)

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D

A sample of 75 information systems managers had an average hourly income of $40.75 with a standard deviation of $7.00.If we want to determine a 95% confidence interval for the average hourly income of the population, the value of t is


A) 1.96.
B) 1.645.
C) 1.28.
D) 1.993.

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

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In a random sample of 144 observations, pˉ\bar { p } = .6.The 95% confidence interval for p is


A) .52 to .68.
B) .14 to .20.
C) .55 to .65.
D) .50 to .70.

E) C) and D)
F) A) and C)

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A

In interval estimation, as the sample size becomes larger, the interval estimate


A) becomes narrower.
B) becomes wider.
C) remains the same, because the mean is not changing.
D) gets closer to 1.96.

E) A) and C)
F) A) and D)

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For the interval estimation of μ when σ is known and the sample is large, the proper distribution to use is the


A) normal distribution.
B) t distribution with n degrees of freedom.
C) t distribution with n + 1 degrees of freedom.
D) t distribution with n - 1 degrees of freedom.

E) A) and C)
F) C) and D)

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In order to estimate the average electric usage per month, a sample of 81 houses was selected and the electric usage was determined.Assume a population standard deviation of 450 kilowatt-hours.The standard error of the mean is


A) 450.
B) 81.
C) 500.
D) 50.

E) B) and C)
F) C) and D)

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In developing an interval estimate, if the population standard deviation is unknown


A) it is impossible to develop the interval estimate.
B) the standard deviation is arrived at using the range.
C) the sample standard deviation must be used.
D) it is assumed that the population standard deviation is 1.

E) All of the above
F) B) and C)

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In a random sample of 100 observations, pˉ\bar { p } = .2.The 95.44% confidence interval for p is


A) .122 to .278.
B) .164 to .236.
C) .134 to .266.
D) .120 to .280.

E) C) and D)
F) B) and D)

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In a sample of 400 voters, 360 indicated they favor the incumbent governor.The 95% confidence interval of voters not favoring the incumbent is


A) .871 to .929.
B) .120 to .280.
C) .765 to .835.
D) .071 to .129.

E) B) and C)
F) A) and D)

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