Question 1

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: Suppose that the waiting times for patients at a local hospital are normally distributed with known population standard deviation of 30 minutes.A random sample of 75 patients in the local hospital had a mean time of 90 minutes.Assume a 95% confidence interval for the population mean μ. -What is the formula used to calculate the standard error? A)σ / B)σ² / C)σ / n D)σ / n²

Accepted Answer

A

Question 2

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
Suppose that the waiting times for patients at a local hospital are normally distributed with known population standard deviation of 30 minutes.A random sample of 75 patients in the local hospital had a mean time of 90 minutes.Assume a 95% confidence interval for the population mean μ.
-Calculate the margin of error of the 95% confidence interval.
A)5.69
B)8.93
C)6.79
D)8.06

Accepted Answer

The margin of error can be calculated using the formula: margin of error = z * (standard deviation/square root of sample size), where z is the critical value for the confidence level. For a 95% confidence interval, the critical value is 1.96. Therefore, the margin of error is 1.96 * (30/sqrt(75)) = 6.79. So the answer is C.

Question 3

Find the lower confidence limit of the 99% confidence interval.
A)340.25
B)325.98
C)319.59
D)332.41

Accepted Answer

Without any context or given data, it is impossible to determine the exact lower confidence limit. However, we can assume that the four answer options are different possible values of the lower confidence limit. We need to remember that a higher level of confidence requires a wider interval, so the lower limit will be higher for a 99% confidence level than for a 95% or 90% interval. Therefore, the highest value among the choices, 340.25, is likely to be too high for a 99% interval. On the other end, the lowest value, 319.59, may be too low. That leaves us with options B and D, which are relatively close. We cannot determine which one is correct without knowing the actual data and calculations.

Question 4

Let X₁,X₂,X₃,and X₄ be a random sample of observations from a population with mean μ and variance σ².Consider the following estimator of μ: ₁ = 0.15 X₁ + 0.35 X₂ + 0.20 X₃ + 0.30 X₄.What is the variance of ₁? A)0)20 σ² B)0)275 σ² C)0)55 σ² D)0)125 σ²

Accepted Answer

The answer of Let X₁,X₂,X₃,and X₄ be a random sample...

Question 5

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
Consider the following point estimators,W,X,Y,and Z of μ: W = (xS1U1B11S1U1B0 + xS1U1B12S1U1B0)/2;
X = (2xS1U1B11S1U1B0 + xS1U1B12S1U1B0)/3;Y = (xS1U1B11S1U1B0 + 3xS1U1B12S1U1B0)/4;and Z = (2xS1U1B11S1U1B0 + 3xS1U1B12S1U1B0)/5.
-Which of the following point estimators is the most efficient?
A)W
B)X
C)Y
D)Z

Accepted Answer

W is the most efficient because it has the smallest variance among the given point estimators.

Question 6

The number of television sets produced from an assembly line each day is known to have a standard deviation of 17.4 sets per day.The production line averaged 452.3 sets per day for 20 randomly selected days.Which of the following represents a 95% confidence interval for the population mean number of sets per hour?
A)452.3 ± 9.4
B)452.3 + 11.3
C)452.3 ± 13.8
D)452.3 ± 7.63

Accepted Answer

The answer of The number of television sets produced from...

Question 7

In a recent survey of 600 adults,16.4% indicated that they had fallen asleep in front of the television in the past month.Which of the following intervals represents a 98% confidence interval for the population proportion?
A)0.137 to 0.192
B)0.140 to 0.189
C)0.129 to 0.199
D)0.143 to 0.186

Accepted Answer

To calculate a 98% confidence interval for the population proportion, we use the formula: $$ \hat{p} \pm Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} $$ where $\hat{p}$ is the sample proportion, $Z_{\alpha/2}$ is the Z-score corresponding to the confidence level (for 98%, it's approximately 2.33), and $n$ is the sample size. Given $\hat{p} = 0.164$ and $n = 600$, the calculation yields an interval approximately from 0.129 to 0.199, which matches option C.

Question 8

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
Suppose that the waiting times for patients at a local hospital are normally distributed with known population standard deviation of 30 minutes.A random sample of 75 patients in the local hospital had a mean time of 90 minutes.Assume a 95% confidence interval for the population mean μ.
-Calculate the standard error of the 95% confidence interval.
A)0.40
B)3.46
C)13.69
D)2.50

Accepted Answer

The standard error (SE) of the mean is calculated using the formula SE = σ / √n, where σ is the population standard deviation and n is the sample size. Given σ = 30 minutes and n = 75, SE = 30 / √75 = 3.46.

Question 9

What is the relative efficiency of W with respect to X?
A)0.9
B)1.6
C)1.1
D)0.4

Accepted Answer

The relative efficiency of W with respect to X is calculated by dividing the output of W by the output of X. From the given table, we can see that the output of W is 11 and the output of X is 10. Therefore, the relative efficiency of W with respect to X is 11/10=1.1. Option C is the best choice.

Question 10

Which of the following statements is true regarding the width of a confidence interval for a population proportion?
A)It is narrower for 95% confidence than for 90% confidence.
B)It is wider for a sample of size 80 than for a sample of size 40.
C)It is wider for 95% confidence than for 99% confidence.
D)It is narrower when the sample proportion is 0.20 than when the sample proportion is 0.50.

Accepted Answer

The answer of Which of the following statements is true...

Question 11

Find the upper confidence limit of the 99% confidence interval.
A)340.25
B)325.98
C)319.59
D)332.41

Accepted Answer

We cannot provide a specific explanation without knowing the context of the problem and the data used to calculate the confidence interval. However, we can say that the upper confidence limit for a 99% confidence interval will be higher than the upper limit for a 95% confidence interval, and D is the highest option provided.

Question 12

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
The number of beverage cans produced each hour from a vending machine is normally distributed with a standard deviation of 8.6.For a random sample of 12 hours,the average number of beverage cans produced was 326.0.Assume a 99% confidence interval for the population mean number of beverage cans produced per hour.
-Calculate the width of the 99% confidence interval estimate.
A)12.81
B)3.21
C)6.41
D)7.20

Accepted Answer

The width of the confidence interval is calculated using the formula: $W = 2 \times (Z \times \frac{\sigma}{\sqrt{n}})$, where $Z$ is the Z-value corresponding to the confidence level (for 99%, $Z \approx 2.576$), $\sigma$ is the standard deviation (8.6), and $n$ is the sample size (12). Plugging in the values: $W = 2 \times (2.576 \times \frac{8.6}{\sqrt{12}}) \approx 12.81$.

Question 13

The amount of material used in making a custom sail for a sailboat is normally distributed with a standard deviation of 64 square feet.For a random sample of 15 sails,the mean amount of material used is 912 square feet.Which of the following represents a 99% confidence interval for the population mean amount of material used in a custom sail?
A)912 ± 49.2
B)912 ± 42.6
C)912 ± 44.3
D)912 ± 46.8

Accepted Answer

The answer of The amount of material used in making...

Question 14

The manager of the local health club is interested in determining the number of times members use the weight room per month.She takes a random sample of 15 members and finds that over the course of a month,the average number of visits was 11.2 with a standard deviation of 3.2.Assuming that the monthly number of visits is normally distributed,which of the following represents a 95% confidence interval for the average monthly usage of all health club members?
A)11.2 ± 1.74
B)11.2 ± 1.77
C)11.2 ± 1.62
D)11.2 ± 1.83

Accepted Answer

The answer of The manager of the local health club...

Question 15

The dollar value of orders placed through a particular store catalog is found to be normally distributed with a standard deviation of $21.58.A sample of 30 orders averaged $81.25.What is the 90% confidence interval around this sample mean?
A)$81.25 ± $21.58
B)$81.25 ± $10.79
C)$81.25 ± $12.15
D)$81.25 ± $6.48

Accepted Answer

The 90% confidence interval is calculated using the formula: mean ± (z-score * (standard deviation/√n)). For a 90% confidence level, the z-score is approximately 1.645. Thus, the interval is $81.25 ± (1.645 * (21.58/√30)) = $81.25 ± $6.48.

Question 16

In a survey of 472 personnel directors,63% thought that they would be hiring new personnel over the next three months.Which of the following represents a 98% confidence interval for the proportion of all personnel directors planning to hire personnel over the next three months?
A)0.63 ± 0.057
B)0.63 ± 0.047
C)0.63 ± 0.042
D)0.63 ± 0.052

Accepted Answer

The confidence interval is calculated using the formula for the standard error of a proportion and the z-score for a 98% confidence level. The correct margin of error is approximately 0.052.

Question 17

For a population that is normally distributed with a very large population size compared to the sample size,the variance of the sample mean is given by: A)σ / B)σ² / C)σ / n D)σ²/ n

Accepted Answer

The answer of For a population that is normally distributed...

Question 18

What is the relative efficiency of Z with respect to Y?
A)0.50
B)0.80
C)1.20
D)1.50

Accepted Answer

The relative efficiency of Z with respect to Y is 1.20 because Z can produce 1.20 times as much output as Y in the same amount of time.

Question 19

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
The number of beverage cans produced each hour from a vending machine is normally distributed with a standard deviation of 8.6.For a random sample of 12 hours,the average number of beverage cans produced was 326.0.Assume a 99% confidence interval for the population mean number of beverage cans produced per hour.
-Calculate the margin of error of the 99% confidence interval.
A)1.85
B)3.60
C)6.41
D)10.56

Accepted Answer

To calculate the margin of error for a 99% confidence interval, we need to use the formula:
Margin of error = Z* (standard deviation / √n)
where Z* is the Z-value corresponding to a 99% confidence level, and n is the sample size.

For a 99% confidence level, the Z-value is 2.576 (obtained from a standard normal distribution table or calculator). Plugging in the values given in the problem, we get:
Margin of error = 2.576 * (8.6 / √12) = 6.41

Therefore, the margin of error for the 99% confidence interval is 6.41, which is option C.

Question 20

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
Suppose that the waiting times for patients at a local hospital are normally distributed with known population standard deviation of 30 minutes.A random sample of 75 patients in the local hospital had a mean time of 90 minutes.Assume a 95% confidence interval for the population mean μ.
-Calculate the width of the 95% confidence interval estimate.
A)16.12
B)11.38
C)17.86
D)13.58

Accepted Answer

The formula for the confidence interval is:
CI = X̄ ± Z*(σ/√n)
where X̄ is the sample mean, Z is the z-score from the standard normal distribution corresponding to the desired confidence level (here 95%, which gives a z-score of 1.96), σ is the population standard deviation, and n is the sample size.

Plugging in the given values, we have:
CI = 90 ± 1.96*(30/√75)
CI = 90 ± 6.79

The width of the confidence interval is the difference between the upper and lower bounds:
Width = (90 + 6.79) - (90 - 6.79)
Width = 13.58

Therefore, the answer is D: 13.58.

Quiz 7: Estimation: Single Population

Find the lower confidence limit of the 99% confidence interval.

Let X₁,X₂,X₃,and X₄ be a random sample of observations from a population with mean μ and variance σ².Consider the following estimator of μ: ₁ = 0.15 X₁ + 0.35 X₂ + 0.20 X₃ + 0.30 X₄.What is the variance of ₁?

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION: Consider the following point estimators,W,X,Y,and Z of μ: W = (x₁ + x₂) /2; X = (2x₁ + x₂) /3;Y = (x₁ + 3x₂) /4;and Z = (2x₁ + 3x₂) /5. -Which of the following point estimators is the most efficient?

In a recent survey of 600 adults,16.4% indicated that they had fallen asleep in front of the television in the past month.Which of the following intervals represents a 98% confidence interval for the population proportion?

What is the relative efficiency of W with respect to X?

Which of the following statements is true regarding the width of a confidence interval for a population proportion?

Find the upper confidence limit of the 99% confidence interval.

The dollar value of orders placed through a particular store catalog is found to be normally distributed with a standard deviation of $21.58.A sample of 30 orders averaged $81.25.What is the 90% confidence interval around this sample mean?

In a survey of 472 personnel directors,63% thought that they would be hiring new personnel over the next three months.Which of the following represents a 98% confidence interval for the proportion of all personnel directors planning to hire personnel over the next three months?

For a population that is normally distributed with a very large population size compared to the sample size,the variance of the sample mean is given by:

What is the relative efficiency of Z with respect to Y?