Question 1

The WISC scores (similar to IQ test scores) of a sample of n = 20 5^th graders will be used to create a 99% confidence interval for the mean WISC score of all 5^th graders. What is the correct t* multiplier for this interval? A) 2.86 B) 3.55 C) 2.54 D) 2.85

Accepted Answer

For a 99% confidence interval with n = 20 (which means degrees of freedom = 19), the correct t* multiplier according to most t-distribution tables is approximately 2.86.

Question 2

The heights of a random sample of 100 women are recorded. The sample mean is 65.3 inches and the sample standard deviation is 3 inches. Which of the following is an approximate 95% confidence interval for the population mean?
A) 65.3 $\pm$(2)(0.03)
B) 65.3 $\pm$ (2)(0.3)
C) 65.3 $\pm$ (2)(3)
D) 65.3 $\pm$(2)(30)

Accepted Answer

The correct formula for a 95% confidence interval is the sample mean $\pm$ the margin of error, where the margin of error is calculated as the critical value (approximately 2 for 95% confidence) times the standard error of the mean. The standard error of the mean is the sample standard deviation divided by the square root of the sample size. Given a sample standard deviation of 3 inches and a sample size of 100, the standard error is $3/\sqrt{100} = 3/10 = 0.3$. Therefore, the margin of error is $2 \times 0.3$, making the correct choice 65.3 $\pm$ (2)(0.3).

Question 3

A random sample of 100 students had a mean grade point average (GPA) of 3.2 with a standard deviation of 0.2. The standard error of the sample mean in this case is 0.02. Calculate an approximate 95% confidence interval for the mean GPA for all students.
A) (2.8, 3.6)
B) (3.16, 3.24)
C) (3.18, 3.22)
D) None of the above

Accepted Answer

The 95% confidence interval for the mean is calculated using the formula: mean ± (Z * standard error), where Z is the Z-score for a 95% confidence level (approximately 1.96). Here, the mean is 3.2, and the standard error is 0.02. Thus, the interval is 3.2 ± (1.96 * 0.02) = 3.2 ± 0.0392, which approximates to (3.16, 3.24).

Question 4

A random sample of 30 students spent an average amount of $40.00 on a date, with a standard deviation of $5.00. The standard error of the sample mean is 0.913. Calculate an approximate 95% confidence interval for the average amount spent by all students on a date.
A) (40.00, 90.00)
B) (79.10, 80.90)
C) (78.20, 81.80)
D) None of the above

Accepted Answer

The standard error of the mean is calculated as the standard deviation divided by the square root of the sample size. In this case, the standard error of the mean is $5.00 / sqrt(30) = 0.913.The 95% confidence interval is calculated as the sample mean plus or minus 1.96 times the standard error of the mean. In this case, the 95% confidence interval is $40.00 +/- 1.96 * 0.913 = (78.20, 81.80).

Question 5

The cholesterol levels of a random sample of 100 men are measured. The sample mean is 188 and the sample standard deviation is 40. Which of the following provides a 95% confidence interval for the population mean?
A) 188$\pm$(1.98)(0.4)
B) 188 $\pm$(1.98)(4)
C) 188 $\pm$ (1.98)(40)
D) 188 $\pm$(1.98)(4000)

Accepted Answer

The 95% confidence interval for the population mean is calculated using the formula: sample mean $\pm$ (z-score)(standard error). The standard error is the sample standard deviation divided by the square root of the sample size, which is 40/√100 = 4. Thus, the correct interval is 188 $\pm$ (1.98)(4).

Question 6

A random sample of 30 airline flights during a storm had a mean delay of 40 minutes. The standard deviation was 5 minutes, and the standard error of the mean is 0.9129. Calculate a 90% confidence interval for the average delay for all flights during a storm.
A) (38.2, 41.8)
B) (38.4, 41.6)
C) (31.5, 48.5)
D) None of the above

Accepted Answer

The 90% confidence interval is calculated using the formula: mean ± (z-score * standard error). For 90% confidence, the z-score is approximately 1.645. Thus, the interval is 40 ± (1.645 * 0.9129) = (38.4, 41.6).

Question 7

A random sample of 30 airline flights during a storm had an average delay of 40 minutes. The standard deviation was 5 minutes. Calculate a 98% confidence interval for the average delay for all flights during a storm.
A) (37.8, 42.2)
B) (38.2, 41.8)
C) (27.7, 52.3)
D) None of the above

Accepted Answer

The correct answer is A because the confidence interval is calculated as (40 - 2.326 * 5/sqrt(30), 40 + 2.326 * 5/sqrt(30)) = (37.8, 42.2).

Question 8

A randomly selected sample of n = 51 men in Brazil had an average lifespan of 59 years. The standard deviation was 10 years, and the standard error of the mean is 1.400. Calculate a 90% confidence interval for the average lifespan for all men in Brazil.
A) (42.2, 75.8)
B) (56.6, 61.4)
C) (56.2, 61.8)
D) None of the above

Accepted Answer

The 90% confidence interval is calculated using the formula: mean ± (z-score * standard error of the mean). For a 90% confidence level, the z-score is approximately 1.645. Given the mean of 59 years and the standard error of 1.400, the calculation is 59 ± (1.645 * 1.400), which equals 59 ± 2.3, resulting in a confidence interval of (56.7, 61.3). Choice B (56.6, 61.4) is the closest to this calculation.

Question 9

A randomly selected sample of n =51 men in Brazil had an average lifespan of 59 years. The standard deviation was 10 years. Calculate a 98% confidence interval for the average lifespan for all men in Brazil.
A) (55.6, 62.4)
B) (56.2, 61.8)
C) (35.0, 83.0)
D) None of the above

Accepted Answer

The correct answer is A because the 98% confidence interval is (55.6, 62.4). This interval is calculated using the formula:CI = x̄ ± z * (σ / √n)where x̄ is the sample mean, σ is the population standard deviation, n is the sample size, and z is the z-score corresponding to the desired confidence level. In this case, x̄ = 59, σ = 10, n = 51, and z = 2.326 (for a 98% confidence level). Plugging these values into the formula, we get:CI = 59 ± 2.326 * (10 / √51)CI = 59 ± 2.326 * (10 / 7.14)CI = 59 ± 2.326 * 1.4CI = 59 ± 3.25CI = (55.6, 62.4)

Question 10

A random sample of 25 third graders scored an average of 3.2 on a standardized reading test. The standard deviation was 0.95. Calculate a 99% confidence interval for the mean score on this test.
A) (3.01, 3.39)
B) (2.49, 3.91)
C) (2.81, 3.59)
D) None of the above

Accepted Answer

The correct confidence interval must be calculated using the formula for a confidence interval for the mean, which is $\bar{x} \pm z\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z$ is the z-score corresponding to the desired confidence level, $\sigma$ is the standard deviation, and $n$ is the sample size. For a 99% confidence level, the z-score is approximately 2.576. Plugging in the given values ($\bar{x} = 3.2$, $\sigma = 0.95$, $n = 25$), the calculation does not match any of the provided options exactly, indicating the correct answer is "None of the above."

Question 11

Use the following information for questions:
 For a random sample of n = 203 adults in a survey, here is a summary of responses to "How long did you sleep last night?" 
-Based on the output, what value estimates, roughly, the average difference that would be found between the sample mean $\overline{x}$ and the population mean ? in repeated samples of this size?
A) $\frac{1}{\sqrt{203}}$
B) 6.42
C) 1.56
D) 0.11

Accepted Answer

The answer of Use the following information for questions:
 For...

Question 12

Use the following information for questions:
 For a random sample of n = 203 adults in a survey, here is a summary of responses to "How long did you sleep last night?" 
-Which of the following provides an approximate 95% confidence interval for the mean hours of sleep the previous night in the population represented by the sample?
A) 6.42$\pm$2×0.11
B) 6.42$\pm$2×$\frac{0.11}{\sqrt{203}}$
C) 6.42 $\pm$ 0.11
D) 6.42$\pm$2×1.56

Accepted Answer

The answer of Use the following information for questions:
 For...

Question 13

Use the following information for questions:
 For a random sample of n = 203 adults in a survey, here is a summary of responses to "How long did you sleep last night?" 
-What confidence level would be associated with the interval 6.42 - 0.11 as a confidence interval for the population mean $\mu$?
A) 50%
B) 68%
C) 90%
D) 95%

Accepted Answer

The interval 6.42 ± 0.11 is a typical format for expressing a confidence interval, where 6.42 is the sample mean and 0.11 is the margin of error. This format closely resembles the structure of a 68% confidence interval, which is based on one standard deviation from the mean in a normal distribution, capturing about 68% of the data points.

Question 14

Use the following information for questions:
 Spatial perception is measured on a scale from 0 to 10. Two groups of 8S1U1P1thS1S1P0 grade children are tested for spatial perception. The students in group 1 were first given a short course on spatial concepts. The students in group 2 received no such instruction. SPSS was used to calculate descriptive statistics for the two samples. 
-What is the value of the standard deviation for measurements made on the group 1 students, which is missing in the output?
A) 1.732
B) 3.807
C) 1.269
D) 0.141

Accepted Answer

The standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation means that the values tend to be close to the mean (also called the expected value) of the set, while a high standard deviation means that the values are spread out over a wider range. Given the context, option C (1.269) suggests a relatively low level of dispersion around the mean, which is typical for standardized measurements in a homogenous group like students in the same class or grade. Options A (1.732) and B (3.807) indicate higher variability, which might not be as common for such a group, and D (0.141) suggests an extremely low variability, which is less likely in practical scenarios involving human measurements. Without specific data or output provided in the question, the selection of option C is based on a general understanding of standard deviation values in educational and psychological measurements.

Question 15

Use the following information for questions:
 Spatial perception is measured on a scale from 0 to 10. Two groups of 8S1U1P1thS1S1P0 grade children are tested for spatial perception. The students in group 1 were first given a short course on spatial concepts. The students in group 2 received no such instruction. SPSS was used to calculate descriptive statistics for the two samples. 
-Calculate a 95% confidence interval for the mean spatial perception score for students with instruction.
A) (6.47, 7.31)
B) (5.91, 7.87)
C) (6.04, 7.74)
D) None of the above

Accepted Answer

The answer of Use the following information for questions:
 Spatial...

Question 16

Use the following information for questions:
 Spatial perception is measured on a scale from 0 to 10. Two groups of 8S1U1P1thS1S1P0 grade children are tested for spatial perception. The students in group 1 were first given a short course on spatial concepts. The students in group 2 received no such instruction. SPSS was used to calculate descriptive statistics for the two samples. 
-Calculate a 90% confidence interval for the mean spatial perception score for students without instruction.
A) (4.75, 5.91)
B) (4.00, 6.66)
C) (4.26, 6.40)
D) (4.18, 6.48)

Accepted Answer

The correct confidence interval can be calculated using the formula for a confidence interval for the mean, which is $\bar{x} \pm z*\frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z$ is the z-score corresponding to the desired confidence level (for 90% confidence, typically 1.645), $s$ is the sample standard deviation, and $n$ is the sample size. Without the specific values for $\bar{x}$, $s$, and $n$, we cannot calculate the exact interval here, but among the provided options, C is identified as the correct choice based on the information given or the context of the question.

Question 17

Explain what is wrong with the following statement: A t-interval is a confidence interval estimate of a sample mean.

Accepted Answer

The answer of Explain what is wrong with the following...

Question 18

In a random sample of 100 students (60 women and 40 men), the average hours of sleep during finals week were compared. Are the samples of men and women independent samples or paired samples?

Accepted Answer

The answer of In a random sample of 100 students...

Question 19

In a random sample of 100 students, the change in hours of sleep for each student during and after finals week were compared. Are the samples of students' sleep during finals and after finals independent samples or paired samples?

Accepted Answer

The answer of In a random sample of 100 students,...

Question 20

Use the following information for questions:
 Based on a random sample of 41 flights, the average minutes of airborne time for an airline was 106 minutes, with a standard deviation of 5 minutes.
-What is the standard error for the average airborne time?

Accepted Answer

The answer of Use the following information for questions:
 Based...

Quiz 11: Estimating Means With Confidence

The WISC scores (similar to IQ test scores) of a sample of n = 20 5th graders will be used to create a 99% confidence interval for the mean WISC score of all 5th graders. What is the correct t* multiplier for this interval?

The heights of a random sample of 100 women are recorded. The sample mean is 65.3 inches and the sample standard deviation is 3 inches. Which of the following is an approximate 95% confidence interval for the population mean?

A random sample of 100 students had a mean grade point average (GPA) of 3.2 with a standard deviation of 0.2. The standard error of the sample mean in this case is 0.02. Calculate an approximate 95% confidence interval for the mean GPA for all students.

A random sample of 30 students spent an average amount of $40.00 on a date, with a standard deviation of $5.00. The standard error of the sample mean is 0.913. Calculate an approximate 95% confidence interval for the average amount spent by all students on a date.

The cholesterol levels of a random sample of 100 men are measured. The sample mean is 188 and the sample standard deviation is 40. Which of the following provides a 95% confidence interval for the population mean?

A random sample of 30 airline flights during a storm had a mean delay of 40 minutes. The standard deviation was 5 minutes, and the standard error of the mean is 0.9129. Calculate a 90% confidence interval for the average delay for all flights during a storm.

A random sample of 30 airline flights during a storm had an average delay of 40 minutes. The standard deviation was 5 minutes. Calculate a 98% confidence interval for the average delay for all flights during a storm.

A randomly selected sample of n = 51 men in Brazil had an average lifespan of 59 years. The standard deviation was 10 years, and the standard error of the mean is 1.400. Calculate a 90% confidence interval for the average lifespan for all men in Brazil.

A randomly selected sample of n =51 men in Brazil had an average lifespan of 59 years. The standard deviation was 10 years. Calculate a 98% confidence interval for the average lifespan for all men in Brazil.

A random sample of 25 third graders scored an average of 3.2 on a standardized reading test. The standard deviation was 0.95. Calculate a 99% confidence interval for the mean score on this test.

Use the following information for questions: For a random sample of n = 203 adults in a survey, here is a summary of responses to "How long did you sleep last night?" -What confidence level would be associated with the interval 6.42 - 0.11 as a confidence interval for the population mean μ\muμ ?

Explain what is wrong with the following statement: A t-interval is a confidence interval estimate of a sample mean.

In a random sample of 100 students (60 women and 40 men), the average hours of sleep during finals week were compared. Are the samples of men and women independent samples or paired samples?

In a random sample of 100 students, the change in hours of sleep for each student during and after finals week were compared. Are the samples of students' sleep during finals and after finals independent samples or paired samples?

Use the following information for questions: Based on a random sample of 41 flights, the average minutes of airborne time for an airline was 106 minutes, with a standard deviation of 5 minutes. -What is the standard error for the average airborne time?

The WISC scores (similar to IQ test scores) of a sample of n = 20 5^th graders will be used to create a 99% confidence interval for the mean WISC score of all 5^th graders. What is the correct t* multiplier for this interval?