Question 1

The weights of aluminum castings produced by a process are normally distributed.A random sample of 5 castings is selected; the sample mean weight is 2.21 kilograms; and the sample standard deviation is 0.12 kilograms.The 98% confidence interval for the population mean casting weight is ___.
A)1.76 to 2.66
B)2.01 to 2.41
C)2.08 to 2.34
D)1.93 to 2.49
E)2.49 to 2.67

Accepted Answer

The answer of The weights of aluminum castings produced by...

Question 2

The lengths of steel rods produced by a shearing process are normally distributed.A random sample of 10 rods is selected; the sample mean length is 119.05 centimetres; and the sample standard deviation is 0.10 centimetres.The 90% confidence interval for the population mean rod length is ___.
A)118.99 to 119.11
B)118.87 to 119.23
C)119.00 to 119.10
D)118.89 to 119.21
E)119.21 to 123.87

Accepted Answer

The 90% confidence interval is calculated using the sample mean, sample standard deviation, and the t-distribution for a sample size of 10. The correct interval is 118.99 to 119.11.

Question 3

Brian Vanecek, VP of Operations at Portland Trust Bank, is evaluating the service level provided to walk-in customers.Accordingly, his staff recorded the waiting times for 64 randomly selected walk-in customers and determined that their mean waiting time was 15 minutes.Assume that the population standard deviation is 4 minutes.The 95% confidence interval for the population mean of waiting times is ___.
A)14.02 to 15.98
B)7.16 to 22.84
C)14.06 to 15.94
D)8.42 to 21.58
E)19.80 to 23.65

Accepted Answer

The answer of Brian Vanecek, VP of Operations at Portland...

Question 4

A random sample of 64 items is selected from a population of 400 items.The sample mean is 200.The population standard deviation is 48.From this data, a 95% confidence interval to estimate the population mean can be computed as ___.
A)189.21 to 210.79
B)188.24 to 211.76
C)190.13 to 209.87
D)190.94 to 209.06
E)193.45 to 211.09

Accepted Answer

The answer of A random sample of 64 items is...

Question 5

The table t value associated with the upper 5% of the t distribution and 14 degrees of freedom is ___.
A)2.977
B)2.624
C)2.145
D)1.761

Accepted Answer

The answer of The table t value associated with the...

Question 6

Suppose a random sample of 36 is selected from a population with a standard deviation of 12.If the sample mean is 98, the 99% confidence interval to estimate the population mean is ___.
A)94.08 to 101.92
B)92.85 to 103.15
C)97.35 to 98.65
D)93.34 to 102.66
E)90.20 to 105.00

Accepted Answer

Use the formula: 
CI = x̄ ± z* (σ/√n)
where x̄ = 98, σ = 12, n = 36, and z* for a 99% confidence interval is 2.576. 
Plugging in the values, we get: 
CI = 98 ± 2.576 * (12/√36) 
CI = 98 ± 6.512 
CI = (98 - 6.512, 98 + 6.512) 
CI = (92.85, 103.15) 
Therefore, the 99% confidence interval to estimate the population mean is (92.85, 103.15), which is closest to option B.

Question 7

James Desreumaux, VP of Human Resources of Maritime Boat Manufacturing (MBM), is reviewing the employee training programs of MBM factories.His staff randomly selected personnel files for 100 workers in New Brunswick and determined that their mean training time was 25 hours.Assume that the population standard deviation is 5 hours.The 88% confidence interval for the population mean of training times is ___.
A)17.25 to 32.75
B)24.23 to 25.78
C)24.42 to 25.59
D)19.15 to 30.85
E)21.00 t0 32.00

Accepted Answer

The answer of James Desreumaux, VP of Human Resources of...

Question 8

The z value associated with a two-sided 88% confidence interval is ___.
A)1.28
B)1.555
C)1.17
D)0.88
E)1.90

Accepted Answer

To find the z value associated with a confidence interval, we need to use the standard normal distribution table. Since this is a two-sided confidence interval, we need to split the remaining area (100% - 88% = 12%) into two equal parts, giving us 6% on each tail.

Looking at the standard normal distribution table, the closest value to 0.06 in the body of the table is 0.0611, and the corresponding z-value is 1.56. Therefore, the z value associated with a two-sided 88% confidence interval is approximately 1.555.

Question 9

Brian Vanecek, VP of Operations at Portland Trust Bank, is evaluating the service level provided to walk-in customers.Accordingly, his staff recorded the waiting times for 64 randomly selected walk-in customers and determined that their mean waiting time was 15 minutes.Assume that the population standard deviation is 4 minutes.The 90% confidence interval for the population mean of waiting times is ___.
A)14.27 to 15.73
B)14.18 to 15.82
C)9.88 to 20.12
D)13.86 to 16.14
E)18.12 to 19.87

Accepted Answer

The answer of Brian Vanecek, VP of Operations at Portland...

Question 10

Suppose a random sample of 36 is selected from a population with a standard deviation of 12.If the sample mean is 98, the 90% confidence interval for the population mean is ___.
A)94.71 to 101.29
B)97.45 to 98.55
C)94.08 to 101.92
D)97.35 to 98.65
E)95.00 to 105.00

Accepted Answer

The 90% confidence interval is calculated using the formula: sample mean ± (Z * (population standard deviation / sqrt(sample size))). For a 90% confidence level, Z is approximately 1.645. Thus, the interval is 98 ± (1.645 * (12 / sqrt(36))) = 98 ± 3.29, which is 94.71 to 101.29.

Question 11

The normal distribution is used to test about a population mean for large samples if the population standard deviation is known."Large" is usually defined as ___.
A)at least 10
B)at least 5% of the population size
C)at least 30
D)at least 12
E)at least 100

Accepted Answer

The generally accepted rule of thumb is that a sample size of at least 30 is considered "large enough" to use the normal distribution to test about a population mean, regardless of whether the population standard deviation is known or not. Therefore, choice C is the best answer. Choices A, B, D, and E do not align with this rule of thumb. The rule of thumb is to use the normal distribution for sample sizes of at least 30 or greater, regardless of the population standard deviation. Choice A, 10, is too small of a sample size. Choice B, 5% of the population size, is not a reliable criterion for determining sample size. Choices D and E are arbitrary numbers and do not align with commonly accepted guidelines for sample size.

Question 12

James Desreumaux, VP of Human Resources of Maritime Boat Manufacturing (MBM), is reviewing the employee training programs of MBM factories.His staff randomly selected personnel files for 100 workers in New Brunswick and determined that their mean training time was 25 hours.Assume that the population standard deviation is 5 hours.The 92% confidence interval for the population mean of training times is ___.
A)16.25 to 33.75
B)24.30 to 25.71
C)17.95 to 32.05
D)24.13 to 25.88
E)24.45 to 27.32

Accepted Answer

The confidence interval is calculated using the formula: mean ± (Z * (σ/√n)). For a 92% confidence level, the Z-score is approximately 1.75. Using the given mean of 25, standard deviation of 5, and sample size of 100, the interval is 25 ± (1.75 * (5/√100)), which results in 24.13 to 25.88.

Question 13

A researcher is interested in estimating the mean value for a population.She takes a random sample of 17 items and computes a sample mean of 224 and a sample standard deviation of 32.She decides to construct a 98% confidence interval to estimate the mean.The degrees of freedom associated with this problem are ___.
A)18
B)17
C)16
D)15
E)20

Accepted Answer

The degrees of freedom associated with this problem is n-1, where n is the sample size. Therefore, degrees of freedom = 17-1 = 16.

Question 14

The z value associated with a two-sided 80% confidence interval is ___.
A)1.645
B)1.28
C)0.84
D)0.29
E)2.00

Accepted Answer

To find the z value associated with an 80% confidence interval, we first find the area in the middle 80% of the standard normal distribution table, which is 0.8.

From the table, we can find the z score that gives an area of 0.8 to the left of it, which is 1.28.

However, since this is a two-sided confidence interval, we need to find the z values that correspond to the middle 80% of the distribution, which includes 40% in each tail.

To find the z value for the lower tail, we find the area to the left of the middle 90%, which is (1-0.8)/2 = 0.1. From the table, the z score that gives an area of 0.1 to the left of it is -1.28.

To find the z value for the upper tail, we find the area to the right of the middle 90%, which is also 0.1. From the table, the z score that gives an area of 0.1 to the right of it is 1.28.

Thus, the z values for the 80% confidence interval are -1.28 and +1.28, and the answer is B. A two-sided 80% confidence interval leaves 10% in each tail, which corresponds to a z-score of 1.28.

Question 15

The lengths of steel rods produced by a shearing process are normally distributed.A random sample of 10 rods is selected; the sample mean length is 119.05 centimetres; and the sample standard deviation is 0.10 centimetres.The 95% confidence interval for the population mean rod length is ___.
A)118.99 to 119.11
B)118.82 to 119.28
C)118.98 to 119.12
D)118.85 to 119.25
E)119.89 to 122.12

Accepted Answer

We need to use the formula for a confidence interval for a population mean:

CI = X̄ ± Zα/2 * (σ/√n)

Where:
X̄ = sample mean 
Zα/2 = the critical value from the standard normal distribution for the desired level of confidence (0.95 in this case; so α = 0.025 and we can look up the critical value to be 1.96) 
σ = population standard deviation 
n = sample size

Substituting the given values, we get:

CI = 119.05 ± 1.96 * (0.10/√10) 
CI = 119.05 ± 0.039 
CI = (118.98, 119.12)

Therefore, the 95% confidence interval for the population mean rod length is 118.98 to 119.12 centimeters, which corresponds to choice C.

Question 16

Suppose a random sample of size 64 is selected from a population yielding a sample mean of 26.The population standard deviation is 4.From this information, the 90% confidence interval to estimate the population mean can be computed to be ___.
A)25.36 to 26.64
B)25.92 to 26.08
C)25.18 to 26.82
D)25.90 to 26.10
E)26.00 to 27.80

Accepted Answer

The answer of Suppose a random sample of size 64...

Question 17

The table t value associated with the upper 5% of the t distribution and 12 degrees of freedom is ___.
A)2.179
B)1.782
C)1.356
D)3.055
E)3.330

Accepted Answer

Using a t-table or a calculator, we can find that the t value associated with the upper 5% of the t distribution and 12 degrees of freedom is approximately 1.782. Therefore, the correct answer is B.

Question 18

The table t value associated with the upper 10% of the t distribution and 23 degrees of freedom is ___.
A)1.319
B)1.714
C)2.069
D)1.321
E)2.332

Accepted Answer

The answer of The table t value associated with the...

Question 19

James Desreumaux, VP of Human Resources of Maritime Boat Manufacturing (MBM), is reviewing the employee training programs of MBM factories.His staff randomly selected personnel files for 100 workers in New Brunswick and determined that their mean training time was 25 hours.Assume that the population standard deviation is 5 hours.The 95% confidence interval for the population mean of training times is ___.
A)15.20 to 34.80
B)24.18 to 25.82
C)24.02 to 25.98
D)16.78 to 33.23
E)23.32 to 35.46

Accepted Answer

Since the sample size is greater than 30 and we know the population standard deviation, we can use a z-test. The z-score for a 95% confidence level is 1.96. 
The margin of error can be calculated as:
margin of error = z-score * (standard deviation / square root of sample size)
margin of error = 1.96 * (5 / square root of 100) = 1
Therefore, the 95% confidence interval for the population mean is:
sample mean ± margin of error
25 ± 1 = (24, 26)
So the answer is C) 24.02 to 25.98.

Question 20

A random sample of 64 items is selected from a population of 400 items.The sample mean is 200.The population standard deviation is 48.From this data, a 90% confidence interval to estimate the population mean can be computed as ___.
A)189.21 to 210.79
B)188.24 to 211.76
C)190.13 to 209.87
D)190.94 to 209.06
E)193.45 to 211.09

Accepted Answer

The answer of A random sample of 64 items is...

Quiz 8: Statistical Inference: Estimation for Single Populations

The weights of aluminum castings produced by a process are normally distributed.A random sample of 5 castings is selected; the sample mean weight is 2.21 kilograms; and the sample standard deviation is 0.12 kilograms.The 98% confidence interval for the population mean casting weight is ___.

The lengths of steel rods produced by a shearing process are normally distributed.A random sample of 10 rods is selected; the sample mean length is 119.05 centimetres; and the sample standard deviation is 0.10 centimetres.The 90% confidence interval for the population mean rod length is ___.

A random sample of 64 items is selected from a population of 400 items.The sample mean is 200.The population standard deviation is 48.From this data, a 95% confidence interval to estimate the population mean can be computed as ___.

The table t value associated with the upper 5% of the t distribution and 14 degrees of freedom is ___.

Suppose a random sample of 36 is selected from a population with a standard deviation of 12.If the sample mean is 98, the 99% confidence interval to estimate the population mean is ___.

The z value associated with a two-sided 88% confidence interval is ___.

Suppose a random sample of 36 is selected from a population with a standard deviation of 12.If the sample mean is 98, the 90% confidence interval for the population mean is ___.

The normal distribution is used to test about a population mean for large samples if the population standard deviation is known."Large" is usually defined as ___.

The z value associated with a two-sided 80% confidence interval is ___.

The lengths of steel rods produced by a shearing process are normally distributed.A random sample of 10 rods is selected; the sample mean length is 119.05 centimetres; and the sample standard deviation is 0.10 centimetres.The 95% confidence interval for the population mean rod length is ___.

Suppose a random sample of size 64 is selected from a population yielding a sample mean of 26.The population standard deviation is 4.From this information, the 90% confidence interval to estimate the population mean can be computed to be ___.

The table t value associated with the upper 5% of the t distribution and 12 degrees of freedom is ___.

The table t value associated with the upper 10% of the t distribution and 23 degrees of freedom is ___.

A random sample of 64 items is selected from a population of 400 items.The sample mean is 200.The population standard deviation is 48.From this data, a 90% confidence interval to estimate the population mean can be computed as ___.