Question 1

n=10,&#10; =13.0,  s = 4.5.Find a 95% confidence interval for the mean.&#10;A)(9.80, 16.20)&#10;B)(9.83, 16.17)&#10;C)(10.39, 15.61)&#10;D)(9.83, 16.20)&#10;E)(9.78, 16.22)

Accepted Answer

(9.78, 16.22)

Question 2

A small business ships specialty homemade candies to anywhere in the world.Past records indicate that the weight of orders is normally distributed.Suppose a random sample of 16 orders is selected and each is weighed.The sample mean was found to be 110 grams with a standard deviation of 14 grams.Which of the following statements is true?

A)The sampling distribution for the sample mean follows the t-distribution with 16 degrees of freedom.
B)The sampling distribution for the sample mean follows the t-distribution with 15 degrees of freedom.
C)The mean of the sampling distribution for the sample mean is 110 grams.
D)The sampling distribution for the sample mean is Normal with a mean of 110 grams and standard deviation of 14 grams.
E)The standard error is 14 grams.

Accepted Answer

The sampling distribution for the sample mean follows the t-distribution with n-1 degrees of freedom, where n is the sample size. Here, n = 16, so the degrees of freedom is 15.

Question 3

In a metal fabrication process, metal rods are produced to a specified target length of 15 feet.Suppose that the lengths are normally distributed.A quality control specialist collects a random sample of 16 rods and finds the sample mean length to be 14.8 feet and a standard deviation of 0.65 feet.The standard error of the mean is ________________________ .

A)0.65 ft
B)0.346 ft
C)0.1625 ft
D)0.0098 ft
E)1.625 ft

0.1625 ft

Accepted Answer

The standard error of the mean (SEM) is calculated by dividing the standard deviation by the square root of the sample size: SEM = 0.65 / sqrt(16) = 0.65 / 4 = 0.1625 feet.

Question 4

Analysis of a random sample of 250 Illinois nurses produced a 95% confidence interval for the mean annual salary of $42,846 <$\mu$(Nurse Salary) < $49,686&#10;A)About 95% of the nurses surveyed earn between $42,846 and $49,686.&#10;B)If we took many random samples of Illinois nurses, about 95% of them would produce this confidence interval.&#10;C)We are 95% confident that the interval from $42,846 to $49,686 contains the true mean of all Illinois nurses.&#10;D)About 95% of Illinois nurses earn between $42,846 and $49,686.&#10;E)We are 95% confident that the average nurse salary in the U.S.is between $42,846 and $49,686.

Accepted Answer

A confidence interval provides a range of values within which we can be a certain percentage confident (in this case, 95%) that the true mean (average) of the population falls. Therefore, option C is correct because it correctly interprets the confidence interval as an estimate of where the true mean salary of all Illinois nurses lies, not the salaries of individual nurses or the outcome of repeated samples.

Question 5

Top management at a large software company wishes to estimate the average number of hours its firm's professional employees volunteer in the local community.Based on past similar studies, the standard deviation was found to be 2.22 hours.If top management wants to estimate the average number of hours volunteered per month by their professional staff to within one hour with 99% confidence, how many randomly selected professional employees would they need to sample?

A)19
B)25
C)54
D)44
E)33

Accepted Answer

The answer of Top management at a large software company...

Question 6

A pharmaceutical company wants to answer the question whether it takes longer than 45 seconds for a drug in pill form to dissolve in the gastric juices of the stomach.A sample was taken from patients taken the given drug in pill form and times for the pills to be dissolved were measured.b.State assumptions of the test.

A)The subjects must be part of a medical study group investigating the drug.
B)The subjects in the sample must be randomly selected from a population.
C)B and D are required conditions.
D)The sample data must come from a normally distributed population of observations for the variable under study.
E)None of the above.

Accepted Answer

The assumptions for the test include that the subjects in the sample must be randomly selected from a population (B) and that the sample data must come from a normally distributed population of observations for the variable under study (D).

Question 7

Grandma Gertrude's Chocolates, a family owned business, has an opportunity to supply its product for distribution through a large coffee house chain.However, the coffee house chain has certain specifications regarding cacao content as it wishes to advertise the health benefits (antioxidants) of the chocolate products it sells.In order to determine the mean % cacao in its dark chocolate products, quality inspectors sample 36 pieces.They find a sample mean of 55% with a standard deviation of 4%.
What is the margin of error at 95% confidence?

A)1.13%
B)1.36%
C)0.67%
D)4%
E)1.67%

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Accepted Answer

The answer of Grandma Gertrude's Chocolates, a family owned business,...

Question 8

A manufacturer of cheese filled ravioli supplies a pizza restaurant chain.Based on data collected from its automatic filling process, the amount of cheese inserted into the ravioli is normally distributed.To make sure that the automatic filling process is on target, quality control inspectors take a sample of 25 ravioli.The correct value of t^* to construct a 98% confidence interval for the true mean amount of cheese filling is ________________________ .

A)2.492
B)3.467
C)2.797
D)2.064
E)3.745

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Accepted Answer

The margin of error at a 90% confidence level can be calculated using the formula: $ME = Z \times \frac{\sigma}{\sqrt{n}}$, where $Z$ is the Z-score corresponding to the desired confidence level (for 90% confidence, $Z \approx 1.645$), $\sigma$ is the standard deviation of the sample (1.5 grams), and $n$ is the sample size (25). Plugging in the values: $ME = 1.645 \times \frac{1.5}{\sqrt{25}} = 1.645 \times 0.3 = 0.4935$, which rounds to approximately 0.5 grams, but since the closest option provided is 0.5133 grams, that's the correct choice despite the slight discrepancy in rounding.

Question 9

A sociologist develops a test to measure attitudes about public transportation, and 27 randomly selected subjects are given the test.Their mean score is 76.2 and their standard deviation is 21.4.Construct the 95% confidence interval for the mean score of all such subjects ________________________ .
Please fix alignment

A)(64.2, 88.2)
B)(67.7, 84.7)
C)(64.2, 83.2)
D)(69.2, 83.2)
E)(74.6, 77.8)

Accepted Answer

To construct a 95% confidence interval for the mean score of all such subjects, we use the formula for the confidence interval for a mean, which is $\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, $n$ is the sample size, and $t_{\alpha/2}$ is the t-score corresponding to the desired confidence level and degrees of freedom (df = n - 1). Given $\bar{x} = 76.2$, $s = 21.4$, and $n = 27$, and knowing that the t-score for a 95% confidence interval with 26 degrees of freedom is approximately 2.056 (this value might slightly vary depending on the source, but it's close to 2 for large sample sizes), we calculate the margin of error and then the confidence interval. The correct calculation leads to the interval (67.7, 84.7), which matches choice B.

Question 10

A manufacturer of cheese filled ravioli supplies a pizza restaurant chain.Based on data collected from its automatic filling process, the amount of cheese inserted into the ravioli is normally distributed.To make sure that the automatic filling process is on target, quality control inspectors take a sample of 25 ravioli and measure the weight of cheese filling.They find the 99% confidence interval of 14.16 to 15.84 grams.Which of the following is the correct interpretation?

A)We are 99% confident that the mean weight of cheese filling in all ravioli made by this process is between 14.16 and 15.84 grams.
B)99% of all ravioli made by this process will have cheese filling weights between 14.16 and 15.84 grams.
C)The weight of cheese filling in the ravioli is between 14.16 and 15.84 grams 99% of the time.
D)All of the above.
E)None of the above.

Accepted Answer

The correct interpretation of a confidence interval is that we are a certain percent confident (in this case, 99%) that the true mean (average) of the population (in this case, the mean weight of cheese filling in all ravioli made by this process) falls within the interval. Choices B and C misinterpret the concept of a confidence interval, suggesting it applies to individual observations or to the frequency of observations within the interval, which is incorrect.

Question 11

A small business ships specialty homemade candies to anywhere in the world.Past records indicate that the weight of orders is normally distributed.Suppose a random sample of 16 orders is selected and each is weighed.The sample mean was found to be 110 grams with a standard deviation of 14 grams.The 90% confidence interval for the true mean weight of orders is ________________________ .

A)103.87 to 116.14 grams.
B)86.046 to 133.954 grams.
C)99.5 to 120.5 grams.
D)102.55 to 117.45 grams.
E)103.14 to 116.86 grams.

Accepted Answer

The answer of A small business ships specialty homemade candies...

Question 12

A credit union took a random sample of 40 accounts and yielded the following 90% confidence interval for the mean checking account balance at the institution:
$2183 < $\mu$ (balance) < $3828.Which of the following is the correct interpretation?

A)If we took random samples of checking accounts at this credit union, about nine out of ten of them would produce this confidence interval.
B)About 9 out of 10 people have a checking account balance between $2183 and $3828.
C)We are 90% confident that the mean checking account balance at this credit union is between $2183 and $3828, based on this sample.
D)We are 90% sure that the mean balance for checking accounts in the sample was between $2183 and $3828.
E)We are 90% confident that the mean checking account balance in the U.S.is between $2183 and $3828.

Accepted Answer

The answer of A credit union took a random sample...

Question 13

A manufacturer of cheese filled ravioli supplies a pizza restaurant chain.Based on data collected from its automatic filling process, the amount of cheese inserted into the ravioli is normally distributed.To make sure that the automatic filling process is on target, quality control inspectors take a sample of 25 ravioli.The correct value of t^* to construct a 98% confidence interval for the true mean amount of cheese filling is ________________________ .

A)2.492
B)3.467
C)2.797
D)2.064
E)3.745

Accepted Answer

The answer of A manufacturer of cheese filled ravioli supplies...

Question 14

Grandma Gertrude's Chocolates, a family owned business, has an opportunity to supply its product for distribution through a large coffee house chain.However, the coffee house chain has certain specifications regarding cacao content as it wishes to advertise the health benefits (antioxidants) of the chocolate products it sells.In order to determine the mean % cacao in its dark chocolate products, quality inspectors sample 36 pieces.They find a sample mean of 55% with a standard deviation of 4%.
The 90% confidence interval for the true mean % cacao is ________________________ .

A)53.87% to 56.13%.
B)51% to 59%.
C)53.33% to 56.67%.
D)53.64% to 56.36%
E)54.33% to 55.67%.

Accepted Answer

The answer of Grandma Gertrude's Chocolates, a family owned business,...

Question 15

In a metal fabrication process, metal rods are produced to a specified target length of 15 feet.Suppose that the lengths are normally distributed.A quality control specialist collects a random sample of 16 rods and finds the sample mean length to be 14.8 feet and a standard deviation of 0.65 feet.The 95% confidence interval for the true mean length of rods produced by this process is ________________________ .

A)14.544 to 15.056 ft
B)14.345 to 15.255 ft
C)13.912 to 15.688 ft
D)14.454 to 15.146 ft
E)13.834 to 15.766 ft

Accepted Answer

The answer of In a metal fabrication process, metal rods...

Question 16

Grandma Gertrude's Chocolates, a family owned business, has an opportunity to supply its product for distribution through a large coffee house chain.However, the coffee house chain has certain specifications regarding cacao content as it wishes to advertise the health benefits (antioxidants) of the chocolate products it sells.In order to determine the mean % cacao in its dark chocolate products, quality inspectors sample 36 pieces.They find a sample mean of 55% with a standard deviation of 4%.
What is the margin of error at 95% confidence?

A)1.13%
B)1.36%
C)0.67%
D)4%
E)1.67%

Accepted Answer

The answer of Grandma Gertrude's Chocolates, a family owned business,...

Question 17

A large software development firm recently relocated its facilities.Top management is interested in fostering good relations with their new local community and has encouraged their professional employees to engage in local service activities.They wish to determine the average number of hours the firm's professionals volunteer per month.They find the 95% confidence interval of 14.62 to 15.78.Which of the following is the correct interpretation?

A)We are 95% confident that the average number of hours per month volunteered by professionals employed with this firm is between 14.62 and 15.78.
B)95% of all firm's professional employees volunteer between 14.62 and 15.78 hours per month.
C)The number of hours per month volunteered by the firm's professional employees is between 14.62 and 15.78 95% of the time.
D)All of the above.
E)None of the above.

Accepted Answer

The answer of A large software development firm recently relocated...

Question 18

A small business ships specialty homemade candies to anywhere in the world.Past records indicate that the weight of orders is normally distributed.Suppose a random sample of 16 orders is selected and each is weighed.The sample mean was found to be 110 grams with a standard deviation of 14 grams.The standard error of the mean is ________________________ .

A)14 grams
B)0.875 grams
C).35 grams
D)3.5 grams
E)27.5 grams

Accepted Answer

The answer of A small business ships specialty homemade candies...

Question 19

A manufacturer of cream filled donuts wants to make sure that its automatic filling process is on target.Based on a sample of 25 donuts the mean weight of cream filling is estimated to be 15 grams with a standard deviation of 1.5 grams.However, the quality control inspectors wish to estimate the mean weight of cream filling more accurately so that they can be 99% confident that it is within 0.25 grams of the true mean.How many donuts should they sample?

A)150
B)543
C)212
D)410
E)239

Accepted Answer

The answer of A manufacturer of cream filled donuts wants...

Question 20

In a metal fabrication process, metal rods are produced to a specified target length of 15 feet.Suppose that the lengths are normally distributed.A quality control specialist collects a random sample of 16 rods and finds the sample mean length to be 14.8 feet and a standard deviation of 0.65 feet.Which of the following statements is true?

A)The sampling distribution for the sample mean follows the t-distribution with 16 degrees of freedom.
B)The sampling distribution for the sample mean follows the t-distribution with 15 degrees of freedom.
C)The mean of the sampling distribution for the sample mean is 14.8 feet.
D)The sampling distribution for the sample mean is Normal with a mean of 14.8 feet and standard deviation of .65 feet.
E)The standard error is 0.65 feet.

Accepted Answer

The answer of In a metal fabrication process, metal rods...

Deck 11: Confidence Intervals for Means