Question 1

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 a sample mean weight of 15 grams with a standard deviation of
1)5 grams. What is the margin of error at 90% confidence?
A) 0.3 grams
B) 0.5133 grams
C) 0.8391 grams
D) 1.5 grams
E) 0.06 grams

Accepted Answer

The margin of error (ME) can be calculated using the formula:
ME = z*(S/sqrt(n))
Where z is the z-score for the desired confidence level (90% = 1.645), S is the sample standard deviation (1.5 grams), and n is the sample size (25).
ME = 1.645*(1.5/sqrt(25)) = 0.5133 grams
Therefore, the margin of error at 90% confidence is 0.5133 grams. Choice B is the correct answer.

Question 2

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. State the hypotheses to test this question. A) $\mathrm { H } _ { 0 } : \mu = 45 \mathrm { sec } ; \mathrm { H } _ { \mathrm { A } } : \mu > 45 \mathrm { sec }$ B) $\mathrm { H } _ { 0 } : \mu > 45 \mathrm { sec } ; \mathrm { H } _ { \mathrm { A } } : \mu \leq 45 \mathrm { sec }$ C) $\mathrm { H } _ { 0 } : \mu = 45 \mathrm { sec } ; \mathrm { H } _ { \mathrm { A } } : \mu \leq 45 \mathrm { sec }$ D) $H _ { 0 } : \mu \geq 45 \mathrm { sec } ; H _ { d } : \mu < 45 \mathrm { sec }$ E) $\mathrm { H } _ { 0 } : \mu < 45 \mathrm { sec } ; \mathrm { H } _ { \mathrm { A } } : \mu = 45 \mathrm { sec }$

Accepted Answer

The null hypothesis should state that the mean time of dissolution is 45 seconds, while the alternative hypothesis should state that it takes longer than 45 seconds. Therefore, the correct choice is A. B, C, D, and E have the alternative hypothesis in the wrong direction or incorrect statement.

Question 3

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%.
A. Describe the sampling distribution for the sample mean.
b. What is the standard error?
c. What is the margin of error for 90% confidence?
d. What is the margin of error for 95% confidence?
e. Based on the sample results, find the 90% confidence interval and interpret.
f. Based on the sample results, find the 95% confidence interval and interpret.
g. For a more accurate determination of the mean weight, the quality control inspectors
wish to estimate it within 1% with 95% confidence. How many pieces of dark chocolate
should they sample?

Accepted Answer

a. Describe the sampling distribution for the sample mean.
The sampling distribution for the sample mean can be modeled using the $t$-distribution with 35 degrees of freedom.
b. What is the standard error?
$$S E ( \bar { y } ) = \frac { s } { \sqrt { n } } = \frac { 4 } { \sqrt { 36 } } = 0.67 \%$$ c. What is the margin of error for $90 \%$ confidence?
$$\dot { t } _ { n - 1 } ^ { * } \times S E ( \bar { y } ) = 1.690 \times 0.67 = 1.13 \%$$
d. What is the margin of error for $95 \%$ confidence?
$$\dot { t } _ { n - 1 } ^ { * } \times S E ( \bar { y } ) = 2.030 \times 0.67 = 1.36 \%$$
e. Based on the sample results, find the $90 \%$ confidence interval and interpret.
$$\bar { y } \pm t _ { n - 1 } ^ { * } \times S E ( \bar { y } ) = 55 \pm 1.13$$
The $90 \%$ confidence interval for true $\%$ cacao is $53.87 \%$ to $56.13 \%$. We are $90 \%$ confident that the mean percentage of cacao in Grandma Gertrude's dark chocolate is between $53.87 \%$ and $56.13 \%$.
f. Based on the sample results, find the $95 \%$ confidence interval and interpret.
$$\bar { y } \pm t _ { n - 1 } ^ { * } \times S E ( \bar { y } ) = 55 \pm 1.36$$
The $95 \%$ confidence interval for true $\%$ cacao is $53.64 \%$ to $56.36 \%$. We are $95 \%$ confident that the mean percentage of cacao in Grandma Gertrude's dark chocolate is between $53.64 \%$ and $56.36 \%$.
g. For a more accurate determination of the mean weight, the quality control inspectors wish to estimate it within $1 \%$ with $95 \%$ confidence. How many pieces of dark chocolate should they sample?
$$n = \frac { ( 1.96 ) ^ { 2 } ( 4 ) ^ { 2 } } { 1 ^ { 2 } }$$
$n = 61.46$ or 62 pieces of chocolate

Question 4

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.
A random sample of 24 professionals reported the following number of hours: $$\begin{array} { | l | l | l | l | l | l | l | l | l | l | l | l | } 
\hline 12 & 13 & 14 & 14 & 15 & 15 & 15 & 16 & 16 & 16 & 16 & 16 \
\hline 17 & 17 & 17 & 18 & 18 & 18 & 18 & 19 & 19 & 19 & 20 & 21 \
\hline
\end{array}$$ 
A. Based on the sample results, find the 95% confidence interval and interpret.
b. For a more accurate determination, 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?
c. Suppose 40 professional employees are randomly selected. This sample yields a mean
of 15.2 hours and a standard deviation of 1.8 hours. Find a 95% confidence interval and
interpret.

Accepted Answer

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

Question 5

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 8 patients taking the given drug in pill form and times for the pills
To be dissolved were measured. The mean was 45.212 sec for the sample data with a
Standard error of 0.580. Determine the P-value for this test.
A) 0.181
B) 0.724
C) 0.649
D) 0.362
E) 0.366

Accepted Answer

The answer of A pharmaceutical company wants to answer the...

Question 6

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

For a 98% confidence interval with a sample size of 25 (degrees of freedom = 24), the t* value from the t-distribution table is approximately 2.492.

Question 7

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 a sample mean weight of 15 grams with a standard deviation of
1.5 grams.
A. Describe the sampling distribution for the sample mean.
b. What is the standard error?
c. What is the margin of error for 99% confidence?
d. What is the margin of error for 90% confidence?
e. Based on the sample results, find the 99% confidence interval and interpret.
f. Based on the sample results, find the 90% confidence interval and interpret.
g. For a more accurate determination of the mean weight, the quality control inspectors
wish to estimate it within 0.25 grams with 99% confidence. How many ravioli should
they sample?

Accepted Answer

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

Question 8

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.
A. Describe the sampling distribution for the sample mean.
b. What is the standard error?
c. For 90% confidence, what is the margin of error?
d. Based on the sample results, create the 90% confidence interval and interpret.

Accepted Answer

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

Question 9

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 99% confidence interval of 14.16 to 15.84 grams means that if we repeatedly take samples of 25 ravioli, 99% of the intervals obtained will contain the true mean weight of cheese filling in all ravioli made by this process. Therefore, we can be 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 and C are incorrect because they imply a certainty or probability about all ravioli made by this process, which cannot be inferred from a confidence interval based on a sample. D is incorrect because only A is a valid interpretation. E is incorrect because there is a valid interpretation, A.

Question 10

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

We can use the formula for the confidence interval for a population mean, which is:

CI = X̄ ± z*(σ/√n)

where X̄ is the sample mean, σ is the population standard deviation (unknown), n is the sample size, and z is the critical value for the desired confidence level. Since we want a 95% confidence interval, we can use the z-value for the 97.5th percentile, which is 1.96.

Plugging in the values from the problem, we get:

CI = 14.8 ± 1.96*(0.65/√16) = 14.8 ± 0.403

So the 95% confidence interval for the true mean length of rods produced by this process is (14.8 - 0.403) to (14.8 + 0.403), or 14.397 to 15.203 feet. The closest answer choice is D, 14.454 to 15.146 feet.

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. 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 t-distribution is used when the sample size is small (n < 30) and the population standard deviation is unknown. The degrees of freedom for the t-distribution is n-1, which in this case is 16-1 = 15.

Question 12

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 of the test are that the subjects in the sample must be randomly selected from a population and that the sample data must come from a normally distributed population of observations for the variable under study.

Question 13

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.

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 = σ/√n. Here, σ = 0.65 feet and n = 16. Thus, SEM = 0.65 / √16 = 0.65 / 4 = 0.1625 feet.

Question 14

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.
A. Describe the sampling distribution for the sample mean.
b. What is the standard error?
c. For 95% confidence, what is the margin of error?
d. Based on the sample results, create the 95% confidence interval and interpret.

Accepted Answer

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

Question 15

Insurance companies track life expectancy information to assist in determining the
cost of life insurance policies. Last year the average life expectancy of all policyholders
was 77 years. ABI Insurance wants to determine if their clients now have a longer life
expectancy, on average, so they randomly sample some of their recently paid policies.
The ages of the clients in the sample are shown below. $$\begin{array} { | l | l | l | l | l | l | l | l | l | l | } 
\hline 86 & 75 & 83 & 84 & 81 & 77 & 78 & 79 & 79 & 81 \
\hline 76 & 85 & 70 & 76 & 79 & 81 & 73 & 74 & 72 & 83 \
\hline
\end{array}$$ 
A. Based on the sample results, find the 90% confidence interval and interpret.
b. For more accurate cost determination, ABI Insurance wants to estimate the average life
expectancy to within one year with 95% confidence. How many randomly selected
recently paid policies would they need to sample?
c. Suppose ABI samples 100 recently paid policies. This sample yields a mean of 77.7
years and a standard deviation of 3.6 years. Find a 90% confidence interval and interpret.

Accepted Answer

The answer of Insurance companies track life expectancy information to...

Question 16

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 17

A sample of students from an introductory business course were polled regarding the
number of hours they spent studying for the last exam. All students anonymously
submitted the number of hours on a 3 by 5 card. There were 24 individuals in the one
section of the course polled. The data was used to make inferences regarding the other
students taking the course. The data are shown below: $$\begin{array} { | c | c | c | c | c | c | c | c | c | c | c | c | } 
\hline 4.5 & 22 & 7 & 14.5 & 9 & 9 & 3.5 & 8 & 11 & 7.5 & 18 & 20 \
\hline 7.5 & 9 & 10.5 & 15 & 19 & 2.5 & 5 & 9 & 8.5 & 14 & 20 & 8 \
\hline
\end{array}$$ 
A. Based on the sample results, find the 95% confidence interval.
b. Interpret the results.
c. Do you expect a 90% confidence interval to be wider or narrower and why?
d. A previous study of a large cross-section of students in this course showed that
students studies 12 hours per week. Are the results of this current study statistically
different than the assumption of 12 hours per week studying? Define the hypotheses and
compare the t-value to the critical t value, evaluating at α = 0.05.

Accepted Answer

The answer of A sample of students from an introductory...

Question 18

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 sampling distribution for the sample mean follows the t-distribution with 15 degrees of freedom because the sample size is 16, and degrees of freedom for a t-distribution is calculated as n-1, where n is the sample size.

Question 19

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 as the following: $$\begin{array} { | l | l | l | l | l | l | l | l | } 
\hline 42.7 & 43.4 & 44.6 & 45.1 & 45.6 & 45.9 & 46.8 & 47.6 \
\hline
\end{array}$$ 
a. State the hypotheses to test this question.
b. State the assumptions of the test.
c. Perform the test and determine if it is statistically significant at α = 0.05.
d. Interpret the result.

Accepted Answer

The answer of A pharmaceutical company wants to answer the...

Question 20

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 standard error of the mean is calculated by dividing the standard deviation by the square root of the sample size: 14 / √16 = 3.5 grams.

Quiz 11: Confidence Intervals and Hypothesis Tests for Means