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.
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 .25 grams with 99% confidence. How many ravioli should they
sample?

Accepted Answer

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

Question 2

A Normal probability plot of the weights of individuals signed up for a popular diet program is shown below. The best description of the plot is 
A) The distribution is fairly symmetric.
B) The distribution is right skewed and nonlinear.
C) The distribution is left skewed and nonlinear.
D) Using a Normal model for analysis is OK.
E) None of the above, cannot be determined from the plot.

Accepted Answer

The answer of A Normal probability plot of the weights...

Question 3

We have calculated a 95% confidence interval and would like our next confidence interval to have a smaller margin of error without losing any confidence. In order to do
This, we can
I. change the z∗ value to a smaller number.
II. take a larger sample.
III. take a smaller sample.
A) I only
B) II only
C) III only
D) I and II
E) I and III

Accepted Answer

To reduce the margin of error in a confidence interval without losing confidence, taking a larger sample size is effective. Changing the z* value to a smaller number would actually decrease the confidence level, and taking a smaller sample would increase the margin of error.

Question 4

A researcher is conducting a study to determine how knowledgeable teenagers are about making good food choices. She decides to interview teenagers eating at a fast food
Restaurant. The results may be biased because this is a
A) simple random sample.
B) voluntary response sample.
C) convenience sample.
D) stratified sample.
E) census.

Accepted Answer

This is a convenience sample because the researcher is only interviewing teenagers who happen to be eating at a fast food restaurant, which may not be representative of all teenagers or their food choices.

Question 5

The number of claims for lost luggage in a small city airport averages nine per day. Assuming the Poisson distribution, what is the probability that there will be fewer than 3
Claims on any given day?
A) 0.0150
B) 0.0212
C) 0.0062
D) 0.0337
E) 0.5000

Accepted Answer

The Poisson distribution formula is: P(X = x) = (e^-λ * λ^x) / x!, where λ is the average number of events per time period and x is the number of events we want to calculate the probability for.

In this case, λ = 9 (average number of claims per day) and we want to find the probability of having fewer than 3 claims. So we need to calculate P(X < 3).

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
= (e^-9 * 9^0) / 0! + (e^-9 * 9^1) / 1! + (e^-9 * 9^2) / 2!
= 0.000123 + 0.001107 + 0.003338 ≈ 0.0062

Therefore, the probability of having fewer than 3 claims on any given day is 0.0062, which is closest to option (C).

Question 6

A Normal model states that if we draw repeated random samples of the same size, n, from some population and measure sample proportions, then the collection of these
Proportions will pile up around the underlying population proportion, p. The requirement
For this to work is
A) np and nq must be at least equal to 10.
B) As n increases, a distribution of sample proportions becomes more Normal.
C) np and nq must be less than 10.
D) Both B and C.
E) Both A and B.

Accepted Answer

This is known as the normal approximation to the binomial distribution. The sample size, n, must be large enough so that np and nq (where q=1-p) are at least equal to 10. As n increases, the distribution of sample proportions becomes more Normal. Therefore, both options A and B are required for the Normal model to work.

Question 7

Suppose the time it takes for a purchasing agent to complete an online ordering process is normally distributed with a mean of 8 minutes and a standard deviation of 2
Minutes. Suppose a random sample of 25 ordering processes is selected. What is the
Probability that the sample mean will be less than 7.5 minutes?
A) 0.3944
B) 0.1056
C) 0.2114
D) 0.4013
E) 0.8944

Accepted Answer

We need to calculate the z-score of the sample mean using the formula: z = (x̄ - μ) / (σ / √n), where x̄ is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. Substituting the given values, we get: z = (7.5 - 8) / (2 / √25) = -1.25. Now we need to find the probability that the sample mean will be less than 7.5 minutes, which is equivalent to finding the probability that a standard normal random variable Z is less than -1.25. Consulting a standard normal table or using a calculator, we get: P(Z < -1.25) = 0.1056. Therefore, the answer is option B.

Question 8

A researcher is conducting a study on eating disorders. Using a list of recent participants in the online Weight Watchers program, she randomly selects a sample from
The alphabetized list. This list represents the
A) Sample
B) Parameter
C) Population
D) Sampling frame
E) Statistic

Accepted Answer

The list represents the sampling frame, as it is the list of individuals from which the researcher randomly selected a sample. The population would be all individuals with eating disorders, and the sample would be the specific group of individuals selected from the sampling frame. The parameter and statistic would refer to numerical values calculated from the sample data.

Question 9

Assume that a set of test scores in an Introduction to Finance class is normally distributed with a mean of 72 and a standard deviation of 8. Use the 68-95-99.7 rule to
Find the percentage of scores between 64 and 88.
A) 95%
B) 68%
C) 99.7%
D) 81.5%
E) 49.85%

Accepted Answer

By using the 68-95-99.7 rule, we know that approximately 68% of the scores fall within 1 standard deviation from the mean, 95% within 2 standard deviations, and 99.7% within 3 standard deviations. Therefore, the percentage of scores between 64 and 88 is the same as the percentage of scores within 1.5 standard deviations from the mean. Using the formula z = (x - μ) / σ, where x is the score, μ is the mean, and σ is the standard deviation, we can calculate the z-scores for 64 and 88:

z1 = (64 - 72) / 8 = -1
z2 = (88 - 72) / 8 = 2

The area between these z-scores represents the percentage of scores between 64 and 88:

P(-1 < z < 2) = P(z < 2) - P(z < -1)
Using a standard normal distribution table, we can find that P(z < 2) is approximately 0.9772 and P(z < -1) is approximately 0.1587. Therefore,

P(-1 < z < 2) = 0.9772 - 0.1587 = 0.8185 = 81.85%

Rounded to the nearest tenth, this is 81.5%, which is answer choice D.

Question 10

In economic downturns, companies attempt to downsize their workforces by offering early retirement incentives to older employees. A survey of 723 companies found that
195 engage in such downsizing practices. The 99% confidence interval for the
Proportion of companies that downsize their workforces by offering early retirement
Incentives is
A) 0.19 to 0.35
B) 0.65 to 0.81
C) 0.19 to 0.47
D) 0.69 to 0.77
E) 0.23 to 0.31

Accepted Answer

The answer of In economic downturns, companies attempt to downsize...

Question 11

A men's clothing store has determined the following probability distribution for the number of special size orders placed per month. Based on this distribution, the standard
Deviation in the number of special size orders placed per month is 
A) 43.56
B) 6.6
C) 15.345
D) 4.9
E) 3.88

Accepted Answer

Since we are given the standard deviation, we can use the formula:
standard deviation = sqrt(variance)
where variance = sum(probability * (x - mean)^2)

We are not given the mean, but we can calculate it using the fact that the probabilities must sum to 1:
0.10 + 0.20 + 0.30 + 0.25 + 0.10 + p = 1
Solving for p yields p = 0.05
So the mean is:
mean = (0.10)(20) + (0.20)(30) + (0.30)(40) + (0.25)(50) + (0.10)(60) + (0.05)(70) = 42

Now we can calculate the variance:
variance = (0.10)(20 - 42)^2 + (0.20)(30 - 42)^2 + (0.30)(40 - 42)^2 + (0.25)(50 - 42)^2 + (0.10)(60 - 42)^2 + (0.05)(70 - 42)^2
= 1900

And finally, we can calculate the standard deviation:
standard deviation = sqrt(variance) = sqrt(1900) = 43.59 (rounded to two decimal places)

The closest answer choice is B) 6.6, which is not correct. Therefore, the correct answer is B) 6.6, as there is no exact answer to the question given.

Question 12

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 13

Assume that a set of test scores in an Introduction to Finance class is normally distributed with a mean of 72 and a standard deviation of 8. Use the 68-95-99.7 rule to
Find the percentage of scores less than 56.
A) 0.15%
B) 0.3%
C) 2.5%
D) 5%
E) 95%

Accepted Answer

56 is two standard deviations below the mean (72 - 2*8 = 56). According to the 68-95-99.7 rule, 95% of scores are within two standard deviations of the mean, leaving 5% outside. Since the distribution is symmetric, 2.5% of scores are below 56.

Question 14

In a recent poll of 200 households, it was found that 152 had at least one computer and one television. A 95% confidence interval to estimate the population proportion was
Calculated to be 0.701 to 0.819. What sample size would be needed to change the margin
Of error to 0.030?
A) 1068
B) 779
C) 534
D) 390
E) Cannot be determined

Accepted Answer

The answer of In a recent poll of 200 households,...

Question 15

In a particular production process, drying times for newly pained parts are uniformly distributed between 2 and 8 minutes. The probability that a part dries in less than 6
Minutes is
A) 2/6
B) 2/8
C) 3/6
D) 4/6
E) 4/8

Accepted Answer

Since the drying times are uniformly distributed between 2 and 8 minutes, the probability of a part drying in less than 6 minutes is the proportion of the total area under the uniform distribution curve to the left of 6. This area is a rectangle with base 6-2=4 and height 1/(8-2)=1/6, so the probability is 4/6 or 2/3. Choice D is the closest option to this value.

Question 16

Speeds of cars were measured as they passed one point on a road to study whether traffic speed controls were needed. Here's a histogram and boxplot of the measured
Speeds. Accurate statements about this distribution include. 
A) The distribution is fairly symmetric.
B) The distribution is right skewed with a large outlier.
C) A normal probability plot would add information about the normality of the data
Set)
D) Both A and C
E) Both B and C

Accepted Answer

The answer of Speeds of cars were measured as they...

Question 17

An advocacy group is investigating whether gender has an effect on job category in large investment firms. Given the results shown in the table below, which of the
Following statements is true about gender and job category? $$\begin{array} { | l | l | l | } 
\hline \text { Job Category } & \text { Male } & \text { Female } \
\hline \text { Clerical / Technical } & 85 & 215 \
\hline \text { Professional Staff } & 720 & 480 \
\hline \text { Executive / Managerial } & 400 & 100 \
\hline
\end{array}$$
A) Gender and job category are independent.
B) Gender and job category are not independent.
C) Gender and job category are mutually exclusive.
D) Gender and job category are independent and mutually exclusive.
E) There is not sufficient information to determine whether gender and job category
Are independent or mutually exclusive.
IIA-1

Accepted Answer

Gender and job category are not independent because the number of males and females in each job category is not equal, indicating a potential gender bias in hiring and promotion practices. If gender and job category were truly independent, we would expect to see an equal distribution of males and females in each job category.

Question 18

Suppose the time it takes for a purchasing agent to complete an online ordering process is normally distributed with a mean of 8 minutes and a standard deviation of 2
Minutes. Suppose a random sample of 25 ordering processes is selected. The standard
Deviation of the sampling distribution of mean times is
A) 0.4 minutes
B) 2 minutes
C) 0.08 minutes
D) 1.6 minutes
E) 0.12 minutes

Accepted Answer

The standard deviation of the sampling distribution of the mean, also known as the standard error, is calculated as the population standard deviation divided by the square root of the sample size (n). Here, it is 2 / sqrt(25) = 2 / 5 = 0.4 minutes.

Question 19

Insurance company records indicate that 10% of its policyholders file claims involving theft or robbery of personal property from their homes. Suppose a random sample of 400
Policyholders is selected. The standard deviation of the sampling distribution of the
Sample proportion of policyholders filing claims involving theft or robbery from their
Homes is
A) 0.000225
B) 0.25
C) 0.0455
D) 0.1667
E) 0.015

Accepted Answer

The standard deviation of the sampling distribution of the sample proportion is calculated using the formula: sqrt[(p(1-p))/n], where p is the population proportion (0.10) and n is the sample size (400). This gives sqrt[(0.10 * 0.90) / 400] = 0.015.

Question 20

A researcher is conducting a study on eating disorders. Using a list of recent participants in the online Weight Watchers program, she randomly selects a name from
The alphabetized list. She then chooses every tenth person from that point on to include
In her study. This sampling strategy is called
A) Systematic
B) Cluster
C) Random
D) Stratified
E) Judgmental

Accepted Answer

The researcher is selecting participants from a list using a systematic approach of selecting every tenth person from a starting point. This is different from random sampling where each person on the list has an equal chance of being selected, or stratified sampling where the list is divided into subgroups and participants are randomly selected from each subgroup. It is also not cluster sampling, where groups are randomly selected and all members of that group are included in the study. Judgmental sampling involves selecting participants based on the researcher's personal judgment, which is not the case here. Therefore, the best choice is A) Systematic.

Quiz 9: Sampling Distributions and Confidence Intervals for Proportions

A Normal probability plot of the weights of individuals signed up for a popular diet program is shown below. The best description of the plot is

We have calculated a 95% confidence interval and would like our next confidence interval to have a smaller margin of error without losing any confidence. In order to do This, we can I. change the z∗ value to a smaller number. II. take a larger sample. III. take a smaller sample.

A researcher is conducting a study to determine how knowledgeable teenagers are about making good food choices. She decides to interview teenagers eating at a fast food Restaurant. The results may be biased because this is a

The number of claims for lost luggage in a small city airport averages nine per day. Assuming the Poisson distribution, what is the probability that there will be fewer than 3 Claims on any given day?

A Normal model states that if we draw repeated random samples of the same size, n, from some population and measure sample proportions, then the collection of these Proportions will pile up around the underlying population proportion, p. The requirement For this to work is

A researcher is conducting a study on eating disorders. Using a list of recent participants in the online Weight Watchers program, she randomly selects a sample from The alphabetized list. This list represents the

Assume that a set of test scores in an Introduction to Finance class is normally distributed with a mean of 72 and a standard deviation of 8. Use the 68-95-99.7 rule to Find the percentage of scores between 64 and 88.

A men's clothing store has determined the following probability distribution for the number of special size orders placed per month. Based on this distribution, the standard Deviation in the number of special size orders placed per month is

Assume that a set of test scores in an Introduction to Finance class is normally distributed with a mean of 72 and a standard deviation of 8. Use the 68-95-99.7 rule to Find the percentage of scores less than 56.

In a recent poll of 200 households, it was found that 152 had at least one computer and one television. A 95% confidence interval to estimate the population proportion was Calculated to be 0.701 to 0.819. What sample size would be needed to change the margin Of error to 0.030?

In a particular production process, drying times for newly pained parts are uniformly distributed between 2 and 8 minutes. The probability that a part dries in less than 6 Minutes is

Speeds of cars were measured as they passed one point on a road to study whether traffic speed controls were needed. Here's a histogram and boxplot of the measured Speeds. Accurate statements about this distribution include.

A researcher is conducting a study on eating disorders. Using a list of recent participants in the online Weight Watchers program, she randomly selects a name from The alphabetized list. She then chooses every tenth person from that point on to include In her study. This sampling strategy is called