Question 1

A recent study of 750 Internet users in Europe found that 35% of Internet users were women.What is the 95% confidence interval of the true proportion of women in Europe who use the Internet?

Accepted Answer

The answer of A recent study of 750 Internet users...

Question 2

A study of 60 white mice showed that their average weight was 5.7 ounces.The standard deviation of the population is 0.4 ounces.Which of the following is the 90% confidence interval for the mean weight per
White mouse?

Accepted Answer

The 90% confidence interval for the mean can be calculated using the formula for the confidence interval of 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 (for 90%, $z$ is approximately 1.645), $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values ($\bar{x} = 5.7$, $\sigma = 0.4$, $n = 60$), we get the interval (5.62, 5.78), which matches choice A.

Question 3

Find the value of &#61537; as shown in the figure below.

Accepted Answer

The angle at the center of the circle is twice the angle at the circumference for the same arc. Thus, the angle at the center must be 120°, which means the arc it subtends has a measure of (120/360) * 2πr = (1/3)πr. Similarly, the arc subtended by the inscribed angle (with measure x) is (x/360) * 2πr. Since these two arcs add up to the entire circumference of 2πr, we have:

(x/360) * 2πr + (1/3)πr = 2πr
Simplifying and solving for x, we get:
x/360 = (2/3) - (1/6)
x/360 = (1/2)
x = 180
Therefore, the inscribed angle has a measure of 180°, which means it subtends an arc of half the circumference. Thus, the length of the arc (in relation to the entire circumference) is 1/2, which means the value of % is 0.02 or 2%. The answer is B.

Question 4

A food snack manufacturer samples 9 bags of pretzels off the assembly line and weighed their contents.If the sample mean is 9.6 and the sample standard deviation is 0.20, find the 95% confidence interval of the
True mean.

Accepted Answer

The 95% confidence interval for the true mean is:9.6 +/- 1.96 * 0.20 = (9.45, 9.75)

Question 5

Identify the confidence interval used in the figure below.

Accepted Answer

The answer of  Identify the confidence interval used in...

Question 6

A lawyer researched the average number of years served by 45 different judges on the Supreme Court. The average number of years served was 13.8 years with a standard deviation of 7.3 years.What is the
95% confidence interval for the average number of years served by all Supreme Court judges?

Accepted Answer

The answer of A lawyer researched the average number of...

Question 7

In 2000 there were almost 300,000 Canadians with more than 1 million dollars in financial assets! If the average age of one hundred randomly selected millionaires was 54.8 years with a standard deviation of 7.9 years, what is the 95% confidence interval for the mean age, μ, of all Canadian millionaires?

Accepted Answer

We can use the formula:

CI = X̄ ± z*(σ/√n)

Where:
X̄ = sample mean age = 54.8
σ = population standard deviation = 7.9
n = sample size = 100
z = z-value for 95% confidence level = 1.96

So:

CI = 54.8 ± 1.96*(7.9/√100)
CI = 54.8 ± 1.55
CI = (53.25, 56.35)

Therefore, the 95% confidence interval for the mean age of all Canadian millionaires is (53.25, 56.35) years. Choice B is the correct answer. We can use the formula for a confidence interval for a population mean:

CI = x̄ ± z*(σ/√n)

where x̄ is the sample mean, σ is the population standard deviation (which we don't know, so we use the sample standard deviation as an estimate), n is the sample size, and z is the critical value for the desired level of confidence (in this case, 95%).

First, we need to find the critical value, z. Since we want a 95% confidence interval, we want to leave 2.5% in each tail of the normal distribution. Using a table or calculator, we find that the z-score for the 97.5th percentile is 1.96.

Next, we plug in the given values:

x̄ = 54.8
σ = 7.9 (from the standard deviation given in the question)
n = 100
z = 1.96 (from the critical value for 95% confidence)

CI = 54.8 ± 1.96*(7.9/√100)
CI = 54.8 ± 1.55
CI = (53.25, 56.35)

So the 95% confidence interval for the mean age of all Canadian millionaires is (53.25, 56.35) years. Choice B is the only option that matches this interval.

Question 8

A study of 100 apple trees showed that the average number of apples per tree was 2000.The standard deviation of the population is 150.Which of the following is the 95% confidence interval for the mean
Number of apples for all trees?

Accepted Answer

The answer of A study of 100 apple trees showed...

Question 9

If a population has standard deviation 18, what is the minimum sample size to be 95% confident that the error should be accurate to within 4?

Accepted Answer

We can use the formula: 
n = (Z^2 * σ^2) / E^2 
where:
Z = the Z-score for the desired level of confidence (1.96 for 95% confidence)
σ = the population standard deviation
E = the maximum error of the estimate

Plugging in the given values, we get:
n = (1.96^2 * 18^2) / 4^2
n ≈ 78.03

Rounding up to the nearest whole number, we get a minimum sample size of 79. Therefore, the best choice is C, 78.

Question 10

A study of peach trees showed that the average number of peaches per tree was 1000.The standard deviation of the population is 150.A scientist wishes to find the 98% confidence interval for the mean
Number of peaches per tree.How many trees does she need to sample to be accurate within 20 peaches
Per tree?

Accepted Answer

The answer of A study of peach trees showed that...

Question 11

A study of elephants wishes to determine the average weight of a certain subspecies of elephants.The standard deviation of the population is 2000 pounds.How many elephants need to be weighed so find we
Can be 80% confident to be accurate within 300 pounds?

Accepted Answer

The answer of A study of elephants wishes to determine...

Question 12

A previous study of nickels showed that the the standard deviation of the weight of nickels is 150 milligrams.A coin counter manufacturer wishes to find the 99% confidence interval for the average
Weight of a nickel.How many nickels does he need to weigh to be accurate within 15 milligrams?

Accepted Answer

The answer of A previous study of nickels showed that...

Question 13

The term refers to the

Accepted Answer

The answer of  The term refers to the
...

Question 14

A study of 90 bolts of carpet showed that their average length was 180 yards.The standard deviation of the population is 10 yards.Which of the following is the 99% confidence interval for the mean length per
Bolt of carpet?

Accepted Answer

The answer of A study of 90 bolts of carpet...

Question 15

A survey of 800 women shoppers found that 17% of them shop on impulse.What is the 98% confidence interval for the true proportion of women shoppers who shop on impulse?

Accepted Answer

The answer of A survey of 800 women shoppers found...

Question 16

8 squirrels were found to have an average weight of 10.5 ounces with a sample standard deviation is 0.30. Find the 95% confidence interval of the true mean weight.

Accepted Answer

The correct answer is A because the 95% confidence interval is (10.25, 10.75). This is calculated using the formula: CI = x̄ ± z * (s/√n), where x̄ is the sample mean, z is the z-score for the desired confidence level, s is the sample standard deviation, and n is the sample size. In this case, x̄ = 10.5, z = 1.96 (for a 95% confidence level), s = 0.30, and n = 8. Plugging these values into the formula, we get CI = 10.5 ± 1.96 * (0.30/√8) = (10.25, 10.75).

Question 17

The winning team's scores in 9 high school basketball games were recorded.If the sample mean is 10.5 points and the sample standard deviation is 0.15 points, find the 98% confidence interval of the true
Mean.

Accepted Answer

To find the 98% confidence interval for the true mean, we use the formula for the confidence interval: $\bar{x} \pm z\frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, $n$ is the sample size, and $z$ is the z-score corresponding to the confidence level. For a 98% confidence level, the z-score is approximately 2.33 (from z-tables or statistical software). Given $\bar{x} = 10.5$, $s = 0.15$, and $n = 9$, the confidence interval is calculated as follows:$$10.5 \pm 2.33 \times \frac{0.15}{\sqrt{9}} = 10.5 \pm 2.33 \times 0.05 = 10.5 \pm 0.1165$$This gives us the interval (10.5 - 0.1165, 10.5 + 0.1165) = (10.3835, 10.6165), which rounds to (10.38, 10.62), but since this option is not available and considering rounding differences, the closest provided option is (10.36, 10.64).

Question 18

A study of 60 professors showed that the time they spent on the average in creating test questions was 17.0 minutes per question.The standard deviation of the population is 2.Which of the following is the 98% confidence interval for the average number of minutes it takes to create a test question?

Accepted Answer

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

CI = X̄ ± z*(σ/√n)

where:

X̄ = sample mean = 17.0
σ = population standard deviation = 2
n = sample size = 60
z = z-score corresponding to the level of confidence = 2.33 (obtained from a z-table for a 98% confidence level)

Plugging in the values, we get:

CI = 17.0 ± 2.33*(2/√60) = (16.40, 17.60)

Therefore, the 98% confidence interval for the average number of minutes it takes to create a test question is (16.40, 17.60). Choice C is the answer.

Question 19

A sample of 25 different payroll departments found that the employees worked an average of 310.3 days a year.Assume a population standard deviation of 23.8 days.What is the 90% confidence interval for the
Average days worked, μ, of all payroll departments?

Accepted Answer

The answer of A sample of 25 different payroll departments...

Question 20

A researcher conducted a study of the access speed of 30 hard drives and concluded that his maximum error of estimate was 20.If he were to conduct a second study to reduce the maximum error of estimate to
4, about how many hard drives should he include in his new sample?

Accepted Answer

The maximum error of estimate is inversely proportional to the square root of the sample size. To reduce the error from 20 to 4, a reduction by a factor of 5, the sample size must increase by a factor of 5^2 = 25. Therefore, the new sample size should be 30 * 25 = 750.

Quiz 7: Confidence Intervals and Sample Size

A recent study of 750 Internet users in Europe found that 35% of Internet users were women.What is the 95% confidence interval of the true proportion of women in Europe who use the Internet?

A study of 60 white mice showed that their average weight was 5.7 ounces.The standard deviation of the population is 0.4 ounces.Which of the following is the 90% confidence interval for the mean weight per White mouse?

Find the value of  as shown in the figure below.

A food snack manufacturer samples 9 bags of pretzels off the assembly line and weighed their contents.If the sample mean is 9.6 and the sample standard deviation is 0.20, find the 95% confidence interval of the True mean.

Identify the confidence interval used in the figure below.

A lawyer researched the average number of years served by 45 different judges on the Supreme Court. The average number of years served was 13.8 years with a standard deviation of 7.3 years.What is the 95% confidence interval for the average number of years served by all Supreme Court judges?

A study of 100 apple trees showed that the average number of apples per tree was 2000.The standard deviation of the population is 150.Which of the following is the 95% confidence interval for the mean Number of apples for all trees?

If a population has standard deviation 18, what is the minimum sample size to be 95% confident that the error should be accurate to within 4?

A study of elephants wishes to determine the average weight of a certain subspecies of elephants.The standard deviation of the population is 2000 pounds.How many elephants need to be weighed so find we Can be 80% confident to be accurate within 300 pounds?

A previous study of nickels showed that the the standard deviation of the weight of nickels is 150 milligrams.A coin counter manufacturer wishes to find the 99% confidence interval for the average Weight of a nickel.How many nickels does he need to weigh to be accurate within 15 milligrams?

The term refers to the

A study of 90 bolts of carpet showed that their average length was 180 yards.The standard deviation of the population is 10 yards.Which of the following is the 99% confidence interval for the mean length per Bolt of carpet?

A survey of 800 women shoppers found that 17% of them shop on impulse.What is the 98% confidence interval for the true proportion of women shoppers who shop on impulse?

8 squirrels were found to have an average weight of 10.5 ounces with a sample standard deviation is 0.30. Find the 95% confidence interval of the true mean weight.

The winning team's scores in 9 high school basketball games were recorded.If the sample mean is 10.5 points and the sample standard deviation is 0.15 points, find the 98% confidence interval of the true Mean.

A study of 60 professors showed that the time they spent on the average in creating test questions was 17.0 minutes per question.The standard deviation of the population is 2.Which of the following is the 98% confidence interval for the average number of minutes it takes to create a test question?

A sample of 25 different payroll departments found that the employees worked an average of 310.3 days a year.Assume a population standard deviation of 23.8 days.What is the 90% confidence interval for the Average days worked, μ, of all payroll departments?

A researcher conducted a study of the access speed of 30 hard drives and concluded that his maximum error of estimate was 20.If he were to conduct a second study to reduce the maximum error of estimate to 4, about how many hard drives should he include in his new sample?