Question 1

A report states that 40% of home owners had a vegetable garden. How large a sample is needed to estimate the true proportion of home owners who have vegetable gardens to within 3% with 99% confidence?

Accepted Answer

We use the formula n = (z^2 * p * q) / E^2, where z = 2.576 (from the z-table for 99% confidence), p = 0.4, q = 0.6, and E = 0.03. Plugging in the values, we get n = (2.576^2 * 0.4 * 0.6) / 0.03^2 = 1775.745. Since we need a whole number for the sample size, we round up to 1776. Therefore, choice C is the best option.

Question 2

It was found that in a sample of 90 teenage boys, 70% of them have received speeding tickets. What is the 90% confidence interval of the true proportion of teenage boys who have received speeding tickets?

Accepted Answer

The answer of It was found that in a sample...

Question 3

The rounding rule for a confidence interval for a proportion is to round off to three decimal places.

Accepted Answer

The rounding rule for a confidence interval for a proportion is to round off to three decimal places. This is because proportions are usually expressed as percentages, which are typically rounded to the nearest hundredth or two decimal places. Rounding to three decimal places ensures that the confidence interval is reported with sufficient precision, without exaggerating the level of accuracy.

Question 4

A report states that 44% of home owners have a vegetable garden. How large a sample is needed to estimate the true proportion of home owners who have vegetable gardens to within 3 percentage points with 92% confidence?

Accepted Answer

The answer of A report states that 44% of home...

Question 5

A random sample of 70 printers discovered that 25 of them were being used in small businesses . Find the 95% limit for the population proportion of printers that are used in small businesses.

Accepted Answer

The answer of A random sample of 70 printers discovered...

Question 6

The Academy of Orthopedic Surgeons states that 80% of women wear shoes that are too small for their feet. A researcher wants to be 98% confident that this proportion is within 3 percentage points of the true proportion. How large a sample is necessary?

Accepted Answer

The sample size needed is 966. This is because the margin of error is 3 percentage points, and the confidence level is 98%. The formula for calculating the sample size is:n = (Z^2 * p * q) / e^2where:n is the sample sizeZ is the z-score for the desired confidence levelp is the proportion of women who wear shoes that are too small for their feetq is the complement of p (1 - p)e is the margin of errorPlugging in the values given in the question, we get:n = (2.326^2 * 0.8 * 0.2) / 0.03^2n = 966.04Therefore, the sample size needed is 966.

Question 7

A retailer wants to estimate with 99% confidence the number of people who shop at his store. A previous study showed that 24% of those interviewed had shopped at his store. He wishes to be accurate within 3% of the true proportion. The minimum sample size necessary would be 1,100.

Accepted Answer

To estimate the minimum sample size necessary for a given confidence level and margin of error, the formula is $n = \frac{Z^2 \cdot p \cdot (1-p)}{E^2}$, where $Z$ is the Z-score corresponding to the desired confidence level (for 99% confidence, $Z \approx 2.576$), $p$ is the estimated proportion (0.24 in this case), and $E$ is the margin of error (0.03). Plugging these values in gives a different number than 1,100, indicating that the statement is false.

Question 8

John Davis, a manager of a supermarket, wants to estimate the proportion of customers who use food stamps at his store. He has no initial estimate of what the sample proportion will be. How large a sample is required to estimate the true proportion to within 3 percentage points with 98% confidence?

Accepted Answer

The answer of John Davis, a manager of a supermarket,...

Question 9

In a sample of 855 bartenders, 48.0% report hearing complaints from patrons about their jobs. If the margin of error for the proportion of bartenders hearing job complaints is 4.4 percentage points, what is the degree of confidence used?

Accepted Answer

The answer of In a sample of 855 bartenders, 48.0%...

Question 10

A college believes that 22% of applicants to that school have parents who have remarried. How large a sample is needed to estimate the true proportion of students who have parents who have remarried to within 5 percentage points with 99% confidence?

Accepted Answer

The answer of A college believes that 22% of applicants...

Question 11

The symbol for the sample proportion is ______.

Accepted Answer

The answer of The symbol for the sample proportion is...

Question 12

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

We can use the formula for the confidence interval for a proportion:
p ± z*sqrt((p*(1-p))/n)
where p is the sample proportion, z* is the z-score corresponding to the level of confidence (in this case 2.33 for 98%), and n is the sample size.

Plugging in the values, we get:
0.17 ± 2.33*sqrt((0.17*(1-0.17))/800)

Calculating this out, we get the confidence interval of (0.135, 0.205), which corresponds to option C.

Question 13

In a sample of 40 mice, a biologist found that 42% were able to run a maze in 30 seconds or less. Find the 90% limit for the population proportion of mice who can run a maze in 30 seconds or less.

Accepted Answer

To find the 90% limit for the population proportion, we can use the formula:
p ± z*sqrt(p*(1-p)/n) 
where p is the sample proportion, z is the z-score corresponding to the desired level of confidence (90%), and n is the sample size.

First, we need to find the z-score for the 90% confidence level. This can be found using a standard normal distribution table or a calculator, and it is approximately 1.645.

Next, we plug in the values we know from the problem:
p = 0.42 (sample proportion)
z = 1.645 (z-score for 90% confidence)
n = 40 (sample size)

p ± z*sqrt(p*(1-p)/n) 
= 0.42 ± 1.645*sqrt(0.42*0.58/40) 
= 0.42 ± 0.132 
= 0.288 to 0.552

Therefore, the 90% limit for the population proportion of mice who can run a maze in 30 seconds or less is 29.1% to 54.9%. The closest answer choice is C.

Question 14

A chi-square distribution is negatively skewed.

Accepted Answer

The answer of A chi-square distribution is negatively skewed....

Question 15

A quality control expert wants to estimate the proportion of defective components that are being manufactured by his company. A sample of 300 components showed that 20 were defective. How large a sample is needed to estimate the true proportion of defective components to within 2.5 percentage points with 99% confidence?

Accepted Answer

The answer of A quality control expert wants to estimate...

Question 16

If  is equal to 0.85, then  is equal to ______.

Accepted Answer

The only difference between the two given values is in the 12th character. Changing that character changes the value from 0.85 to 0.15. Therefore, the answer is D.

Question 17

A sample of 400 racing cars showed that 80 of them cost over $700,000. What is the 99% confidence interval for the true proportion of racing cars that cost over $700,000?

Accepted Answer

The answer of A sample of 400 racing cars showed...

Question 18

A recent poll of 700 people who work indoors found that 278 smoke. If the researchers want to be 98% confident of their results to within 3.5 percentage points, how large a sample is necessary?

Accepted Answer

To determine the sample size, we need to use the formula:
n = (Z^2 * p * q) / E^2
where:
Z = Z-value for the desired level of confidence (2.33 for 98% confidence)
p = proportion of smokers in the sample (278/700 ≈ 0.397)
q = proportion of non-smokers in the sample (1 - p ≈ 0.603)
E = maximum error margin (0.035)

Plugging in the values, we get:
n = (2.33^2 * 0.397 * 0.603) / 0.035^2
n ≈ 1061.96

Rounding up to the nearest whole number, we get a sample size of 1062. Therefore, the best choice is B.

Question 19

The Pizza Shop wanted to determine what proportion of its customers ordered only cheese pizza. Out of 80 customers surveyed, 15 ordered only cheese pizza. What is the 99% confidence interval of the true proportion of customers who order only cheese pizza?

Accepted Answer

The answer of The Pizza Shop wanted to determine what...

Question 20

A random sample of 100 voters found that 46% were going to vote for a certain candidate. Find the 95% limit for the population proportion of voters who will vote for that candidate.

Accepted Answer

The 95% confidence interval for the population proportion is:0.46 +/- 1.96 * sqrt(0.46 * 0.54 / 100) = 0.46 +/- 0.099 = (0.362, 0.558)

Quiz 7: Confidence Intervals and Sample Size