College Statistics Quiz: Confidence Level And Margin Of Error

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Confidence Level And Margin Of ErrorQuestion 1 of 20

Two independent polls were conducted to estimate the proportion of voters favoring a ballot measure. Poll A, with 1000 voters, resulted in a 95% confidence interval of [0.52, 0.58]. Poll B, with 600 voters, resulted in a 95% confidence interval of [0.47, 0.55]. Based on these intervals, which of the following is the most appropriate conclusion?

The evidence suggests the proportion of voters favoring the measure is higher for the population represented by Poll A.

The true proportion must be within the overlapping region of [0.52, 0.55], as this is the only range of values supported by both polls.

Poll A's estimate is more accurate than Poll B's because it was based on a larger sample size, resulting in a narrower interval.

The results do not provide convincing statistical evidence of a difference between the two population proportions because the intervals overlap.

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College Statistics Quiz

College Statistics Quiz: Confidence Level And Margin Of Error

Practice Confidence Level And Margin Of Error in College Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Confidence Level And Margin Of Error, giving you a quick way to practice the rules, question types, and explanations that matter most for College Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The evidence suggests the proportion of voters favoring the measure is higher for the population represented by Poll A.
The true proportion must be within the overlapping region of [0.52, 0.55], as this is the only range of values supported by both polls.
Poll A's estimate is more accurate than Poll B's because it was based on a larger sample size, resulting in a narrower interval.
The results do not provide convincing statistical evidence of a difference between the two population proportions because the intervals overlap. (correct answer)

Explanation: When two confidence intervals at the same confidence level overlap, it means there is a range of plausible values for the parameter that are common to both samples. In this case, any value between 0.52 and 0.55 is plausible for both population proportions. Since it is plausible that the two population proportions are equal (e.g., both could be 0.53), we cannot conclude there is a statistically significant difference between them.

Question 2

A researcher constructs a confidence interval for a population mean based on a sample of size $n=20$ . The population is known to be approximately normal, but the population standard deviation $\sigma$ is unknown. To achieve a 90% confidence level, what is the correct critical value to use in the margin of error calculation?

The z-value for which 5% of the standard normal distribution is in the right tail, $z^*=1.645$ .
The z-value for which 2.5% of the standard normal distribution is in the right tail, $z^*=1.96$ .
The t-value from a t-distribution with 20 degrees of freedom that corresponds to 90% confidence.
The t-value from a t-distribution with 19 degrees of freedom that corresponds to 90% confidence. (correct answer)

Explanation: When the population standard deviation

\sigma

is unknown and the sample size is relatively small, the Student's t-distribution should be used instead of the standard normal (z) distribution. The degrees of freedom for the t-distribution in this context are

df = n - 1

. Since

n=20

, the degrees of freedom are

20 - 1 = 19

. For a 90% confidence interval, we need to find the t-value that separates the middle 90% of the distribution from the 10% in the tails, which is based on

df=19

Question 3

A 95% confidence interval for a population mean is calculated from sample data to be (15.2, 18.8). If the researcher had decided to use a 99% confidence level with the same sample data, which of the following statements about the new interval would be true?

The new interval would be narrower and would be centered at 17.0.
The new interval would become wider to increase the probability that the true population mean falls within it.
The new interval would be wider and would remain centered at 17.0. (correct answer)
The center of the new interval would shift to a higher value to reflect the increased confidence.

Explanation: Increasing the confidence level from 95% to 99% means we require greater certainty that our method captures the true mean. This requires a larger critical value (

z^*

t^*

), which in turn increases the margin of error and makes the interval wider. Since the same sample data is used, the point estimate, which is the center of the interval, remains unchanged. The original center is

(15.2 + 18.8) / 2 = 17.0

. Thus, the new interval will be wider and still centered at 17.0.

Question 4

Based on a single random sample, a researcher computes two confidence intervals for a population proportion. Interval A is (0.62, 0.78) and Interval B is (0.60, 0.80). The sample proportion is the same for both. Which of the following statements must be true?

The sample size used for Interval A was larger than the sample size used for Interval B.
Interval B has a higher confidence level than Interval A. (correct answer)
Interval A has a higher confidence level than Interval B.
The population proportion is more likely to be in Interval B than in Interval A.

Explanation: Both intervals are centered at the same point estimate,

\hat{p} = (0.62+0.78)/2 = 0.70

. The width of Interval A is

0.78 - 0.62 = 0.16

, and the width of Interval B is

0.80 - 0.60 = 0.20

. Since the intervals are based on the same sample (and thus the same sample size and proportion), the only factor that can change the width is the confidence level. A wider interval corresponds to a higher confidence level because a larger critical value is used to create a larger margin of error. Therefore, Interval B must have been constructed with a higher confidence level than Interval A.

Question 5

A study is conducted to test the null hypothesis $H_0: \mu = 100$ against the alternative $H_A: \mu \neq 100$ . Using a sample of data, a 95% confidence interval for $\mu$ is calculated to be [102, 108]. Based on this interval, what can be concluded about the p-value for the hypothesis test?

The p-value is less than 0.05. (correct answer)
The p-value is greater than 0.05.
The p-value is equal to 0.05.
The p-value cannot be estimated without knowing the sample mean and standard deviation.

Explanation: There is a direct correspondence between a two-sided hypothesis test and a confidence interval. If the (1 -

\alpha

) confidence interval does not contain the value specified in the null hypothesis (

\mu_0

), then we would reject

H_0

at the

\alpha

significance level. Here, the 95% confidence interval is [102, 108], and the null value is

\mu_0 = 100

. Since 100 is not in the interval, we would reject

H_0

\alpha = 0.05

. Rejecting

H_0

\alpha = 0.05

means that the p-value must be less than 0.05.

Question 6

A political polling agency reports that the approval rating of a mayor is 52% with a 95% confidence interval of [0.48, 0.56]. A student interprets this finding by stating, "This means that if we were to conduct this poll 100 times, the true approval rating of the mayor would fall between 0.48 and 0.56 in about 95 of those polls." Why is this student's interpretation incorrect?

The interpretation assumes the sample size was the same in all 100 hypothetical polls, which may not be true.
The interpretation incorrectly suggests the true parameter is a random variable that changes from poll to poll. (correct answer)
The interpretation is flawed because the margin of error, which is 4%, is too large for a reliable conclusion.
The interpretation confuses the 95% confidence level with the 52% sample approval rating.

Explanation: The core concept of frequentist confidence intervals is that the true population parameter (the mayor's actual approval rating) is a fixed, unknown constant. It is the confidence interval itself that is random, varying from sample to sample. The 95% confidence level refers to the long-run success rate of the method used to construct the interval. The student's statement incorrectly implies that the true parameter is a variable that 'falls' into the interval in 95% of cases. The correct interpretation is that we expect about 95% of all possible intervals constructed this way to capture the single, true parameter.

Question 7

A pharmaceutical company reports a 95% confidence interval for the mean reduction in systolic blood pressure for patients taking a new drug. The interval is [8 mmHg, 12 mmHg]. The company's press release states, "We are 95% confident that a patient who takes our new drug will experience a blood pressure reduction between 8 and 12 mmHg." Why is this statement a misinterpretation of the statistical result?

The statement is incorrect because for any given interval, the true mean reduction is either in it or not; probability is not involved.
The interval estimates the mean reduction for a population of patients, not the specific reduction for any single individual. (correct answer)
The confidence level should be at least 99% for any claims related to medical treatments, so the statement is unreliable.
The statement confuses the confidence level of 95% with the percentage of patients who will experience this exact range of reduction.

Explanation: A confidence interval for a population mean estimates the range of plausible values for the average outcome across the entire population. It does not describe the range of outcomes for an individual. The variation among individuals is much larger than the uncertainty in the estimate of the mean. The press release incorrectly applies an inference about a population parameter (the mean) to a prediction for an individual observation.

Question 8

A marketing team wants to estimate the proportion of consumers who prefer a new packaging design. They want the estimate to be within $\pm 4\%$ with 95% confidence. For the purpose of determining the required sample size, which value of the population proportion $p$ should they use in their calculations to be the most conservative—that is, to ensure the sample size is large enough to achieve the desired margin of error?

0.04, the value of the desired margin of error.
A value obtained from a small pilot study, as this provides the best available estimate of $p$ .
0.95, the value corresponding to the confidence level.
0.50, as this value maximizes the product $p(1-p)$ and thus yields the largest required sample size. (correct answer)

Explanation: The formula for the sample size required to estimate a population proportion is

n = (\frac{z^*}{MOE})^2 p(1-p)

. The term

p(1-p)

represents the variance of a Bernoulli trial. To ensure the sample size is large enough to achieve the desired margin of error regardless of the true proportion, we must use the value of

p

that maximizes this variance term. The function

f(p) = p(1-p)

is maximized when

p = 0.5

. Using

p=0.5

is the most conservative approach, as it will yield the largest possible required sample size for a given MOE and confidence level.

Question 9

A confidence interval for a population parameter is constructed from a sample, resulting in a margin of error of M. If a new study is conducted with double the original sample size, what will be the new margin of error, assuming the confidence level and the sample standard deviation are the same?

$M / 2$
$M / \sqrt{2}$ (correct answer)
$M \times \sqrt{2}$
$M / 4$

Explanation: The margin of error (MOE) is inversely proportional to the square root of the sample size,

n

. The formula is of the form

MOE = (critical\ value) \times (standard\ error)

, where the standard error has

\sqrt{n}

in the denominator. If the original sample size is

n

and the new sample size is

2n

, the new margin of error

M_{new}

will be related to the old margin of error

M_{old}

by the factor

1/\sqrt{2}

. Therefore,

M_{new} = M_{old} / \sqrt{2}

Question 10

A 95% confidence interval for the proportion of students at a university who own a car is [0.54, 0.66]. If the researchers were to repeat the study with a new sample four times as large as the original, what would be the expected new margin of error, assuming the new sample proportion is the same as the original?

0.015
0.030 (correct answer)
0.060
0.120

Explanation: First, calculate the original margin of error (MOE). The width of the interval is

0.66 - 0.54 = 0.12

. The MOE is half the width, so

MOE = 0.12 / 2 = 0.06

. The margin of error is inversely proportional to the square root of the sample size (

n

). If the sample size is multiplied by 4, the new MOE will be the old MOE divided by

\sqrt{4} = 2

. Therefore, the expected new MOE is

0.06 / 2 = 0.03

Question 11

A research team wants to estimate the mean daily screen time of teenagers. Their initial study used a sample of 400 teenagers and yielded a 95% confidence interval with a margin of error of 0.2 hours. They want to conduct a new study on a similar population to reduce the margin of error to 0.1 hours, but they also want to increase the confidence level to 99%. Approximately how large a sample size will they need for the new study? (Use $z^*_{95\%} = 1.96$ and $z^*_{99\%} = 2.576$ )

800
1600
691
2765 (correct answer)

Explanation: The margin of error (MOE) is given by

MOE = z^* \frac{\sigma}{\sqrt{n}}

. To find the new sample size

n_{new}

, we can set up a ratio:

\frac{MOE_{new}}{MOE_{old}} = \frac{z^*_{new} / \sqrt{n_{new}}}{z^*_{old} / \sqrt{n_{old}}}

. Rearranging to solve for

n_{new}

gives

n_{new} = n_{old} \times (\frac{z^*_{new}}{z^*_{old}})^2 \times (\frac{MOE_{old}}{MOE_{new}})^2

. Plugging in the values:

n_{new} = 400 \times (\frac{2.576}{1.96})^2 \times (\frac{0.2}{0.1})^2 = 400 \times (1.314)^2 \times (2)^2 = 400 \times 1.727 \times 4 \approx 2764.1

. The closest answer is 2765.

Question 12

A national research center conducted a large, well-designed telephone survey of 2,500 randomly selected adults and reported a 95% confidence interval for the proportion who support a new environmental policy. The reported margin of error was $\pm 2\%$ . Which of the following potential sources of error is accounted for by this margin of error?

Errors introduced because some respondents may not have answered truthfully about their opinion on the policy.
Errors introduced because many of the adults selected for the sample could not be reached or refused to participate.
Errors resulting from the random variation inherent in selecting a sample from the population instead of conducting a census. (correct answer)
Errors due to the survey excluding adults who do not have a telephone and could not be selected for the sample.

Explanation: The margin of error in a confidence interval specifically quantifies the uncertainty due to random sampling error—the variability that arises because we are observing a sample rather than the entire population. It does not account for systematic errors or biases, such as response bias (A), nonresponse bias (B), or undercoverage/selection bias (D).

Question 13

A bootstrap procedure is used to generate 1,000 plausible values for a population median from an original sample. A 90% confidence interval for the median is constructed using the percentile method, which involves arranging these 1,000 values in order and trimming a certain percentage of values from both ends of the distribution. How many values should be trimmed from each end to form the 90% confidence interval?

100
90
50 (correct answer)
10

Explanation: A 90% confidence interval should capture the central 90% of the bootstrap distribution. This leaves 10% of the values in the tails, to be excluded from the interval. Because the interval is symmetric, this 10% is split evenly between the lower and upper tails. Therefore, 5% of the values should be trimmed from the lower end and 5% from the upper end. With 1,000 bootstrap values, this corresponds to trimming

0.05 \times 1000 = 50

values from each end.

Question 14

A student calculates a 95% confidence interval for the mean height of plants based on a sample of 25 plants. The interval is [30 cm, 38 cm]. Later, she discovers that one of the measurements was recorded incorrectly as an outlier of 50 cm when it should have been 40 cm. How will correcting this single data point affect the center and width of the confidence interval?

The center of the interval will shift to the left, and the width of the interval will increase.
The center of the interval will shift to the right, and the width of the interval will decrease.
The center of the interval will remain the same, but the width of the interval will decrease.
The center of the interval will shift to the left, and the width of the interval will decrease. (correct answer)

Explanation: The confidence interval is calculated as

\bar{x} \pm t^* \frac{s}{\sqrt{n}}

. Correcting the data point from 50 cm to 40 cm will lower the sum of the values, thus decreasing the sample mean (

\bar{x}

). This will shift the center of the interval to the left. The original value of 50 cm was an outlier, meaning it was far from the other data points. Removing an outlier and replacing it with a value closer to the mean will decrease the variability of the data, thus decreasing the sample standard deviation (

s

). A smaller

s

leads to a smaller margin of error, making the interval narrower (decreasing its width).

Question 15

A pollster is planning a survey to estimate the proportion of residents in a state who support a new law. She requires a margin of error of no more than 3% at a 95% confidence level. A colleague suggests that because the state has a very large population (over 10 million), she will need a much larger sample size than if she were surveying a small city (population 50,000). How should the pollster respond to this suggestion?

The colleague is correct; the required sample size grows in direct proportion to the size of the population being surveyed.
The colleague is partially correct; a larger population exhibits more natural variability, which requires a larger sample to estimate accurately.
The colleague is incorrect; as long as the sample is a small fraction of the total population, the population size has a negligible effect on the required sample size. (correct answer)
The colleague is incorrect; surveying the small city would require a larger sample size because a larger fraction of the population must be sampled for the same precision.

Explanation: The formulas for calculating sample size for a given margin of error depend on the confidence level, the estimated proportion, and the desired margin of error, but not the population size, provided the population is large. The precision of an estimate from a random sample depends on the absolute size of the sample, not the sample size relative to the population size. The effect of population size is only introduced through the Finite Population Correction (FPC) factor, which is negligible unless the sample size is more than 5-10% of the population size.

Question 16

A regulatory agency tests a product to determine if the mean amount of a chemical, $\mu$ , is different from the safety limit of 20 ppm. They construct a 99% confidence interval for $\mu$ based on a sample. The resulting interval is [14.5 ppm, 20.0 ppm]. What is the most appropriate conclusion for the agency to draw at a significance level of $\alpha = 0.01$ ?

Conclude that the mean level is significantly less than 20 ppm because the vast majority of the interval lies below 20.
Reject the null hypothesis that $\mu = 20$ ppm, as the value 20 is only an endpoint and not within the interior of the interval.
Fail to reject the null hypothesis that $\mu = 20$ ppm, as this value is a plausible value for the mean contained within the confidence interval. (correct answer)
Conclude that the evidence is inconclusive and that another sample must be taken to make a decision.

Explanation: The significance level for a two-sided test,

\alpha = 0.01

, corresponds to a

1 - \alpha = 0.99

, or 99%, confidence interval. A confidence interval represents the range of plausible values for the population parameter. Since the null hypothesis value,

\mu = 20

ppm, is included in the interval [14.5, 20.0] (as an endpoint), it is considered a plausible value. Therefore, the agency does not have sufficient evidence to reject the null hypothesis at the 0.01 significance level.

Question 17

A study compares the mean recovery times for two treatments, A and B. A 95% confidence interval for the difference in population means ( $\mu_A - \mu_B$ ) is found to be [-2.5 days, 1.5 days]. Which of the following is a valid interpretation of this interval?

There is a 95% probability that the true difference in mean recovery time for the two populations is between -2.5 and 1.5 days.
Since the interval contains zero, there is no statistically significant evidence of a difference in mean recovery times between the two treatments. (correct answer)
Treatment B is likely more effective, as the interval contains more values that are negative than positive.
For 95% of patients, the difference in recovery time when using Treatment A versus Treatment B will be between -2.5 and 1.5 days.

Explanation: A confidence interval for the difference between two means,

\mu_A - \mu_B

, provides a range of plausible values for that difference. If the interval contains the value 0, it means that

\mu_A - \mu_B = 0

(or

\mu_A = \mu_B

) is a plausible scenario. Therefore, we cannot conclude that there is a statistically significant difference between the two population means at the corresponding significance level (

\alpha = 0.05

for a 95% CI).

Question 18

0.015
0.030 (correct answer)
0.060
0.120