Medium Confidence Interval Practice Questions

Understanding how to calculate and interpret a confidence interval is a cornerstone of inferential statistics. It allows us to use data from a sample to make an educated guess about a larger population. This guide provides a clear explanation of the concept, worked-through examples, and a series of medium-difficulty practice questions to help you master the calculation and interpretation of a confidence interval.

Concept Explanation

A confidence interval is a range of values, derived from sample statistics, that is likely to contain the value of an unknown population parameter. Instead of providing a single number as an estimate (a point estimate), a confidence interval provides a range, acknowledging the uncertainty inherent in sampling. The construction of a confidence interval involves a point estimate, a critical value determined by the confidence level, and the standard error of the point estimate.

The general formula for a confidence interval is:

Confidence Interval = Point Estimate ± Margin of Error

Where the Margin of Error is calculated as:

Margin of Error = Critical Value × Standard Error

Key Components:

• Point Estimate: This is the single best guess for the population parameter. For a population mean (μ), the point estimate is the sample mean (x̄). For a population proportion (p), the point estimate is the sample proportion (p̂).

• Confidence Level: This is the probability that the procedure used to generate the confidence interval will produce an interval that contains the true population parameter. Common levels are 90%, 95%, and 99%. A 95% confidence level means that if we were to take 100 different samples and compute a confidence interval for each, we would expect about 95 of those intervals to contain the true population parameter.

• Critical Value: This value is determined by the confidence level and the sampling distribution. For large samples (n > 30) or when the population standard deviation (σ) is known, we use a Z-critical value from the standard normal distribution. For small samples with an unknown σ, we use a t-critical value from the t-distribution. You can find more practice with these values in our Z-Score Practice Questions.

• Standard Error: This measures the variability of the point estimate. It is the standard deviation of the sampling distribution. For a mean, the standard error is σ/√n (if σ is known) or s/√n (if σ is unknown, where s is the sample standard deviation).

Understanding how to calculate a confidence interval is crucial for many statistical applications, including hypothesis testing, where it can be used to determine if a result is statistically significant.

Solved Examples of Confidence Intervals

Solved examples of confidence intervals demonstrate how to apply the formula to real data to estimate a population parameter. By following these step-by-step solutions, you can learn the process of identifying variables, finding the critical value, calculating the margin of error, and constructing the final interval.

Example 1: 95% Confidence Interval for a Mean (σ Known)

A researcher wants to estimate the average height of a certain species of plant. They take a random sample of 49 plants and find the sample mean height to be 30 cm. From previous studies, the population standard deviation is known to be 3.5 cm. Calculate a 95% confidence interval for the true mean height of the plants.

1. Identify the given information:

   • Sample mean (x̄) = 30 cm

   • Population standard deviation (σ) = 3.5 cm

   • Sample size (n) = 49

   • Confidence level = 95%

2. Find the critical value (Zα/2): For a 95% confidence level, the α is 1 - 0.95 = 0.05. We look for the Z-score that corresponds to an area of α/2 = 0.025 in each tail. The Z-critical value is 1.96.

3. Calculate the Standard Error (SE):

   SE = σ / √n = 3.5 / √49 = 3.5 / 7 = 0.5 cm

4. Calculate the Margin of Error (ME):

   ME = Critical Value × SE = 1.96 × 0.5 = 0.98 cm

5. Construct the Confidence Interval:

   CI = x̄ ± ME = 30 ± 0.98

   The confidence interval is (29.02, 30.98).

6. Interpretation: We are 95% confident that the true mean height of this species of plant is between 29.02 cm and 30.98 cm.

Example 2: 99% Confidence Interval for a Proportion

A city council commissions a survey to determine the proportion of residents who support a new recycling program. Out of 500 residents surveyed, 320 say they support the program. Construct a 99% confidence interval for the proportion of all residents who support the program.

1. Identify the given information:

   • Number of successes (x) = 320

   • Sample size (n) = 500

   • Sample proportion (p̂) = x / n = 320 / 500 = 0.64

   • Confidence level = 99%

2. Find the critical value (Zα/2): For a 99% confidence level, α = 1 - 0.99 = 0.01. α/2 = 0.005. The Z-critical value corresponding to this is 2.576.

3. Calculate the Standard Error (SE) for a proportion:

   SE = √[p̂(1-p̂) / n] = √[0.64(1-0.64) / 500] = √[0.64(0.36) / 500] = √[0.2304 / 500] ≈ 0.02146

4. Calculate the Margin of Error (ME):

   ME = Critical Value × SE = 2.576 × 0.02146 ≈ 0.0553

5. Construct the Confidence Interval:

   CI = p̂ ± ME = 0.64 ± 0.0553

   The confidence interval is (0.5847, 0.6953).

6. Interpretation: We are 99% confident that the true proportion of residents who support the new recycling program is between 58.47% and 69.53%.

Example 3: 90% Confidence Interval for a Mean (σ Unknown)

An educational researcher wants to estimate the average score on a new standardized test. A random sample of 20 students takes the test, yielding a sample mean of 82 and a sample standard deviation of 8. Calculate a 90% confidence interval for the true mean test score.

1. Identify the given information:

   • Sample mean (x̄) = 82

   • Sample standard deviation (s) = 8

   • Sample size (n) = 20

   • Confidence level = 90%

2. Find the critical value (tα/2): Since the population standard deviation (σ) is unknown and the sample size is small (n < 30), we use the t-distribution. The degrees of freedom (df) are n - 1 = 20 - 1 = 19. For a 90% confidence level, α = 0.10, so α/2 = 0.05. Using a t-distribution table or calculator for df=19 and α/2=0.05, the t-critical value is 1.729. For more information on this distribution, see this excellent resource from Statistics How To.

3. Calculate the Standard Error (SE):

   SE = s / √n = 8 / √20 ≈ 8 / 4.472 ≈ 1.789

4. Calculate the Margin of Error (ME):

   ME = Critical Value × SE = 1.729 × 1.789 ≈ 3.093

5. Construct the Confidence Interval:

   CI = x̄ ± ME = 82 ± 3.093

   The confidence interval is (78.907, 85.093).

6. Interpretation: We are 90% confident that the true mean score for all students on this test is between 78.91 and 85.09.

Practice Questions

These practice questions allow you to test your ability to calculate and interpret a confidence interval in various scenarios. Use the formulas and methods from the explanation and examples above.

1. (Easy) A statistician collects a random sample of 100 observations from a population with a known standard deviation of 15. The sample mean is 150. Calculate a 95% confidence interval for the population mean.

2. (Easy) In a survey of 400 likely voters, 220 said they would vote for Candidate A. Construct a 90% confidence interval for the proportion of all likely voters who will vote for Candidate A.

3. (Medium) A biologist measures the weight of 64 trout from a lake. The sample has a mean weight of 2.5 kg. The population standard deviation of trout weight in this lake is known to be 0.8 kg. What is the 99% confidence interval for the mean weight of all trout in the lake?

4. (Medium) A quality control inspector tests a sample of 25 light bulbs from a production line. The sample has a mean lifetime of 1020 hours with a sample standard deviation of 40 hours. Calculate a 95% confidence interval for the mean lifetime of all bulbs produced. (Hint: σ is unknown).

5. (Medium) A research firm reports that a 95% confidence interval for the average daily screen time of teenagers is (5.8 hours, 6.4 hours). Which of the following statements is a correct interpretation of this interval?

• A) 95% of all teenagers have a daily screen time between 5.8 and 6.4 hours.

• B) We are 95% confident that the sample mean screen time is between 5.8 and 6.4 hours.

• C) We are 95% confident that the true average daily screen time for all teenagers is between 5.8 and 6.4 hours.

• D) There is a 95% probability that the true average screen time is exactly 6.1 hours.

6. (Medium) A marketing team wants to estimate the proportion of customers who are satisfied with a product. They want their estimate to be within a margin of error of ±3% with 95% confidence. Assuming they have no prior estimate for the proportion (p̂), what is the minimum sample size required?

7. (Hard) Two different manufacturing processes (A and B) are used to create a component. A sample of 50 components from process A has a mean strength of 115 MPa, while a sample of 60 components from process B has a mean strength of 112 MPa. The population standard deviations are known to be 8 MPa for process A and 7 MPa for process B. Calculate a 95% confidence interval for the difference between the mean strengths of the two processes (μA - μB).

8. (Medium) A 95% confidence interval for the mean salary of data analysts in a city is calculated to be [$85,000,$95,000]. If the researcher had chosen to calculate a 99% confidence interval using the same sample data, how would the new interval compare to the old one?

9. (Hard) A pharmaceutical company is testing a new drug. They measure the blood pressure reduction in 16 patients. The sample mean reduction is 12 mmHg, with a sample standard deviation of 4 mmHg. Construct a 99% confidence interval for the true mean blood pressure reduction.

10. (Medium) A pollster finds that 45% of 1200 people surveyed approve of the mayor's performance. The margin of error for the 95% confidence interval is reported as ±2.83%. If the pollster wanted to reduce the margin of error to ±2%, approximately how many people would they need to survey, assuming the approval rating remains the same?

Answers & Explanations

The answers and explanations section provides detailed, step-by-step solutions for each practice question, clarifying the calculation process and interpretation.

1. Answer: (147.06, 152.94)

1. Data: x̄ = 150, σ = 15, n = 100, Confidence Level = 95%.

2. Critical Value (Zα/2): For 95% confidence, Zα/2 = 1.96.

3. Standard Error (SE): SE = σ / √n = 15 / √100 = 1.5.

4. Margin of Error (ME): ME = 1.96 × 1.5 = 2.94.

5. Confidence Interval: CI = 150 ± 2.94 = (147.06, 152.94).

2. Answer: (0.509, 0.591)

1. Data: n = 400, x = 220, Confidence Level = 90%.

2. Sample Proportion (p̂): p̂ = 220 / 400 = 0.55.

3. Critical Value (Zα/2): For 90% confidence, Zα/2 = 1.645.

4. Standard Error (SE): SE = √[p̂(1-p̂) / n] = √[0.55(0.45) / 400] ≈ 0.02487.

5. Margin of Error (ME): ME = 1.645 × 0.02487 ≈ 0.0409.

6. Confidence Interval: CI = 0.55 ± 0.0409 = (0.5091, 0.5909).

3. Answer: (2.2424 kg, 2.7576 kg)

1. Data: x̄ = 2.5 kg, σ = 0.8 kg, n = 64, Confidence Level = 99%.

2. Critical Value (Zα/2): For 99% confidence, Zα/2 = 2.576.

3. Standard Error (SE): SE = σ / √n = 0.8 / √64 = 0.8 / 8 = 0.1.

4. Margin of Error (ME): ME = 2.576 × 0.1 = 0.2576.

5. Confidence Interval: CI = 2.5 ± 0.2576 = (2.2424, 2.7576).

4. Answer: (1003.44 hours, 1036.56 hours)

1. Data: x̄ = 1020, s = 40, n = 25, Confidence Level = 95%. Since σ is unknown and n < 30, we use the t-distribution.

2. Degrees of Freedom (df): df = n - 1 = 24.

3. Critical Value (tα/2): For 95% confidence and df=24, the t-critical value is 2.064.

4. Standard Error (SE): SE = s / √n = 40 / √25 = 40 / 5 = 8.

5. Margin of Error (ME): ME = 2.064 × 8 = 16.512.

6. Confidence Interval: CI = 1020 ± 16.512 = (1003.488, 1036.512). Rounded: (1003.49, 1036.51).

5. Answer: C

The correct interpretation of a 95% confidence interval is that we are 95% confident that the interval contains the true population parameter (in this case, the true average daily screen time). Option A is incorrect because the interval describes the mean, not the range for 95% of individuals. Option B is incorrect because we know the sample mean exactly; the confidence is about the true population mean. Option D is incorrect because the interval gives a range of plausible values, not a probability for a single value.

6. Answer: 1068

1. Data: Margin of Error (ME) = 0.03, Confidence Level = 95%.

2. Critical Value (Zα/2): For 95% confidence, Zα/2 = 1.96.

3. Estimate for p̂: When p̂ is unknown, we use 0.5 to maximize the required sample size, which provides the most conservative estimate.

4. Sample Size Formula: n = p̂(1-p̂) * (Zα/2 / ME)²

5. Calculation: n = 0.5(1-0.5) * (1.96 / 0.03)² = 0.25 * (65.333)² ≈ 0.25 * 4268.44 ≈ 1067.11.

6. Conclusion: Since we cannot survey a fraction of a person, we always round up to the next whole number. The minimum sample size required is 1068.

7. Answer: (-0.33 MPa, 6.33 MPa)

1. Data: x̄A=115, nA=50, σA=8; x̄B=112, nB=60, σB=7. Confidence Level = 95%.

2. Point Estimate (Difference): x̄A - x̄B = 115 - 112 = 3.

3. Critical Value (Zα/2): For 95% confidence, Zα/2 = 1.96.

4. Standard Error of the Difference: SE = √[(σA²/nA) + (σB²/nB)] = √[(8²/50) + (7²/60)] = √[(64/50) + (49/60)] = √[1.28 + 0.8167] ≈ √2.0967 ≈ 1.448.

5. Margin of Error (ME): ME = 1.96 × 1.448 ≈ 2.838.

6. Confidence Interval: CI = (x̄A - x̄B) ± ME = 3 ± 2.838 = (0.162, 5.838). Note: My initial calculation had an error. Let's re-calculate SE: SE = sqrt( (64/50) + (49/60) ) = sqrt(1.28 + 0.81666) = sqrt(2.09666) = 1.44798. ME = 1.96 * 1.44798 = 2.838. CI = 3 ± 2.838 = (0.162, 5.838). As the interval is entirely positive, it suggests Process A has a higher mean strength.

8. Answer: The 99% confidence interval would be wider.

To be more confident (99% vs 95%) that the interval contains the true mean, you need to cast a wider net. This is reflected in the critical value; the Z-critical value for 99% confidence (2.576) is larger than for 95% confidence (1.96). A larger critical value results in a larger margin of error and thus a wider confidence interval.

9. Answer: (8.94 mmHg, 15.06 mmHg)

1. Data: x̄ = 12, s = 4, n = 16, Confidence Level = 99%. Use t-distribution as σ is unknown.

2. Degrees of Freedom (df): df = n - 1 = 15.

3. Critical Value (tα/2): For 99% confidence and df=15, the t-critical value is 2.947.

4. Standard Error (SE): SE = s / √n = 4 / √16 = 4 / 4 = 1.

5. Margin of Error (ME): ME = 2.947 × 1 = 2.947.

6. Confidence Interval: CI = 12 ± 2.947 = (9.053, 14.947).

10. Answer: Approximately 2401 people.

1. Goal: Reduce ME from 0.0283 to 0.02.

2. Formula: n = p̂(1-p̂) * (Zα/2 / ME)². We know p̂=0.45 and Zα/2=1.96 (for 95% confidence).

3. Calculation for new sample size: n = 0.45(1-0.45) * (1.96 / 0.02)² = 0.45(0.55) * (98)² = 0.2475 * 9604 = 2376.99.

4. Conclusion: We round up to the next whole number. The required sample size is 2377. (Note: A common shortcut is n_new ≈ n_old * (ME_old / ME_new)², which gives 1200 * (0.0283/0.02)² ≈ 1200 * (1.415)² ≈ 1200 * 2.002 ≈ 2402. Both methods yield a similar result.)

Quick Quiz

Frequently Asked Questions

What is the difference between a confidence interval and a confidence level?

A confidence interval is the actual range of values (e.g., 25 to 35) calculated from sample data that is believed to contain the population parameter. The confidence level is a probability (e.g., 95%) that expresses how confident we are in the statistical method used; it represents the long-run success rate of the method in capturing the true parameter.

What does a 95% confidence interval actually mean?

A 95% confidence interval means that if you were to take many random samples from the same population and construct a confidence interval from each sample, about 95% of those intervals would contain the true, unknown population parameter. It is a statement about the reliability of the procedure, not a direct probability statement about one specific interval. You can explore this concept further on the Wikipedia page for Confidence Intervals.

How does sample size affect the confidence interval?

Increasing the sample size decreases the width of the confidence interval, assuming all other factors remain constant. A larger sample provides more information about the population, reducing the uncertainty in the point estimate. This leads to a smaller standard error and, consequently, a smaller margin of error, resulting in a more precise (narrower) interval.

Can a confidence interval be 100%?

Theoretically, to achieve a 100% confidence interval, the interval would have to be infinitely wide (from negative infinity to positive infinity for a mean), which would be statistically useless. It would tell you that the true parameter is *some* number, which you already knew. Therefore, in practice, 100% confidence intervals are not used as they provide no practical information.

When should I use a t-distribution instead of a z-distribution?

You should use a t-distribution when calculating a confidence interval for a population mean if the population standard deviation (σ) is unknown and you are using the sample standard deviation (s) as an estimate. The t-distribution accounts for the additional uncertainty introduced by estimating σ from the sample data, and it is especially important when the sample size is small (typically n < 30).

What is a margin of error?

The margin of error is the value that is added to and subtracted from the point estimate to create the confidence interval. It represents the "radius" of the interval and quantifies the maximum expected difference between the point estimate and the true population parameter for a given confidence level. It is calculated by multiplying the critical value by the standard error.

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