Q1: A researcher calculates a 95% confidence interval for a population mean and obtains (12.5, 17.3). What does this interval represent? (a) There is a 95% probability that the true population mean lies between 12.5 and 17.3 (b) 95% of all sample means will fall between 12.5 and 17.3 (c) We are 95% confident that the interval from 12.5 to 17.3 contains the true population mean (d) The sample mean is 95% likely to be 14.9
Solution:
Ans: (c) Explanation: A confidence interval is interpreted as our level of confidence that the interval captures the true population parameter. Option (a) is incorrect because the population mean is fixed, not random. Option (b) confuses confidence intervals with sampling distributions. Option (d) misinterprets what confidence means in this context.
Q2: What happens to the width of a confidence interval when the sample size increases, assuming all else remains constant? (a) The interval becomes wider (b) The interval becomes narrower (c) The interval stays the same (d) The interval doubles in width
Solution:
Ans: (b) Explanation: As sample size increases, the standard error decreases because it is calculated as \(\sigma/\sqrt{n}\). Since the margin of error is proportional to the standard error, a smaller standard error produces a narrower confidence interval, giving us more precision.
Q3: A 90% confidence interval for a population proportion is calculated as (0.42, 0.58). What is the point estimate for the population proportion? (a) 0.16 (b) 0.50 (c) 0.08 (d) 0.90
Solution:
Ans: (b) Explanation: The point estimate is the center of the confidence interval. To find it, calculate \((0.42 + 0.58)/2 = 1.00/2 = 0.50\). The point estimate represents our best single guess for the true population proportion.
Q4: Which of the following would result in a wider confidence interval for a population mean? (a) Increasing the sample size (b) Decreasing the confidence level from 99% to 95% (c) Increasing the confidence level from 90% to 99% (d) Decreasing the population standard deviation
Solution:
Ans: (c) Explanation: A higher confidence level requires a larger critical value (z* or t*), which increases the margin of error and thus makes the interval wider. Options (a), (b), and (d) all result in narrower intervals.
Q5: A statistician constructs a 95% confidence interval for a mean and gets (45, 55). What is the margin of error? (a) 10 (b) 5 (c) 50 (d) 2.5
Solution:
Ans: (b) Explanation: The margin of error is half the width of the confidence interval. The width is \(55 - 45 = 10\), so the margin of error is \(10/2 = 5\). This means the point estimate is 50, and we add/subtract 5 to get the interval.
Q6: If a 95% confidence interval for a population mean is (22, 38), which of the following statements is correct? (a) The probability that the population mean is 30 is 0.95 (b) 95% of all data values fall between 22 and 38 (c) If we repeated this process many times, about 95% of intervals would contain the true mean (d) The sample mean must be exactly 30
Solution:
Ans: (c) Explanation: The correct interpretation of a 95% confidence level is that if we repeated the sampling process many times, approximately 95% of the constructed intervals would capture the true population mean. Option (a) treats the parameter as random, (b) confuses the interval with data spread, and (d) is true but doesn't address confidence.
Q7: Which critical value (z*) is used for a 99% confidence interval? (a) 1.645 (b) 1.96 (c) 2.326 (d) 2.576
Solution:
Ans: (d) Explanation: For a 99% confidence interval, we need the z-value that captures 99% of the area under the standard normal curve, leaving 0.5% in each tail. This critical value is approximately 2.576. The value 1.96 corresponds to 95%, and 1.645 to 90%.
Q8: A sample of 100 students has a mean test score of 78 with a standard deviation of 10. What is the margin of error for a 95% confidence interval? (Use z* = 1.96) (a) 1.96 (b) 19.6 (c) 10 (d) 0.196
Solution:
Ans: (a) Explanation: The margin of error is calculated as \(z^* \times \frac{s}{\sqrt{n}}\). Substituting values: \(1.96 \times \frac{10}{\sqrt{100}} = 1.96 \times \frac{10}{10} = 1.96 \times 1 = 1.96\). The margin of error represents how much we add and subtract from the sample mean to construct the interval.
## Section B: Fill in the Blanks Q9: The point estimate for a population mean is the __________ of the sample.
Solution:
Ans: mean Explanation: The sample mean serves as the best point estimate for the population mean. It is the center of our confidence interval.
Q10: A confidence interval is calculated as point estimate ± __________.
Solution:
Ans: margin of error Explanation: The general formula for a confidence interval is the point estimate plus or minus the margin of error, which accounts for sampling variability.
Q11: The value that multiplies the standard error to create the margin of error is called the __________ value.
Solution:
Ans: critical Explanation: The critical value (z* or t*) depends on the desired confidence level and determines how many standard errors we extend from the point estimate.
Q12: If all other factors remain constant, increasing the confidence level from 90% to 95% will make the confidence interval __________.
Solution:
Ans: wider Explanation: A higher confidence level requires a larger critical value, which increases the margin of error and makes the interval wider to ensure greater confidence in capturing the true parameter.
Q13: The quantity \(\frac{s}{\sqrt{n}}\) is called the __________ error.
Solution:
Ans: standard Explanation: The standard error measures the variability of the sample mean and is calculated by dividing the sample standard deviation by the square root of the sample size.
Q14: A 95% confidence interval means that if we constructed 100 such intervals from different samples, approximately __________ of them would contain the true population parameter.
Solution:
Ans: 95 Explanation: The confidence level of 95% indicates that in repeated sampling, about 95 out of 100 confidence intervals would successfully capture the true population parameter.
## Section C: Word Problems Q15: A high school principal wants to estimate the average number of hours students spend on homework per week. A random sample of 64 students yields a mean of 12 hours with a standard deviation of 4 hours. Construct a 95% confidence interval for the true mean hours spent on homework. (Use z* = 1.96)
Solution:
Ans: Standard error = \(s/\sqrt{n} = 4/\sqrt{64} = 4/8 = 0.5\) Margin of error = \(z^* \times SE = 1.96 \times 0.5 = 0.98\) Confidence interval = \(12 \pm 0.98 = (11.02, 12.98)\) Final Answer: The 95% confidence interval is (11.02, 12.98) hours
Q16: A survey of 400 voters found that 240 support a new city ordinance. Calculate a 95% confidence interval for the true proportion of voters who support the ordinance. (Use z* = 1.96)
Solution:
Ans: Sample proportion: \(\hat{p} = 240/400 = 0.6\) Standard error: \(SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.6 \times 0.4}{400}} = \sqrt{\frac{0.24}{400}} = \sqrt{0.0006} = 0.0245\) Margin of error: \(1.96 \times 0.0245 = 0.048\) Confidence interval: \(0.6 \pm 0.048 = (0.552, 0.648)\) Final Answer: The 95% confidence interval is (0.552, 0.648) or (55.2%, 64.8%)
Q17: A restaurant owner samples 50 customers and finds the average meal cost is $28 with a standard deviation of $6. If she wants to be 90% confident, what is the margin of error for estimating the true average meal cost? (Use z* = 1.645)
Solution:
Ans: Standard error: \(SE = s/\sqrt{n} = 6/\sqrt{50} = 6/7.071 = 0.849\) Margin of error: \(z^* \times SE = 1.645 \times 0.849 = 1.396\) Final Answer: The margin of error is approximately $1.40
Q18: A biologist measures the wingspan of 36 butterflies and calculates a 95% confidence interval of (4.2 cm, 5.8 cm). What was the sample mean wingspan?
Solution:
Ans: The sample mean is the midpoint of the confidence interval. Sample mean = \((4.2 + 5.8)/2 = 10.0/2 = 5.0\) Final Answer: The sample mean wingspan was 5.0 cm
Q19: A company wants to estimate the mean commute time of its employees. They need the margin of error to be no more than 2 minutes with 95% confidence. If the population standard deviation is known to be 12 minutes, what minimum sample size is needed? (Use z* = 1.96)
Solution:
Ans: Formula: \(n = \left(\frac{z^* \times \sigma}{E}\right)^2\) where E is the desired margin of error. \(n = \left(\frac{1.96 \times 12}{2}\right)^2 = \left(\frac{23.52}{2}\right)^2 = (11.76)^2 = 138.30\) Round up to ensure the margin of error requirement is met. Final Answer: The minimum sample size needed is 139 employees
Q20: A quality control inspector samples 100 light bulbs and finds that 8 are defective. Construct a 99% confidence interval for the true proportion of defective light bulbs. (Use z* = 2.576)
Solution:
Ans: Sample proportion: \(\hat{p} = 8/100 = 0.08\) Standard error: \(SE = \sqrt{\frac{0.08 \times 0.92}{100}} = \sqrt{\frac{0.0736}{100}} = \sqrt{0.000736} = 0.0271\) Margin of error: \(2.576 \times 0.0271 = 0.0698\) Confidence interval: \(0.08 \pm 0.0698 = (0.0102, 0.1498)\) Final Answer: The 99% confidence interval is (0.0102, 0.1498) or (1.02%, 14.98%)
Extra Questions, Previous Year Questions with Solutions, Summary, Semester Notes, mock tests for examination, Exam, Worksheet (with Solutions): Confidence Intervals, pdf , Worksheet (with Solutions): Confidence Intervals, Sample Paper, practice quizzes, shortcuts and tricks, MCQs, Important questions, study material, Objective type Questions, Free, past year papers, video lectures, Worksheet (with Solutions): Confidence Intervals, ppt, Viva Questions;
