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Andrei's avatarAndrei
June 23, 2026
6 min read
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Point and Interval Estimation Workbook: 10 Exercises with Full Solutions

Ten point and interval estimation exercises with full worked solutions, from standard error and confidence intervals for means and proportions to paired samples and sample size. Try them, then check.

This workbook accompanies the Point and Interval Estimation lesson and is built for hands-on practice with a calculator. The ten exercises follow the lesson’s path, beginning with the ideas of a point estimate and standard error, then building confidence intervals for a mean with known and unknown variance, for a proportion, for the difference between two means, and for paired data, before finishing with sample size determination. Each exercise targets one technique so you can pinpoint exactly where your method needs sharpening.

Work through every exercise before reading the solutions, and write out each step rather than skipping to the answer. The solutions name the formula, show the arithmetic, and interpret the result, so the aim is to compare your reasoning with the standard approach. Use the z critical values from the lesson throughout: 1.6449 for 90% confidence, 1.9600 for 95%, and 2.5758 for 99%. Where a t-value is needed, it is supplied inside the exercise.

Part One: The Exercises

Exercise 1 (Estimators and bias). Explain the difference between a point estimate and a confidence interval. State which population parameters the sample mean and the sample proportion estimate, and explain in one sentence what it means for an estimator to be unbiased.

Exercise 2 (Standard error and margin of error). A sample of 100 observations is drawn from a population with known standard deviation 10. For a 95% confidence interval for the mean, compute the standard error and the margin of error.

Exercise 3 (Confidence interval for a mean, variance known). A sample of 64 steel rods has a mean length of 500 mm, and the population standard deviation is known to be 16 mm. Construct a 95% confidence interval for the true mean length.

Exercise 4 (Effect of the confidence level). Using the same data as Exercise 3, construct a 99% confidence interval for the mean length, and explain why it is wider than the 95% interval.

Exercise 5 (Confidence interval for a mean, variance unknown). A sample of 25 monthly electricity bills has a mean of £50 and a sample standard deviation of £10. Construct a 95% confidence interval for the true mean bill, using the critical value t_{0.025,,24} = 2.064.

Exercise 6 (Confidence interval for a proportion). In a survey of 400 customers, 120 said they prefer online support over phone support. Construct a 95% confidence interval for the true proportion who prefer online support.

Exercise 7 (Difference between two means, variances known). Two production lines are compared. Line 1 yields a sample of 36 items with mean strength 85 and known standard deviation 12. Line 2 yields 49 items with mean strength 80 and known standard deviation 14. Construct a 95% confidence interval for the difference in mean strength, and state whether the difference is statistically significant.

Exercise 8 (Paired samples). Six employees are timed on a task before and after training. The reductions in time, in minutes, are 3, 5, 1, 4, 2, and 3. Construct a 95% confidence interval for the mean reduction, using t_{0.025,,5} = 2.571, and state whether training made a significant difference.

Exercise 9 (Sample size for a mean). A researcher wants to estimate a mean to within a tolerance of 1 unit with 95% confidence, and knows the population standard deviation is 4. Determine the minimum sample size required.

Exercise 10 (Sample size for a proportion). A pollster wants to estimate a population proportion to within 4 percentage points with 95% confidence, with no prior estimate of the proportion available. Determine the minimum sample size required.

Part Two: Worked Solutions

Solution 1. A point estimate is a single number computed from sample data that serves as the best available guess for an unknown population parameter, while a confidence interval is a range around that estimate, formed as the point estimate plus or minus a margin of error, that expresses how precise the estimate is. The sample mean estimates the population mean, and the sample proportion estimates the population proportion. An estimator is unbiased when its expected value equals the parameter it estimates, meaning it neither systematically overshoots nor undershoots in the long run.

Solution 2. The standard error measures the variability of the estimator and, for a mean, divides the standard deviation by the square root of the sample size.

SE=σn=10100=1SE = \frac{\sigma}{\sqrt{n}} = \frac{10}{\sqrt{100}} = 1

The margin of error then scales this by the critical value for the chosen confidence level.

ME=zα/2×SE=1.9600×1=1.96\text{ME} = z_{\alpha/2} \times SE = 1.9600 \times 1 = 1.96

The standard error is 1 and the margin of error is 1.96.

Solution 3. With the population standard deviation known, use the z-interval for a single mean.

x‾±zα/2,σn=500±1.9600×1664\bar{x} \pm z_{\alpha/2},\frac{\sigma}{\sqrt{n}} = 500 \pm 1.9600 \times \frac{16}{\sqrt{64}}

The standard error is 16 divided by 8, which is 2, so the margin of error is 1.9600 times 2, which is 3.92.

500±3.92=(496.08, 503.92)500 \pm 3.92 = (496.08,\ 503.92)

The 95% confidence interval for the mean length is from 496.08 mm to 503.92 mm.

Solution 4. The only change is the critical value, which rises to 2.5758 at the 99% level.

500±2.5758×2=500±5.15=(494.85, 505.15)500 \pm 2.5758 \times 2 = 500 \pm 5.15 = (494.85,\ 505.15)

The 99% interval, from 494.85 mm to 505.15 mm, is wider than the 95% interval. This happens because demanding greater confidence that the interval captures the true mean forces the interval to cover more ground, so higher confidence is bought at the cost of lower precision.

Solution 5. Because the population standard deviation is unknown and estimated by the sample standard deviation, use the t-interval for a single mean on 24 degrees of freedom.

x‾±tα/2,,n−1,sn=50±2.064×1025 \bar{x} \pm t_{\alpha/2,,n-1},\frac{s}{\sqrt{n}} = 50 \pm 2.064 \times \frac{10}{\sqrt{25}}

The standard error is 10 divided by 5, which is 2, so the margin of error is 2.064 times 2, which is 4.128.

50±4.128=(45.87, 54.13)50 \pm 4.128 = (45.87,\ 54.13)

The 95% confidence interval for the mean bill is from £45.87 to £54.13.

Solution 6. Use the z-interval for a single proportion, with the sample proportion 120/400 = 0.30.

p±zα/2p(1−p)n=0.30±1.9600×(0.30)(0.70)400p \pm z_{\alpha/2}\sqrt{\frac{p(1-p)}{n}} = 0.30 \pm 1.9600 \times \sqrt{\frac{(0.30)(0.70)}{400}}

The standard error is the square root of 0.21 divided by 400, which is 0.0229, so the margin of error is 1.9600 times 0.0229, which is 0.0449.

0.30±0.0449=(0.255, 0.345)0.30 \pm 0.0449 = (0.255,\ 0.345)

The 95% confidence interval for the true proportion is from about 0.26 to 0.34.

Solution 7. With both population standard deviations known, use the z-interval for the difference between two means. The point estimate of the difference is 85 minus 80, which is 5.

(x‾1−x‾2)±zα/2σ12n1+σ22n2=5±1.9600×14436+19649(\bar{x}_1 – \bar{x}2) \pm z{\alpha/2}\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}} = 5 \pm 1.9600 \times \sqrt{\frac{144}{36}+\frac{196}{49}}

The quantity under the root is 4 plus 4, which is 8, whose square root is 2.828, so the margin of error is 1.9600 times 2.828, which is 5.54.

5±5.54=(−0.54, 10.54)5 \pm 5.54 = (-0.54,\ 10.54)

The 95% confidence interval runs from -0.54 to 10.54. Because this interval includes zero, the difference between the two lines is not statistically significant at the 5% level.

Solution 8. For paired data, work with the column of differences and apply a one-sample t-interval. The six reductions have a mean found by summing and dividing.

d‾=3+5+1+4+2+36=186=3\bar{d} = \frac{3 + 5 + 1 + 4 + 2 + 3}{6} = \frac{18}{6} = 3

The deviations from this mean are 0, 2, -2, 1, -1, and 0, whose squares sum to 10, giving the variance and standard deviation of the differences.

sd2=∑(di−d‾)2n−1=105=2⇒sd=1.414 s_d^2 = \frac{\sum (d_i – \bar{d})^2}{n-1} = \frac{10}{5} = 2 \quad\Rightarrow\quad s_d = 1.414

The interval then uses the paired-sample formula on 5 degrees of freedom.

d‾±tα/2,,n−1,sdn=3±2.571×1.4146=3±1.485=(1.52, 4.48)\bar{d} \pm t_{\alpha/2,,n-1},\frac{s_d}{\sqrt{n}} = 3 \pm 2.571 \times \frac{1.414}{\sqrt{6}} = 3 \pm 1.485 = (1.52,\ 4.48)

The 95% confidence interval for the mean reduction is from 1.52 to 4.48 minutes. Because this interval lies entirely above zero, training produced a statistically significant reduction in time.

Solution 9. Use the sample size formula for estimating a mean, rearranged so the margin of error is at most the tolerance.

n≥zα/22,σ2e2=(1.9600)2(4)2(1)2=3.8416×161=61.47n \geq \frac{z_{\alpha/2}^2,\sigma^2}{e^2} = \frac{(1.9600)^2 (4)^2}{(1)^2} = \frac{3.8416 \times 16}{1} = 61.47

Because the sample size must be a whole number and the inequality must hold, round up to 62. A sample of at least 62 observations is required.

Solution 10. Use the sample size formula for a proportion, and because no prior estimate is available, set the proportion to 0.5, which maximises the required size and is therefore the safe choice.

n≥zα/22,p(1−p)e2=(1.9600)2(0.5)(0.5)(0.04)2=3.8416×0.250.0016=600.25n \geq \frac{z_{\alpha/2}^2,p(1-p)}{e^2} = \frac{(1.9600)^2 (0.5)(0.5)}{(0.04)^2} = \frac{3.8416 \times 0.25}{0.0016} = 600.25

Rounding up gives a required sample of at least 601 respondents.

How to Get the Most From This Workbook

Across all of these, a single template repeats: the estimate, plus or minus a critical value times a standard error. What changes from problem to problem is only the standard error and whether the critical value comes from the z or the t distribution. Use z when the population variance is known or the sample is large, and use t when the variance is estimated from a small sample. Two interpretation habits are worth locking in as well. A confidence interval that contains zero signals no significant difference, and raising the confidence level always widens the interval while raising the sample size always narrows it. Drill the template until choosing the right standard error is the only real decision left, and interval estimation becomes routine.

See you soon.

Andrei's avatarAndrei
June 23, 2026
6 min read
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