This workbook accompanies the Hypothesis Testing lesson and is built for practice with pen, paper, and a calculator. The ten exercises follow the lesson’s structure, starting with how to set up hypotheses and reason about errors, then moving through critical values and p-values, and finally working through the main test procedures: a single mean with known and unknown variance, a single proportion, two proportions, two means, and a paired-samples test. Each one isolates a specific skill so you can see exactly where your method needs work.
Attempt every exercise before reading the solutions, and write out each step rather than jumping to the answer. The solutions name the formula, show the arithmetic, and state the decision at the conventional significance levels, so the point is to compare your reasoning against the standard approach. Throughout, use the standard normal critical values from the lesson: for a two-tailed test the boundaries are 1.6449 at 10%, 1.9600 at 5%, and 2.5758 at 1%, and for a one-tailed test they are 1.2816 at 10%, 1.6449 at 5%, and 2.3263 at 1%. Where a t-test is needed, the relevant critical values are given inside the solution.
Part One: The Exercises
Exercise 1 (Formulating hypotheses). A manufacturer claims that the mean lifetime of its batteries is 500 hours. A consumer group suspects the true mean is lower. State the null and alternative hypotheses, and identify whether the test is two-tailed, upper-tailed, or lower-tailed.
Exercise 2 (Type I and Type II errors). A clinical trial tests whether a new drug is more effective than a placebo, with the null hypothesis that it has no effect. Describe what a Type I error and a Type II error mean in this context, and explain what the power of the test represents.
Exercise 3 (Critical value decision). A two-tailed test of a mean produces a test statistic of z = -2.50, where the statistic follows a standard normal distribution under the null hypothesis. Using the critical values above, decide whether to reject the null hypothesis at the 5% and 1% significance levels.
Exercise 4 (P-value). A two-tailed test produces a test statistic of z = 2.10. Given that the area in the upper tail beyond 2.10 is 0.0179, compute the p-value and state the conclusion at the 5% significance level.
Exercise 5 (Single mean, variance known). A bottling line is supposed to fill containers to a mean of 500 ml, with a known population standard deviation of 8 ml. A sample of 64 containers has a mean of 497.5 ml. Test, at the 5% and 1% levels, whether the mean fill differs from 500 ml.
Exercise 6 (Single mean, variance unknown). A teacher claims the mean exam score in a course is 70. A sample of 16 students has a mean of 74 with a sample standard deviation of 8, and you wish to test whether the true mean is greater than 70. Carry out the test, using the upper-tailed t critical values on 15 degrees of freedom of 1.753 at 5% and 2.602 at 1%.
Exercise 7 (Single proportion). A candidate claims to have the support of 50% of voters. A poll of 400 voters finds 184 supporters. Test, at the 5% and 10% levels, whether the true level of support differs from 50%.
Exercise 8 (Difference between two proportions). Before an advertising campaign, 80 of 200 surveyed people were aware of a brand. After the campaign, 100 of a separate 200 people were aware of it. Test, at the 5% and 1% levels, whether the campaign increased awareness.
Exercise 9 (Difference between two means, variances known). Two filling machines are compared. Machine A produces a sample of 50 bottles with mean 250 ml and known standard deviation 5 ml. Machine B produces 50 bottles with mean 247 ml and known standard deviation 5 ml. Test, at the 1% level, whether the two machines fill to different mean volumes.
Exercise 10 (Paired samples). Six people record their resting heart rate before and after a six-week training program. The reductions (before minus after) in beats per minute are 4, 6, 2, 8, 5, and 5. Test, at the 5% and 1% levels, whether the program reduced resting heart rate, using upper-tailed t critical values on 5 degrees of freedom of 2.015 at 5% and 3.365 at 1%.
Part Two: Worked Solutions
Solution 1. The null hypothesis is the default position of no change from the claimed value, and the alternative captures the consumer group’s suspicion that the mean is lower.
Because the alternative points in one direction only, namely that the mean is below 500, this is a lower-tailed test.
Solution 2. A Type I error is rejecting a true null hypothesis, which here means concluding the drug works when in fact it has no real effect, a false positive that could put an ineffective treatment into use. A Type II error is failing to reject a false null hypothesis, which here means concluding the drug has no effect when it actually does work, a false negative that discards a useful treatment. The power of the test is the probability of correctly detecting a real effect.
Power is the chance of rejecting the null hypothesis when the alternative is genuinely true.
Solution 3. For a two-tailed test the decision depends on the absolute value of the statistic compared with the critical boundary. The absolute value of the statistic is 2.50. At 5% the boundary is 1.9600, and since 2.50 exceeds it, the null hypothesis is rejected. At 1% the boundary is 2.5758, and since 2.50 does not exceed it, the null hypothesis is not rejected. The result is therefore moderately significant, being significant at 5% but not at 1%.
Solution 4. For a two-tailed test the p-value doubles the tail area beyond the observed statistic, because extreme values in either direction count against the null hypothesis.
Since the p-value of 0.0358 is smaller than 0.05, the result is significant at the 5% level and the null hypothesis is rejected.
Solution 5. Because the population standard deviation is known, use the z-test for a single mean.
The test is two-tailed because the claim is that the mean simply differs from 500. The absolute value 2.50 exceeds the 5% boundary of 1.9600, so the null hypothesis is rejected at 5%. It does not exceed the 1% boundary of 2.5758, so it is not rejected at 1%. The evidence that the fill volume differs from 500 ml is moderately significant.
Solution 6. Because the population standard deviation is unknown and estimated from the sample, use the t-test for a single mean on 15 degrees of freedom.
The test is upper-tailed because the claim is that the mean exceeds 70. The statistic 2.00 exceeds the 5% critical value of 1.753, so the null hypothesis is rejected at 5%. It does not exceed the 1% critical value of 2.602, so it is not rejected at 1%. There is moderately significant evidence that the mean score is above 70.
Solution 7. Use the z-test for a single proportion, with the sample proportion 184/400 = 0.46 and the standard error computed under the null value of 0.50.
The test is two-tailed because the claim is that support differs from 50%. The absolute value 1.60 does not exceed the 5% boundary of 1.9600, nor the 10% boundary of 1.6449, so the null hypothesis is not rejected at either level. There is no significant evidence that support differs from 50%.
Solution 8. Use the z-test for the difference between two proportions, first pooling the two samples because the null hypothesis assumes a common underlying proportion. The two sample proportions are 80/200 = 0.40 and 100/200 = 0.50.
The test statistic then uses this pooled proportion in the standard error.
The test is lower-tailed in this direction, because increased awareness means the after-proportion exceeds the before-proportion, making the difference negative. The statistic -2.01 is beyond the 5% one-tailed boundary of -1.6449, so the null hypothesis is rejected at 5%. It is not beyond the 1% one-tailed boundary of -2.3263, so it is not rejected at 1%. There is moderately significant evidence that the campaign raised awareness.
Solution 9. Because both population standard deviations are known, use the z-test for the difference between two means.
The test is two-tailed because the question is whether the machines differ. The absolute value 3.00 exceeds the 1% boundary of 2.5758, so the null hypothesis is rejected at 1%. The evidence that the two machines fill to different mean volumes is highly significant.
Solution 10. For paired data, reduce the two measurements to a single column of differences and apply a one-sample t-test to those differences. The six reductions have a mean found by summing and dividing.
The deviations from this mean are -1, 1, -3, 3, 0, and 0, whose squares sum to 20, giving the sample variance and standard deviation of the differences.
The test statistic then follows the one-sample t formula on 5 degrees of freedom.
The test is upper-tailed because a reduction in heart rate means a positive mean difference. The statistic 6.12 exceeds both the 5% critical value of 2.015 and the 1% critical value of 3.365, so the null hypothesis is rejected at both levels. There is highly significant evidence that the training program reduced resting heart rate.
How to Get the Most From This Workbook
Once you have worked through these, notice that every test follows the same three-step rhythm: state the hypotheses and the tail direction, compute a test statistic of the form (estimate minus hypothesized value) divided by a standard error, and compare that statistic against the critical value or convert it to a p-value. The formula for the standard error changes with the scenario, a single mean, a proportion, a difference of means, but the logic never does. The fastest way to master the exam is to drill that rhythm until choosing the right standard error becomes the only real decision, because everything else is the same machine applied again and again.
See you soon.
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