Medium Hypothesis Testing Practice Questions

Hypothesis testing is a fundamental pillar of inferential statistics, allowing us to use sample data to make educated decisions about an entire population. Whether you're a student, a researcher, or a data analyst, mastering this skill is crucial for drawing valid conclusions from data. This guide provides a clear explanation of the core concepts, worked-out examples, and a range of medium-difficulty practice questions to solidify your understanding of hypothesis testing.

Concept Explanation

Hypothesis testing is a statistical method used to make decisions or draw conclusions about a population based on evidence from a sample. It formalizes the process of evaluating a claim or hypothesis by determining how likely it is that the observed sample results occurred by random chance. At its core, hypothesis testing helps us decide if there is enough evidence to reject a default assumption about a population parameter (like the mean or proportion).

Key Components of a Hypothesis Test

• Null Hypothesis (H₀): This is the default assumption or the status quo. It represents a statement of no effect, no difference, or no relationship. We conduct the test assuming the null hypothesis is true. For example, H₀: μ = 100.

• Alternative Hypothesis (H₁ or Hₐ): This is the claim we are trying to find evidence for. It is the opposite of the null hypothesis and can be one-tailed (e.g., μ > 100 or μ < 100) or two-tailed (e.g., μ ≠ 100).

• Significance Level (α): This is a pre-determined threshold that represents the probability of making a Type I error (rejecting the null hypothesis when it is actually true). Common values are 0.05, 0.01, and 0.10.

• Test Statistic: This is a standardized value calculated from the sample data. It measures how far our sample statistic (e.g., sample mean) deviates from the population parameter assumed in the null hypothesis. Common test statistics include the z-score and the t-score. You can practice calculating one of these with our Z-Score Practice Questions with Answers.

• P-value: The p-value is the probability of obtaining a test statistic at least as extreme as the one observed, assuming the null hypothesis is true. A small p-value (typically ≤ α) suggests that our observed data is unlikely under the null hypothesis, providing evidence against it.

• Conclusion: Based on the p-value and significance level, we make a decision. If p ≤ α, we reject the null hypothesis in favor of the alternative. If p > α, we fail to reject the null hypothesis.

How to Perform Hypothesis Testing

1. State the Hypotheses: Clearly define the null (H₀) and alternative (H₁) hypotheses.

2. Set the Significance Level (α): Choose an appropriate alpha level for your test.

3. Choose the Correct Test and Calculate the Test Statistic: Select the appropriate statistical test (e.g., z-test for means when population standard deviation is known, t-test when it is unknown) based on your data and assumptions.

4. Determine the P-value: Calculate the p-value associated with your test statistic. This often involves using a z-table, t-table, or statistical software.

5. Make a Decision: Compare the p-value to the significance level (α).

6. Interpret the Results: State your conclusion in the context of the original problem.

Solved Examples

Working through examples is the best way to understand the process of hypothesis testing. Here are a few solved problems of medium difficulty.

Example 1: One-Sample Z-Test for a Mean

A car manufacturer claims that a new model gets an average of 35 miles per gallon (mpg). A consumer advocacy group believes the actual mileage is lower. They test a random sample of 49 cars and find a sample mean of 34.2 mpg. Assume the population standard deviation is known to be 2.8 mpg. Using a significance level of α = 0.05, is there enough evidence to support the advocacy group's claim?

1. State the Hypotheses:

   • Null Hypothesis (H₀): μ = 35 mpg (The manufacturer's claim is correct).

   • Alternative Hypothesis (H₁): μ < 35 mpg (The average mileage is lower than claimed). This is a left-tailed test.

2. Set the Significance Level: α = 0.05.

3. Calculate the Test Statistic: We use a z-test because the population standard deviation (σ) is known.

   The formula for the z-test statistic is: Z = (x̄ - μ₀) / (σ / √n)

   Plugging in the values:

   x̄ = 34.2 (sample mean)

   μ₀ = 35 (hypothesized population mean)

   σ = 2.8 (population standard deviation)

   n = 49 (sample size)

   Z = (34.2 - 35) / (2.8 / √49) = -0.8 / (2.8 / 7) = -0.8 / 0.4 = -2.0

4. Determine the P-value: We need to find the probability of getting a z-score of -2.0 or less. Using a standard normal (z) table or calculator, P(Z ≤ -2.0) = 0.0228.

5. Make a Decision and Conclude:

   Our p-value (0.0228) is less than our significance level (0.05). Therefore, we reject the null hypothesis.

   Conclusion: At the 5% significance level, there is sufficient evidence to support the claim that the average mileage of the new car model is less than 35 mpg.

Example 2: One-Sample T-Test for a Mean

A high school principal claims that the average student studies for 8 hours per week. A student council member believes this is an overestimation. They survey a random sample of 25 students and find their average study time is 7.1 hours per week, with a sample standard deviation of 2.5 hours. Test the principal's claim at a significance level of α = 0.10.

1. State the Hypotheses:

   • Null Hypothesis (H₀): μ = 8 hours.

   • Alternative Hypothesis (H₁): μ < 8 hours.

2. Set the Significance Level: α = 0.10.

3. Calculate the Test Statistic: The population standard deviation is unknown, so we use a one-sample t-test. The degrees of freedom (df) are n - 1 = 25 - 1 = 24.

   The formula for the t-test statistic is: t = (x̄ - μ₀) / (s / √n)

   x̄ = 7.1, μ₀ = 8, s = 2.5, n = 25

   t = (7.1 - 8) / (2.5 / √25) = -0.9 / (2.5 / 5) = -0.9 / 0.5 = -1.8

4. Determine the P-value: Using a t-distribution table or calculator with df = 24 for a one-tailed test, the p-value for t = -1.8 falls between 0.025 and 0.05. For example, a t-value of -1.711 corresponds to a tail probability of 0.05, and our t-value is more extreme. A precise calculator gives a p-value of approximately 0.042.

5. Make a Decision and Conclude:

   Our p-value (≈0.042) is less than our significance level (0.10). Therefore, we reject the null hypothesis.

   Conclusion: At the 10% significance level, there is enough evidence to conclude that the average student study time is less than 8 hours per week.

Example 3: One-Sample Z-Test for a Proportion

A recent poll found that 62% of voters in a city supported a new public transit initiative. A local politician wants to test if the support has changed. They commission a new poll of 500 voters, and 325 of them express support. Using α = 0.05, is there evidence that the proportion of support has changed?

1. State the Hypotheses:

   • Null Hypothesis (H₀): p = 0.62 (The proportion of support has not changed).

   • Alternative Hypothesis (H₁): p ≠ 0.62 (The proportion of support has changed). This is a two-tailed test.

2. Set the Significance Level: α = 0.05.

3. Calculate the Test Statistic: First, find the sample proportion (p̂).

   p̂ = 325 / 500 = 0.65

   Now, calculate the z-test statistic for proportions: Z = (p̂ - p₀) / √((p₀ * (1 - p₀)) / n)

   Z = (0.65 - 0.62) / √((0.62 * 0.38) / 500) = 0.03 / √(0.2356 / 500) = 0.03 / √0.0004712 ≈ 0.03 / 0.0217 ≈ 1.38

4. Determine the P-value: Since this is a two-tailed test, we need to find the probability in both tails: P(Z < -1.38) + P(Z > 1.38). From a z-table, P(Z > 1.38) ≈ 0.0838. So, the two-tailed p-value is 2 * 0.0838 = 0.1676.

5. Make a Decision and Conclude:

   Our p-value (0.1676) is greater than our significance level (0.05). Therefore, we fail to reject the null hypothesis.

   Conclusion: At the 5% significance level, there is not sufficient evidence to conclude that the proportion of voter support for the transit initiative has changed from 62%.

Practice Questions

Now it's your turn. Use the principles of hypothesis testing to solve these problems. Questions are grouped by approximate difficulty.

Easy

1. The national average score on a standardized test is 1500 with a population standard deviation of 200. An administrator at a particular school district claims their students score higher than the national average. A random sample of 100 students from this district has a mean score of 1550. At α = 0.05, is there evidence to support the administrator's claim?

2. A coffee shop claims that its large latte contains an average of 16 fluid ounces. A consumer watchdog group suspects the shop is under-filling the cups. They purchase 36 lattes and find the sample mean is 15.8 ounces with a sample standard deviation of 0.6 ounces. Is there significant evidence at the α = 0.01 level to suggest the lattes are being under-filled?

3. Historically, 15% of applicants for a competitive programming job pass the initial coding screen. This year, the company redesigned the screening test. Out of the first 400 applicants who took the new test, 72 passed. Is there evidence at the α = 0.05 level that the pass rate for the new test is different from the historical rate?

Medium

4. A pharmaceutical company developed a new drug to reduce blood pressure. They claim it reduces systolic blood pressure by an average of 10 mmHg. In a clinical trial with 50 patients, the average reduction was 8.5 mmHg with a standard deviation of 4 mmHg. Is there sufficient evidence to reject the company's claim that the average reduction is 10 mmHg? Use a significance level of 0.05 and perform a two-tailed test.

5. A website claims that its users spend an average of 20 minutes per session. To test this, you collect data from 150 sessions and calculate a sample mean of 21.5 minutes and a sample standard deviation of 8 minutes. You calculate a test statistic of t = 2.30. What is the p-value for a two-tailed test? Based on this p-value and an α of 0.05, what is your conclusion?

6. A city's environmental agency states that the mean carbon monoxide level in the air is 4.9 parts per million (ppm). A concerned citizens' group collects 30 air samples over a month and finds a sample mean of 5.4 ppm and a sample standard deviation of 1.1 ppm. They believe the mean level is higher than what the agency states. At α = 0.01, do the data support the citizens' group? For more on how samples relate to populations, see our guide on sampling distributions.

7. An e-commerce company believes that more than 25% of its website visitors add an item to their cart. In a sample of 1200 visitors, 324 added an item to their cart. Perform a hypothesis test at the α = 0.05 significance level to check this belief.

Hard

8. A researcher is testing two different fertilizers on corn yield. Fertilizer A was used on 40 plots, resulting in a mean yield of 160 bushels/acre with a standard deviation of 10. Fertilizer B was used on 50 plots, resulting in a mean yield of 165 bushels/acre with a standard deviation of 12. Assuming unequal variances, is there a significant difference in the mean yields between the two fertilizers at α = 0.05?

9. A study reports a p-value of 0.045 for a one-tailed hypothesis test with H₁: μ > 50 and α = 0.05. Separately, another researcher conducts the same study but formulates it as a two-tailed test (H₁: μ ≠ 50) using the same data. What would the p-value be for the two-tailed test, and what would be the conclusion at α = 0.05?

Answers & Explanations

Below are the detailed solutions to the practice questions on hypothesis testing.

1. Answer & Explanation: Yes, there is evidence to support the administrator's claim.

1. Hypotheses: H₀: μ = 1500; H₁: μ > 1500.

2. Significance Level: α = 0.05.

3. Test Statistic (z-test): Z = (1550 - 1500) / (200 / √100) = 50 / (200 / 10) = 50 / 20 = 2.5.

4. P-value: For a right-tailed test, we find P(Z > 2.5). From a z-table, this is 1 - P(Z < 2.5) = 1 - 0.9938 = 0.0062.

5. Conclusion: Since the p-value (0.0062) is less than α (0.05), we reject the null hypothesis. There is significant evidence to conclude that the students in this district score higher than the national average.

2. Answer & Explanation: Yes, there is significant evidence the lattes are under-filled.

1. Hypotheses: H₀: μ = 16; H₁: μ < 16.

2. Significance Level: α = 0.01.

3. Test Statistic (t-test): Population σ is unknown. df = 36 - 1 = 35. t = (15.8 - 16) / (0.6 / √36) = -0.2 / (0.6 / 6) = -0.2 / 0.1 = -2.0.

4. P-value: For a left-tailed t-test with df=35, the p-value for t = -2.0 is approximately 0.026. (Using a calculator for precision).

5. Conclusion: Oops, let's re-check the decision. Since the p-value (0.026) is greater than α (0.01), we fail to reject the null hypothesis. There is not significant evidence at the 1% level to conclude the lattes are under-filled. Correction: Rerunning with the critical value approach: The critical t-value for α=0.01 and df=35 (one-tail) is approx -2.438. Since our t-statistic (-2.0) is not less than the critical value (-2.438), we fail to reject H₀. The initial conclusion was incorrect. The data do not support the claim at the strict 0.01 level.

3. Answer & Explanation: Yes, there is evidence the pass rate is different.

1. Hypotheses: H₀: p = 0.15; H₁: p ≠ 0.15.

2. Significance Level: α = 0.05.

3. Test Statistic (z-test for proportion): Sample proportion p̂ = 72 / 400 = 0.18. Z = (0.18 - 0.15) / √((0.15 * 0.85) / 400) = 0.03 / √(0.1275 / 400) = 0.03 / 0.01785 ≈ 1.68.

4. P-value: For a two-tailed test, p-value = 2 * P(Z > 1.68) = 2 * (1 - 0.9535) = 2 * 0.0465 = 0.093.

5. Conclusion: Wait, let's re-evaluate. Since the p-value (0.093) is greater than α (0.05), we fail to reject the null hypothesis. There is not sufficient evidence to conclude that the pass rate for the new test is different from the historical rate. The conclusion above was hasty.

4. Answer & Explanation: No, there is not sufficient evidence to reject the company's claim.

1. Hypotheses: H₀: μ = 10; H₁: μ ≠ 10.

2. Significance Level: α = 0.05.

3. Test Statistic (t-test): df = 50 - 1 = 49. t = (8.5 - 10) / (4 / √50) = -1.5 / (4 / 7.071) = -1.5 / 0.5657 ≈ -2.65.

4. P-value: For a two-tailed t-test with df=49 and t=-2.65, the p-value is 2 * P(T < -2.65) ≈ 2 * 0.0055 = 0.011.

5. Conclusion: Since the p-value (0.011) is less than α (0.05), we reject the null hypothesis. There is sufficient evidence to conclude that the average blood pressure reduction is not 10 mmHg.

5. Answer & Explanation: The p-value is approx. 0.0228. We reject the null hypothesis.

1. Hypotheses: H₀: μ = 20; H₁: μ ≠ 20.

2. Significance Level: α = 0.05.

3. Test Statistic (t-test): Given as t = 2.30. The degrees of freedom are df = 150 - 1 = 149.

4. P-value: We need to find the two-tailed p-value for t = 2.30 with df = 149. Using a t-distribution calculator, the p-value is approximately 0.0228.

5. Conclusion: Since the p-value (0.0228) is less than α (0.05), we reject the null hypothesis. There is significant evidence to conclude that the average session time is different from 20 minutes.

6. Answer & Explanation: Yes, the data support the citizens' group's concern.

1. Hypotheses: H₀: μ = 4.9; H₁: μ > 4.9.

2. Significance Level: α = 0.01.

3. Test Statistic (t-test): df = 30 - 1 = 29. t = (5.4 - 4.9) / (1.1 / √30) = 0.5 / (1.1 / 5.477) = 0.5 / 0.2008 ≈ 2.49.

4. P-value: For a right-tailed t-test with df=29 and t=2.49, the p-value is found using a t-table or calculator. It is approximately 0.009.

5. Conclusion: Since the p-value (0.009) is less than α (0.01), we reject the null hypothesis. There is strong evidence to support the claim that the mean carbon monoxide level is higher than 4.9 ppm.

7. Answer & Explanation: Yes, there is evidence that more than 25% of visitors add an item to their cart.

1. Hypotheses: H₀: p = 0.25; H₁: p > 0.25.

2. Significance Level: α = 0.05.

3. Test Statistic (z-test for proportion): Sample proportion p̂ = 324 / 1200 = 0.27. Z = (0.27 - 0.25) / √((0.25 * 0.75) / 1200) = 0.02 / √(0.1875 / 1200) = 0.02 / √0.00015625 = 0.02 / 0.0125 = 1.6.

4. P-value: For a right-tailed test, P(Z > 1.6) = 1 - P(Z < 1.6) = 1 - 0.9452 = 0.0548.

5. Conclusion: Since the p-value (0.0548) is greater than α (0.05), we fail to reject the null hypothesis. There is not sufficient evidence at the 5% significance level to conclude that more than 25% of visitors add an item to their cart. It is very close, but does not meet the threshold.

8. Answer & Explanation: Yes, there is a significant difference in mean yields.

1. Hypotheses: H₀: μ₁ = μ₂ (no difference); H₁: μ₁ ≠ μ₂ (there is a difference).

2. Significance Level: α = 0.05.

3. Test Statistic (two-sample t-test, unequal variances): The formula is t = (x̄₁ - x̄₂) / √((s₁²/n₁) + (s₂²/n₂)). t = (160 - 165) / √((10²/40) + (12²/50)) = -5 / √((100/40) + (144/50)) = -5 / √(2.5 + 2.88) = -5 / √5.38 = -5 / 2.319 ≈ -2.156.

4. P-value: The degrees of freedom calculation (Welch-Satterthwaite equation) is complex, but statistical software would yield df ≈ 93. For a two-tailed test with t = -2.156 and df ≈ 93, the p-value is approx. 0.033.

5. Conclusion: Since the p-value (0.033) is less than α (0.05), we reject the null hypothesis. There is significant evidence to conclude that there is a difference in the mean corn yields between the two fertilizers.

9. Answer & Explanation: The minimum sample size required is n ≈ 246.

1. Setup: This requires the sample size formula for a one-sided proportion test: n = [(Zα√(p₀(1-p₀)) + Zβ√(p₁(1-p₁))) / (p₁ - p₀)]².

2. Values:

   • α = 0.05 → Zα = 1.645 (for one-tailed test)

   • β = 0.20 → Zβ = 0.84 (for 80% power)

   • p₀ = 0.50 (null hypothesis proportion)

   • p₁ = 0.55 (true proportion we want to detect)

9. Answer & Explanation: The two-tailed p-value would be 0.09, and the researcher would fail to reject the null hypothesis.

1. Understanding P-values: The given one-tailed p-value of 0.045 represents the area in one tail of the distribution (in this case, the right tail since H₁: μ > 50).

2. Calculating Two-Tailed P-value: A two-tailed test considers the possibility of an extreme result in either direction. Therefore, to get the two-tailed p-value from a one-tailed result, you simply double it (assuming the test statistic is not zero). p-value (two-tailed) = 2 * p-value (one-tailed) = 2 * 0.045 = 0.09.

3. Conclusion: The researcher is using α = 0.05. Since the new p-value (0.09) is greater than α (0.05), the researcher would fail to reject the null hypothesis. This illustrates how the choice of a one-tailed vs. two-tailed test can change the conclusion, a key concept discussed by sources like Wikipedia's article on the topic.

Quick Quiz

Frequently Asked Questions

What is the difference between a null and alternative hypothesis?

The null hypothesis (H₀) is the default assumption of no effect, no change, or no difference. The alternative hypothesis (H₁) is the claim you are trying to find evidence for, representing a change, an effect, or a difference. All testing is done under the assumption that the null hypothesis is true, and we look for strong evidence to reject it in favor of the alternative.

What does a p-value of 0.03 mean?

A p-value of 0.03 means there is a 3% chance of observing your sample data (or something more extreme) if the null hypothesis were actually true. Since this probability is quite low, it suggests that the observed data is unlikely to have occurred by random chance alone, providing evidence against the null hypothesis.

What is a Type I error versus a Type II error?

A Type I error occurs when you reject a true null hypothesis (a "false positive"). The probability of committing a Type I error is equal to the significance level (α). A Type II error occurs when you fail to reject a false null hypothesis (a "false negative"). The probability of a Type II error is denoted by β. There is a trade-off between these two types of errors; decreasing the chance of one generally increases the chance of the other. For a deeper dive, Penn State's Statistics department offers excellent explanations.

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

You should use a t-test when the population standard deviation (σ) is unknown and you must estimate it using the sample standard deviation (s). A z-test is appropriate when the population standard deviation is known or when you have a very large sample size (often cited as n > 30), as the t-distribution approaches the normal distribution. This same logic applies when constructing confidence intervals.

What is a significance level (alpha)?

The significance level, denoted by α (alpha), is the pre-determined probability of rejecting the null hypothesis when it is actually true. It acts as the threshold for statistical significance. If the p-value of your test is less than or equal to α, you reject the null hypothesis; otherwise, you fail to reject it. Common choices for α are 0.05 (5%), 0.01 (1%), and 0.10 (10%).

What's the difference between one-tailed and two-tailed tests?

A one-tailed test is used when the alternative hypothesis specifies a direction (e.g., μ > 50 or μ < 50). It tests for an effect in only one direction. A two-tailed test is used when the alternative hypothesis does not specify a direction (e.g., μ ≠ 50). It tests for an effect in either direction, an increase or a decrease.

Want unlimited practice questions like these?

Generate AI-powered questions with step-by-step solutions on any topic.

Try Question Generator Free →

Enjoyed this article?

Share it with others who might find it helpful.

Share Article