Q1: A researcher wants to test whether the average height of 9th graders has increased from the previously known average of 165 cm. What are the correct null and alternative hypotheses? (a) \(H_0: \mu = 165\), \(H_a: \mu \neq 165\) (b) \(H_0: \mu = 165\), \(H_a: \mu > 165\) (c) \(H_0: \mu > 165\), \(H_a: \mu = 165\) (d) \(H_0: \mu \neq 165\), \(H_a: \mu = 165\)
Solution:
Ans: (b) Explanation: The null hypothesis \(H_0\) always states that there is no change, so \(\mu = 165\). Since the researcher wants to test if height has increased, the alternative hypothesis \(H_a\) should be \(\mu > 165\), making this a one-tailed test. Option (a) would be for a two-tailed test, while options (c) and (d) incorrectly place the equality in the alternative hypothesis.
Q2: In hypothesis testing, a Type I error occurs when: (a) We reject the null hypothesis when it is actually true (b) We fail to reject the null hypothesis when it is actually false (c) We accept the alternative hypothesis when it is true (d) We correctly reject the null hypothesis
Solution:
Ans: (a) Explanation: A Type I error is also called a false positive. It occurs when we incorrectly reject a true null hypothesis. Option (b) describes a Type II error, while options (c) and (d) describe correct decisions, not errors.
Q3: If the significance level is set at \(\alpha = 0.05\), what does this mean? (a) There is a 5% chance the null hypothesis is true (b) We are willing to accept a 5% chance of making a Type I error (c) There is a 95% chance our conclusion is wrong (d) The p-value must be exactly 0.05
Solution:
Ans: (b) Explanation: The significance level \(\alpha\) represents the maximum probability of making a Type I error that we are willing to accept. Setting \(\alpha = 0.05\) means we accept a 5% risk of rejecting the null hypothesis when it is actually true. Option (a) incorrectly interprets \(\alpha\) as the probability of \(H_0\) being true, option (c) confuses confidence level with error rate, and option (d) incorrectly states the p-value must equal \(\alpha\).
Q4: A p-value of 0.03 in a hypothesis test means: (a) The probability that the null hypothesis is true is 3% (b) The probability of observing data as extreme as ours, assuming the null hypothesis is true, is 3% (c) There is a 97% chance our alternative hypothesis is correct (d) We must reject the alternative hypothesis
Solution:
Ans: (b) Explanation: The p-value is the probability of obtaining test results at least as extreme as the observed results, assuming the null hypothesis is true. A p-value of 0.03 means there is a 3% probability of seeing such extreme data if \(H_0\) is true. Option (a) is a common misconception; the p-value does not tell us the probability that \(H_0\) is true. Option (c) incorrectly interprets the p-value as supporting \(H_a\), and option (d) is incorrect because a small p-value leads us to reject \(H_0\), not \(H_a\).
Q5: A test statistic is calculated to be \(z = 2.1\). For a two-tailed test with \(\alpha = 0.05\), the critical values are \(z = \pm 1.96\). What decision should be made? (a) Fail to reject the null hypothesis (b) Reject the null hypothesis (c) Accept the alternative hypothesis as proven (d) Reduce the significance level
Solution:
Ans: (b) Explanation: Since the test statistic \(z = 2.1\) falls outside the range of \(-1.96\) to \(1.96\) (it exceeds the positive critical value), it falls in the rejection region. Therefore, we reject the null hypothesis. Option (a) would be correct only if the test statistic fell within the critical values. Option (c) uses incorrect terminology; we never "prove" or "accept" hypotheses in statistics. Option (d) is not a valid decision based on test results.
Q6: Which of the following would increase the power of a hypothesis test? (a) Decreasing the sample size (b) Increasing the significance level from 0.01 to 0.05 (c) Using a two-tailed test instead of a one-tailed test (d) Decreasing the effect size
Solution:
Ans: (b) Explanation: The power of a test is the probability of correctly rejecting a false null hypothesis (1 - probability of Type II error). Increasing \(\alpha\) from 0.01 to 0.05 makes it easier to reject \(H_0\), thus increasing power. Option (a) decreases power by reducing information, option (c) decreases power by splitting the rejection region, and option (d) makes the effect harder to detect, decreasing power.
Q7: In testing \(H_0: \mu = 50\) versus \(H_a: \mu \neq 50\), a researcher obtains a p-value of 0.12 with \(\alpha = 0.05\). The correct conclusion is: (a) Reject \(H_0\); the population mean is definitely not 50 (b) Reject \(H_0\); there is strong evidence against it (c) Fail to reject \(H_0\); there is insufficient evidence to conclude the mean differs from 50 (d) Accept \(H_0\); the population mean equals 50
Solution:
Ans: (c) Explanation: Since the p-value (0.12) is greater than the significance level \(\alpha\) (0.05), we fail to reject the null hypothesis. This means there is insufficient evidence to conclude that the mean differs from 50. Option (a) and (b) incorrectly reject \(H_0\) when p > \(\alpha\). Option (d) uses incorrect terminology; we never "accept" \(H_0\), we only fail to reject it.
Q8: A one-tailed test is most appropriate when: (a) We want to test for any difference from the null hypothesis (b) We are only interested in detecting a change in one specific direction (c) The sample size is small (d) The population standard deviation is unknown
Solution:
Ans: (b) Explanation: A one-tailed test is used when the research question specifically asks whether a parameter is greater than or less than a specific value, but not both. We use it when we are only interested in detecting a change in one direction. Option (a) describes a two-tailed test, while options (c) and (d) relate to test conditions, not the choice between one-tailed and two-tailed tests.
## Section B: Fill in the Blanks
Q9: The __________ hypothesis is the statement being tested in a hypothesis test, typically representing no change or no effect.
Solution:
Ans: null Explanation: The null hypothesis (\(H_0\)) is the default assumption in hypothesis testing that represents no change, no difference, or no effect.
Q10: The probability of making a Type II error is denoted by the symbol __________.
Solution:
Ans: \(\beta\) (or beta) Explanation:Beta (\(\beta\)) represents the probability of a Type II error, which occurs when we fail to reject a false null hypothesis.
Q11: If the p-value is less than the significance level \(\alpha\), we __________ the null hypothesis.
Solution:
Ans: reject Explanation: When the p-value is less than \(\alpha\), the evidence against the null hypothesis is strong enough to reject it.
Q12: The __________ region is the range of values for the test statistic that leads to rejection of the null hypothesis.
Solution:
Ans: rejection (or critical) Explanation: The rejection region (also called the critical region) contains all values of the test statistic that would cause us to reject the null hypothesis.
Q13: A __________ test is used when we want to determine if a parameter is different from a specified value in either direction.
Solution:
Ans: two-tailed Explanation: A two-tailed test checks for differences in both directions (greater than or less than) and is used when the alternative hypothesis is \(\neq\).
Q14: The complement of the significance level, expressed as \(1 - \alpha\), is called the __________ level.
Solution:
Ans: confidence Explanation: The confidence level is \(1 - \alpha\). For example, if \(\alpha = 0.05\), the confidence level is 0.95 or 95%.
## Section C: Word Problems
Q15: A school claims that the average time students spend on homework per night is 90 minutes. A teacher believes it is less than this. She surveys 36 students and finds a sample mean of 85 minutes with a sample standard deviation of 12 minutes. Using a significance level of 0.05, she calculates a test statistic of \(z = -2.5\). The critical value for this one-tailed test is \(z = -1.645\). Should the teacher reject the school's claim?
Solution:
Ans: Step 1: Identify the hypotheses: \(H_0: \mu = 90\), \(H_a: \mu <> Step 2: Compare the test statistic to the critical value: \(z = -2.5 <> Step 3: Since the test statistic falls in the rejection region (it is more extreme than the critical value), we reject the null hypothesis. Final Answer: Yes, the teacher should reject the school's claim. There is sufficient evidence at the 0.05 significance level to conclude that students spend less than 90 minutes on homework per night.
Q16: A manufacturer claims that their batteries last an average of 500 hours. A consumer group tests 50 batteries and finds a sample mean of 485 hours with a standard deviation of 40 hours. They conduct a hypothesis test at \(\alpha = 0.01\). The calculated p-value is 0.008. What conclusion should the consumer group make?
Solution:
Ans: Step 1: Compare the p-value to the significance level: \(0.008 <> Step 2: Since the p-value is less than \(\alpha\), we reject the null hypothesis. Step 3: The null hypothesis was that \(\mu = 500\) hours. Final Answer: The consumer group should reject the manufacturer's claim. There is sufficient evidence at the 0.01 significance level to conclude that the batteries do not last an average of 500 hours.
Q17: A researcher is testing whether a new teaching method improves test scores. She sets \(\alpha = 0.05\) and conducts a hypothesis test. After calculations, she obtains a p-value of 0.08. If she rejects the null hypothesis, what type of error might she be making, and what is the probability of making this error given her significance level?
Solution:
Ans: Step 1: Since the p-value (0.08) is greater than \(\alpha\) (0.05), the correct decision is to fail to reject the null hypothesis. Step 2: If she incorrectly rejects the null hypothesis anyway, she would be making a Type I error. Step 3: The probability of a Type I error is equal to \(\alpha\). Final Answer: She should not reject the null hypothesis based on the p-value. However, if she were to reject it, she would be making a Type I error, and the probability of this error is 0.05 or 5%.
Q18: A company claims that at least 80% of customers are satisfied with their service. A survey of 200 customers shows that 150 are satisfied. At a significance level of 0.05, the test statistic is calculated to be \(z = -1.77\). For a one-tailed test with \(\alpha = 0.05\), the critical value is \(z = -1.645\). What conclusion should be drawn?
Solution:
Ans: Step 1: Set up hypotheses: \(H_0: p \geq 0.80\), \(H_a: p <> Step 2: Compare test statistic to critical value: \(z = -1.77 <> Step 3: The test statistic falls in the rejection region. Final Answer: Reject the null hypothesis. There is sufficient evidence at the 0.05 significance level to conclude that less than 80% of customers are satisfied with the service.
Q19: A medical researcher wants to test if a new drug lowers blood pressure. The null hypothesis states that the mean reduction is 0 mmHg, and the alternative states it is greater than 0. After testing 40 patients, she finds a sample mean reduction of 5 mmHg with a standard deviation of 8 mmHg. She calculates a test statistic of \(z = 3.95\). For \(\alpha = 0.01\), the critical value is \(z = 2.33\). What is her conclusion, and what does this mean in practical terms?
Solution:
Ans: Step 1: Compare test statistic to critical value: \(z = 3.95 > 2.33\) Step 2: The test statistic falls in the rejection region. Step 3: Reject the null hypothesis. Final Answer: Reject the null hypothesis. There is strong evidence at the 0.01 significance level that the new drug does lower blood pressure. The mean reduction in blood pressure is significantly greater than 0 mmHg.
Q20: A quality control inspector wants to test if the mean diameter of ball bearings is 10 mm. She uses a two-tailed test with \(\alpha = 0.05\). After measuring 64 ball bearings, she calculates a test statistic of \(z = 1.5\). The critical values are \(z = \pm 1.96\). Additionally, she calculates the p-value to be 0.13. Based on both the critical value method and the p-value method, what should she conclude?
Solution:
Ans: Critical Value Method: The test statistic \(z = 1.5\) falls between \(-1.96\) and \(1.96\), so it does not fall in the rejection region. Therefore, fail to reject \(H_0\). P-value Method: The p-value (0.13) is greater than \(\alpha\) (0.05), so fail to reject \(H_0\). Final Answer: Fail to reject the null hypothesis using both methods. There is insufficient evidence at the 0.05 significance level to conclude that the mean diameter differs from 10 mm.
The document Worksheet (with Solutions): Significance Tests (Hypothesis Testing) is a part of the Grade 9 Course Statistics & Probability.
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