CIs & Hypothesis Testing - IB Mathematics Applications and Interpretation

A pharmaceutical company has developed a new drug to reduce blood pressure. The current standard drug reduces systolic blood pressure by an average of 15 mmHg. The company wants to test if the new drug is more effective than the standard drug. Let $\mu$ be the true mean reduction in systolic blood pressure for the new drug. How should the null and alternative hypotheses be formulated?
Answer: $H_0: \mu = 15$ , $H_1: \mu > 15$ The null hypothesis ( $H_0$ ) is a statement of no effect or no difference, and it always contains the equality. In this case, it's that the new drug has the same effectiveness as the old one, so $H_0: \mu = 15$ . The alternative hypothesis ( $H_1$ ) is what the researchers want to find evidence for, which is that the new drug is more effective. A greater reduction in blood pressure means a value larger than 15. Therefore, the alternative hypothesis is $H_1: \mu > 15$ . This is a one-tailed (right-tailed) test. A describes a two-tailed test, which would check if the drug is simply different (more or less effective). C and D incorrectly place the inequality in the null hypothesis.
A city planner wants to test if the mean commute time for a certain suburb is different from the national average of 28 minutes. A 95% confidence interval for the mean commute time in the suburb is calculated from a sample of 100 commuters and is found to be [28.8, 31.2] minutes. What is the conclusion of a two-tailed hypothesis test with $H_0: \mu = 28$ at a significance level of $\alpha = 0.05$ ?
Answer: Reject $H_0$ because the value 28 is not contained within the 95% confidence interval. There is a direct relationship between a two-tailed hypothesis test and a confidence interval. A two-tailed test with significance level $\alpha$ will reject $H_0: \mu = \mu_0$ if and only if $\mu_0$ is not in the $(1-\alpha)$ confidence interval. Here, $\alpha = 0.05$ , so we consider the $1-0.05=0.95$ , or 95%, confidence interval. The hypothesized value is $\mu_0 = 28$ . The 95% confidence interval is [28.8, 31.2]. Since 28 is not inside this interval, we reject the null hypothesis. A and D provide incorrect reasoning. The decision is based on whether the value is in the interval, not how narrow the interval is or how close the bounds are. C is incorrect because the conclusion can be directly inferred from the confidence interval.
An automotive company claims that its new hybrid car model has a mean fuel efficiency of at least 50 miles per gallon (mpg). A consumer agency tests a sample of 40 cars and finds a sample mean of 48.9 mpg. The agency wants to test the company's claim. They perform a hypothesis test and find that if they had incorrectly performed a two-tailed test ( $H_1: \mu \neq 50$ ), the p-value would be 0.024. What is the correct p-value for the appropriate one-tailed test ( $H_1: \mu < 50$ )?
Answer: 0.012 The company's claim is $\mu \geq 50$ . The agency wants to find evidence against this, so the alternative hypothesis is $H_1: \mu < 50$ . This is a one-tailed (left-tailed) test. The sample mean (48.9) is in the direction of the alternative hypothesis. For a symmetric distribution like the t-distribution, the p-value for a one-tailed test is half the p-value for a two-tailed test, provided the sample result is in the direction specified by $H_1$ . Therefore, the correct p-value is $0.024 / 2 = 0.012$ . B is the p-value for the incorrect two-tailed test. C would be double the two-tailed p-value, which is incorrect. D would be $1 - 0.012$ , which would be the p-value if the sample mean had been above 50.
A farmer tests two new types of fertilizer, FertiMax and GrowFast, on separate, randomly assigned plots of corn. The yield in kg is recorded. For 20 plots using FertiMax, the mean yield was 450 kg with a standard deviation of 30 kg. For 22 plots using GrowFast, the mean yield was 475 kg with a standard deviation of 35 kg. A two-sample t-test is conducted to determine if there is a significant difference in mean yields. What is the approximate p-value for this test?
Answer: 0.015 This scenario involves two independent samples, so a two-sample t-test is appropriate. The hypotheses are $H_0: \mu_1 = \mu_2$ and $H_1: \mu_1 \neq \mu_2$ (a two-tailed test). Using a graphing calculator's two-sample t-test function with the following inputs: $\bar{x}_1 = 450$ , $s_1 = 30$ , $n_1 = 20$ , $\bar{x}_2 = 475$ , $s_2 = 35$ , $n_2 = 22$ , and selecting the 'unpooled' option, the calculator gives a p-value of approximately 0.015. A, 0.008, is approximately the p-value for a one-tailed test (p/2), which would be incorrect as the question asks for a 'significant difference', implying a two-tailed test. C is an incorrect calculation. D would be derived from $1-p$ , which is not how p-values are interpreted in this context.
A company manufactures bolts and uses a hypothesis test to check if the mean diameter meets the specification of 8.00 mm. The null hypothesis is $H_0: \mu = 8.00$ . They set the significance level $\alpha$ to 0.01. What is the meaning of this significance level?
Answer: It is the probability of concluding the mean diameter is not 8.00 mm when it actually is 8.00 mm. The significance level, $\alpha$ , is the probability of making a Type I error. A Type I error is the rejection of a true null hypothesis. In this context, the null hypothesis is that the mean diameter is 8.00 mm. Rejecting this hypothesis means concluding the mean diameter is not 8.00 mm. Therefore, $\alpha = 0.01$ is the probability of concluding the mean is not 8.00 mm when, in fact, it is. A is an incorrect interpretation of probability in this context. C describes the probability of a Type II error ( $\beta$ ). D confuses the significance level of a hypothesis test with a statement about the proportion of individual items meeting a specification.
A psychologist is studying reaction times for a specific task. They collect data from a sample of 8 participants. The data contains a significant outlier and the distribution appears to be heavily skewed. A one-sample t-test is performed to compare the mean reaction time to a known value. The p-value is 0.04. What is the most significant concern about this conclusion?
Answer: The presence of an outlier and skewness in a small sample may invalidate the results of the t-test, which relies on an assumption of normality. The t-test relies on the assumption that the underlying population is normally distributed, or that the sample size is large enough for the Central Limit Theorem to apply (typically $n \geq 30$ ). With a very small sample size ( $n=8$ ), the test is not robust to violations of normality. The presence of a significant outlier and heavy skewness suggests the normality assumption is violated, which can make the p-value from the t-test unreliable. A is incorrect; t-tests are specifically for small samples. B is incorrect because it ignores the violation of the test's assumptions. D is incorrect because the problem describes a one-sample scenario, and changing the test type doesn't fix the data distribution issue.
A clinical trial is conducted to see if a new medication reduces cholesterol levels. After the trial, a hypothesis test yields a p-value of 0.048. The pre-determined significance level for the study was $\alpha = 0.05$ . Which of the following is the most appropriate and nuanced conclusion?
Answer: The p-value of 0.048 is less than 0.05, so the null hypothesis is rejected. This provides statistically significant evidence that the medication is effective. Statistical conclusions should be stated with appropriate language. Since p (0.048) < $\alpha$ (0.05), the result is statistically significant, and we reject the null hypothesis. Option B correctly uses the phrases 'statistically significant evidence' and 'reject the null hypothesis'. A and D use overly strong words like 'proven' and 'accept...as fact'. Hypothesis testing provides evidence, not proof. C confuses statistical significance with practical significance. A result is statistically significant if p < $\alpha$ , regardless of how close it is to the threshold. Whether the effect is large enough to be practically meaningful is a separate question.
A new medical test is developed to screen for a certain disease. The null hypothesis is that the patient does not have the disease. A clinical trial is conducted, and for a particular patient, the test result is negative, leading the doctors to fail to reject the null hypothesis. If a Type II error occurred, what is the situation?
Answer: The patient has the disease, but the test result was negative. A Type II error occurs when we fail to reject a false null hypothesis. The null hypothesis ( $H_0$ ) is 'the patient does not have the disease'. Failing to reject $H_0$ corresponds to a negative test result. For this to be an error, the null hypothesis must be false, which means the patient actually does have the disease. Therefore, a Type II error is when a patient who has the disease gets a negative test result (a false negative). B is a correct negative result. C describes a Type I error (a false positive), where a healthy patient gets a positive result. D is a correct positive result.
An economist wants to compare the mean starting salaries for graduates in two different fields: Engineering and Business. A random sample of 30 recent Engineering graduates has a mean salary of $68,000 with a standard deviation of $5,000. A random sample of 35 recent Business graduates has a mean salary of $64,000 with a standard deviation of $4,500. What is the 99% confidence interval for the difference between the mean salaries ( $\mu_{Eng} - \mu_{Bus}$ )?
Answer: [$817, $7183] This requires a two-sample t-interval for the difference between two means. Using a graphing calculator, enter the statistics for the two samples: Group 1 (Eng): $\bar{x}_1 = 68000, s_1 = 5000, n_1 = 30$ Group 2 (Bus): $\bar{x}_2 = 64000, s_2 = 4500, n_2 = 35$ Set the confidence level to 0.99 and select the 'unpooled' option (as we don't assume equal variances). The calculator will compute the interval for $\mu_1 - \mu_2$ . The resulting 99% confidence interval is approximately [$817, $7183]. C is the 95% confidence interval, a common error if the wrong confidence level is used. B is the interval that would be calculated if the 'pooled' option were incorrectly selected, which assumes the population variances are equal. D is an incorrect calculation.
A manufacturer claims that its chocolate bars weigh an average of 50.0 g. A quality control inspector takes a random sample of 25 bars and finds the sample mean weight is 49.6 g with a sample standard deviation of 1.2 g. A t-test is conducted to test if the true mean weight is different from 50.0 g. The resulting p-value is 0.108. Using a significance level of $\alpha = 0.05$ , what is the correct conclusion?
Answer: Fail to reject $H_0$ . There is insufficient evidence to conclude that the mean weight is different from 50.0 g. The decision rule in hypothesis testing is to compare the p-value to the significance level ( $\alpha$ ). If p-value $\leq \alpha$ , we reject the null hypothesis ( $H_0$ ). If p-value $> \alpha$ , we fail to reject $H_0$ . In this case, the p-value is 0.108, and $\alpha$ is 0.05. Since 0.108 > 0.05, we fail to reject $H_0$ . This means we do not have enough statistical evidence to say the mean weight is different from 50.0 g. A is incorrect because p > $\alpha$ . C is incorrect for the same reason, and it also uses overly strong language ('proves'). D is incorrect because failing to reject $H_0$ does not prove that $H_0$ is true. It only means we lack sufficient evidence to say it's false.
A fitness center designs a new 8-week workout program. To test its effectiveness, they measure the resting heart rate of 12 participants before the program and after it completes. They want to determine if the program significantly reduces resting heart rate. Let $d$ be the difference in heart rate (before - after). The mean of these differences is $\bar{d} = 3.5$ beats per minute and the standard deviation of the differences is $s_d = 2.8$ . What is the value of the t-statistic for the appropriate hypothesis test?
Answer: 4.33 This is a paired-sample design because two measurements are taken on each participant. The hypothesis test is a one-sample t-test on the differences. The null hypothesis is $H_0: \mu_d = 0$ . The formula for the t-statistic is $t = \frac{\bar{d} - \mu_0}{s_d / \sqrt{n}}$ . Here, $\bar{d} = 3.5$ , $\mu_0 = 0$ , $s_d = 2.8$ , and $n = 12$ . Plugging in the values: $t = \frac{3.5}{2.8 / \sqrt{12}} \approx \frac{3.5}{0.808} \approx 4.33$ . A results from incorrectly calculating $t = \bar{d} / s_d = 3.5 / 2.8 = 1.25$ , omitting the $\sqrt{n}$ term. B and D are incorrect calculations.
A coffee shop claims the average volume of its espresso shots is 30.0 mL. A consumer group measures a random sample of 15 shots and finds a mean volume of 29.2 mL with a sample standard deviation of 1.5 mL. A 99% confidence interval is constructed for the true mean volume. Based on this interval, is the coffee shop's claim plausible?
Answer: Yes, the claim is plausible because the 99% confidence interval is approximately [28.05, 30.35], which contains 30.0 mL. Using a graphing calculator or statistical software to find the t-interval for a mean with $\bar{x}=29.2$ , $s=1.5$ , $n=15$ , and a confidence level of 0.99, we get the interval [28.05, 30.35]. Since the claimed mean of 30.0 mL is within this interval, we consider the claim plausible at the 99% confidence level. We do not have sufficient evidence to reject it. B is incorrect because it relies only on the sample mean and ignores sampling variability, which the confidence interval accounts for. C uses an incorrect interval, possibly calculated with a 90% confidence level, which would be [28.52, 29.88]. This interval does not contain 30.0, leading to an incorrect conclusion. D is incorrect because the width of the interval is a result of the data's variability and the high confidence level, but it does not invalidate the conclusion drawn from it.
An environmental scientist calculates a 95% confidence interval for the mean concentration of a pollutant in a lake based on a sample of $n$ measurements. Which of the following changes would most likely result in a narrower confidence interval?
Answer: Increasing the sample size to $2n$ . The width of a confidence interval is determined by the margin of error, which is calculated as $t_{crit} \times \frac{s}{\sqrt{n}}$ . To make the interval narrower, we need to decrease the margin of error. Increasing the sample size $n$ to $2n$ increases the denominator, which decreases the margin of error and thus narrows the interval. A is incorrect because decreasing the sample size would increase the margin of error, making the interval wider. B is incorrect because increasing the confidence level increases the critical value ( $t_{crit}$ ), which makes the interval wider. C is incorrect because changing the sample mean shifts the center of the interval but does not affect its width.
An e-commerce company runs an experiment on its website with 500,000 users to test a new 'buy' button color. The null hypothesis is that the new color has no effect on the click-through rate. The test yields a p-value of 0.0001, and the 95% confidence interval for the increase in click-through rate is [0.01%, 0.03%]. Which is the best interpretation of these results?
Answer: The result is statistically significant, but the magnitude of the effect is very small, so the business impact may not be meaningful. This question tests the difference between statistical significance and practical significance. The very small p-value (<0.05) indicates a statistically significant result; we have strong evidence that the new button color has an effect. However, the confidence interval [0.01%, 0.03%] shows that the size of this effect is very small. An increase of just 0.01% might not be enough to justify the cost of making the change. Therefore, the result is statistically significant but may not be practically important. A ignores the small effect size shown by the confidence interval. B is incorrect; a CI for the difference that does not contain zero is consistent with a small p-value. D is incorrect. A large sample size gives more power to detect even small effects; it does not make the result 'artificial'.
A study is conducted to compare the performance of two types of running shoes, A and B. Ten runners are selected, and each runner completes a 5 km run with shoe A and, on a different day, a 5 km run with shoe B. The times are recorded. A paired t-test is performed to see if there is a significant difference in running times. The resulting p-value is 0.027. Using a 5% significance level, what is the most appropriate conclusion?
Answer: At the 5% significance level, there is sufficient evidence to suggest a significant difference in mean running times between the two shoe types. The significance level is $\alpha = 0.05$ . The p-value is 0.027. Since the p-value (0.027) is less than the significance level (0.05), we reject the null hypothesis. Rejecting the null hypothesis means we have found statistically significant evidence of a difference. Therefore, there is sufficient evidence to suggest a difference in mean running times. A is incorrect because p < $\alpha$ . B is a common misinterpretation of the p-value. The p-value is the probability of observing the data (or more extreme data) if the null hypothesis were true, not the probability that the null hypothesis is true. D is incorrect because the study design, where each runner tests both shoes, is a classic example of a paired design. A paired t-test is the appropriate choice.
A researcher is conducting a one-sample t-test with the hypotheses $H_0: \mu = 100$ and $H_1: \mu > 100$ . The sample mean is $\bar{x} = 105$ and the sample size is $n=25$ . If the sample standard deviation were larger, how would this affect the t-statistic and the p-value, assuming all other values remain constant?
Answer: The t-statistic would decrease, and the p-value would increase. The formula for the t-statistic is $t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$ . The sample standard deviation, $s$ , is in the denominator. If $s$ increases, the value of the entire denominator increases. This causes the value of the t-statistic to decrease. A smaller t-statistic indicates that the sample mean is fewer standard errors away from the null hypothesis value. This means the result is less surprising or less extreme. A less extreme result corresponds to a larger p-value. Therefore, the t-statistic would decrease, and the p-value would increase.
An educational researcher conducts a study to determine if a new teaching method improves student test scores. The null hypothesis is that the new method has no effect on scores. After the study, the researcher rejects the null hypothesis at a 5% significance level. What would a Type I error represent in this context?
Answer: Concluding that the new method improves scores when, in reality, it does not. A Type I error occurs when we reject a true null hypothesis ( $H_0$ ). In this context, $H_0$ is 'the new method has no effect'. Rejecting $H_0$ means concluding the method does have an effect (it improves scores). Therefore, a Type I error is concluding the method improves scores when it actually does not. B describes a Type II error (failing to reject a false $H_0$ ). C describes a correct decision. D relates to the concept of statistical power, which is linked to Type II errors, not Type I.
A paired t-test is used to determine if the fertilizer caused a significant increase in height. What is the p-value for this test?
Answer: 0.0058 This is a paired t-test for an increase, so it is a one-tailed test. First, calculate the differences (After - Before) for each plant: {2, 1, 1, 1, 1, 3, 0}. The null hypothesis is that the mean difference is zero ( $H_0: \mu_d = 0$ ), and the alternative is that the mean difference is greater than zero ( $H_1: \mu_d > 0$ ). Input the list of differences into a calculator and perform a one-sample t-test. The mean of the differences is $\bar{d} \approx 1.286$ and the standard deviation is $s_d \approx 0.951$ . The calculator yields a p-value of approximately 0.0058. B, 0.0116, is the p-value for a two-tailed test, which would be incorrect here. C, 0.0591, is the p-value that would be obtained if an incorrect two-sample (unpaired) t-test were performed on the two lists of data. D is $1 - p$ , which is an incorrect calculation.
A tech company tests a sample of 30 new smartphones and finds a 95% confidence interval for the mean battery life to be [10.2, 11.8] hours. What is the correct interpretation of this interval?
Answer: If many random samples of 30 smartphones were taken, 95% of the calculated confidence intervals would contain the true mean battery life. The correct frequentist interpretation of a 95% confidence interval is that it is a method that, in the long run, produces intervals that capture the true population parameter 95% of the time. It is a statement about the reliability of the method, not the probability of a specific interval containing the parameter. A is incorrect because the interval is for the population mean, not the range of individual data points in the sample. B is a common misconception. In frequentist statistics, the true mean is a fixed, unknown constant, so it does not make sense to talk about the probability of it being in a specific interval. D is an incorrect interpretation that confuses the concept of confidence with time.

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