Skip to main content
Ch. 9 - Inferences from Two Samples
Triola - Elementary Statistics 14th Edition
Triola14th EditionElementary StatisticsISBN: 9780137366446Not the one you use?Change textbook
Chapter 9, Problem 9.1.15a

Can Dogs Detect Malaria? A study was conducted to determine whether dogs could detect malaria from socks worn by malaria patients and socks worn by patients without malaria. Among 175 socks worn by malaria patients, the dogs made correct identifications 123 times. Among 145 socks worn by patients without malaria, the dogs made correct identifications 131 times (based on data presented at an annual meeting of the American Society of Tropical Medicine, by principal investigator Steve Lindsay). Use a 0.05 significance level to test the claim of no difference between the two rates of correct responses.


a. Test the claim using a hypothesis test.

Verified step by step guidance
1
Step 1: Define the null and alternative hypotheses. The null hypothesis (H₀) states that there is no difference between the two rates of correct responses (p₁ = p₂). The alternative hypothesis (H₁) states that there is a difference between the two rates of correct responses (p₁ ≠ p₂).
Step 2: Calculate the sample proportions for each group. For socks worn by malaria patients, the sample proportion is p₁ = 123 / 175. For socks worn by patients without malaria, the sample proportion is p₂ = 131 / 145.
Step 3: Compute the pooled proportion (p̂) using the formula: p̂ = (x₁ + x₂) / (n₁ + n₂), where x₁ and x₂ are the number of correct identifications for each group, and n₁ and n₂ are the total number of socks in each group.
Step 4: Calculate the standard error (SE) for the difference in proportions using the formula: SE = √[p̂(1 - p̂)(1/n₁ + 1/n₂)].
Step 5: Compute the test statistic (z) using the formula: z = (p₁ - p₂) / SE. Then, compare the test statistic to the critical value for a two-tailed test at a significance level of 0.05, or use the p-value approach to determine whether to reject or fail to reject the null hypothesis.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
6m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hypothesis Testing

Hypothesis testing is a statistical method used to make decisions about a population based on sample data. It involves formulating a null hypothesis (H0) that represents no effect or difference, and an alternative hypothesis (H1) that indicates the presence of an effect or difference. The goal is to determine whether the observed data provide sufficient evidence to reject the null hypothesis at a specified significance level, often denoted as alpha (α).
Recommended video:
Guided course
06:21
Step 1: Write Hypotheses

Significance Level

The significance level, commonly set at 0.05, is the threshold for determining whether the results of a hypothesis test are statistically significant. It represents the probability of rejecting the null hypothesis when it is actually true (Type I error). If the p-value obtained from the test is less than the significance level, the null hypothesis is rejected, suggesting that there is a statistically significant difference between the groups being compared.
Recommended video:
Guided course
04:46
Step 4: State Conclusion Example 4

Chi-Square Test

The Chi-Square test is a statistical method used to determine if there is a significant association between categorical variables. In this context, it can be applied to compare the rates of correct identifications by dogs for socks worn by malaria patients versus those worn by non-malaria patients. The test calculates the expected frequencies under the null hypothesis and compares them to the observed frequencies to assess whether any differences are statistically significant.
Recommended video:
Guided course
06:34
Step 2: Calculate Test Statistic
Related Practice
Textbook Question

In Exercises 5–20, assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of n1-1 and n2-1)


Queues Listed on the next page are waiting times (seconds) of observed cars at a Delaware inspection station. The data from two waiting lines are real observations, and the data from the single waiting line are modeled from those real observations. These data are from Data Set 30 “Queues” in Appendix B. The data were collected by the author.


a. Use a 0.01 significance level to test the claim that cars in two queues have a mean waiting time equal to that of cars in a single queue.


87
views
Textbook Question

In Exercises 5–20, assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of n1-1 and n2-1)


Bicycle Commuting A researcher used two different bicycles to commute to work. One bicycle was steel and weighed 30.0 lb; the other was carbon and weighed 20.9 lb. The commuting times (minutes) were recorded with the results shown below (based on data from “Bicycle Weights and Commuting Time,” by Jeremy Groves, British Medical Journal).


a. Use a 0.05 significance level to test the claim that the mean commuting time with the heavier bicycle is the same as the mean commuting time with the lighter bicycle.


128
views
Textbook Question

Second-Hand Smoke Samples from Data Set 15 “Passive and Active Smoke” include cotinine levels measured in a group of smokers ( n = 40, x_bar = 172.48 ng/mL, 119.50 ng/mL ) and a group of nonsmokers not exposed to tobacco smoke ( n = 40, x_bar = 16.35 ng/mL, 62.53 ng/mL ). Cotinine is a metabolite of nicotine, meaning that when nicotine is absorbed by the body, cotinine is produced.


a. Use a 0.05 significance level to test the claim that the variation of cotinine in smokers is greater than the variation of cotinine in nonsmokers not exposed to tobacco smoke.

102
views
Textbook Question

In Exercises 5–20, assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of n1-1 and n2-1)


Color and Creativity Researchers from the University of British Columbia conducted trials to investigate the effects of color on creativity. Subjects with a red background were asked to think of creative uses for a brick; other subjects with a blue background were given the same task. Responses were scored by a panel of judges and results from scores of creativity are given below. Higher scores correspond to more creativity. The researchers make the claim that “blue enhances performance on a creative task.”


a. Use a 0.01 significance level to test the claim that blue enhances performance on a creative task.


" style="" width="460">

111
views
Textbook Question

In Exercises 5–16, use the listed paired sample data, and assume that the samples are simple random samples and that the differences have a distribution that is approximately normal.


Measured and Reported Weights Listed below are measured and reported weights (lb) of random female subjects (from Data Set 4 “Measured and Reported” in Appendix B).


a. Use a 0.05 significance level to test the claim that for females, the measured weights tend to be higher than the reported weights.


92
views
Textbook Question

Confidence Interval Assume that we want to use the sample data in Exercise 1 for constructing a confidence interval to be used for testing the given claim.


a. What is the confidence level that should be used for the confidence interval?


128
views