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Ch. 9 - Inferences from Two Samples
Triola - Elementary Statistics 14th Edition
Triola14th EditionElementary StatisticsISBN: 9780137366446Not the one you use?Change textbook
All textbooksTriola 14th EditionCh. 9 - Inferences from Two SamplesProblem 9.3.1
Chapter 9, Problem 9.3.1

Is Friday the 13th Unlucky? Listed below are numbers of hospital admissions in one region due to traffic accidents on different Fridays falling on the 6th day of a month and the following 13th day of the month (based on data from “Is Friday the 13th Bad for Your Health,” by Scanlon et al., British Medical Journal, Vol. 307). Assume that we want to use a 0.05 significance level to test the claim that the data support the claim that fewer hospital admissions due to traffic accidents occur on Friday the 6th than on the following Friday the 13th. Identify the null hypothesis and alternative hypothesis.


Table comparing hospital admissions on Friday 6th and Friday 13th due to traffic accidents.

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Step 1: Define the null hypothesis (H₀) and the alternative hypothesis (H₁). The null hypothesis states that there is no difference in the number of hospital admissions due to traffic accidents between Friday the 6th and Friday the 13th. Mathematically, H₀: μ₆ = μ₁₃. The alternative hypothesis states that fewer hospital admissions occur on Friday the 6th than on Friday the 13th. Mathematically, H₁: μ₆ < μ₁₃.
Step 2: Identify the type of test to be used. Since we are comparing two sets of paired data (hospital admissions on Friday the 6th and Friday the 13th for the same months), a paired t-test is appropriate for this analysis.
Step 3: Calculate the differences between the paired data points (Friday the 13th admissions minus Friday the 6th admissions). For example, for the first pair, the difference is 13 - 9 = 4. Repeat this for all pairs.
Step 4: Compute the mean and standard deviation of the differences. Use the formulas for the mean (μ) and standard deviation (σ) of a sample. The mean is calculated as the sum of the differences divided by the number of pairs, and the standard deviation is calculated using the formula: σ = sqrt(Σ(dᵢ - μ)² / (n - 1)), where dᵢ represents each difference and n is the number of pairs.
Step 5: Perform the paired t-test. Calculate the t-statistic using the formula: t = μ / (σ / sqrt(n)), where μ is the mean of the differences, σ is the standard deviation of the differences, and n is the number of pairs. Compare the calculated t-statistic to the critical t-value from the t-distribution table at a significance level of 0.05 and degrees of freedom (n - 1). If the calculated t-statistic is less than the critical t-value, reject the null hypothesis; otherwise, fail to reject the null hypothesis.

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Key Concepts

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

Null Hypothesis (H0)

The null hypothesis is a statement that indicates no effect or no difference between groups. In this context, it posits that there is no significant difference in the number of hospital admissions due to traffic accidents between Friday the 6th and Friday the 13th. Formally, it can be stated as H0: μ6th = μ13th, where μ represents the mean number of admissions.
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Step 1: Write Hypotheses

Alternative Hypothesis (H1)

The alternative hypothesis is a statement that contradicts the null hypothesis, suggesting that there is a significant effect or difference. For this scenario, the alternative hypothesis would state that fewer hospital admissions occur on Friday the 6th than on Friday the 13th, expressed as H1: μ6th < μ13th. This hypothesis is what the statistical test aims to support.
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Step 1: Write Hypotheses

Significance Level (α)

The significance level, often denoted as alpha (α), is the threshold for determining whether to reject the null hypothesis. In this case, a significance level of 0.05 indicates that there is a 5% risk of concluding that a difference exists when there is none. If the p-value obtained from the statistical test is less than 0.05, the null hypothesis would be rejected in favor of the alternative hypothesis.
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Step 4: State Conclusion Example 4
Related Practice
Textbook Question

Body Temperatures Listed below are body temperatures from six different subjects measured at two different times in a day (from Data Set 5 “Body Temperatures” in Appendix B).


a. Are the two sets of data independent or dependent? Explain.


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Textbook Question

Robust What does it mean when we say that the F test described in this section is not robust against departures from normality?

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Textbook Question

No Variation in a Sample An experiment was conducted to test the effects of alcohol. Researchers measured the breath alcohol levels for a treatment group of people who drank ethanol and another group given a placebo. The results are given below (based on data from “Effects of Alcohol Intoxication on Risk Taking, Strategy, and Error Rate in Visuomotor Performance,” by Streufert et al., Journal of Applied Psychology, Vol. 77, No. 4). Use a 0.05 significance level to test the claim that the two sample groups come from populations with the same mean.


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Textbook Question

In Exercises 5–8, use (a) randomization and (b) bootstrapping for the indicated exercise from Section 9-1. Compare the results to those obtained in the original exercise.


Exercise 9 in Section 9-1 “Cell Phones and Handedness”


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Textbook Question

Degrees of Freedom For Example 1, we used df=smaller of n1-1 and n2-1 we got df=11 and the corresponding critical value is t=-1.796 (found from Table A-4). If we calculate df using Formula 9-1, we get df=19.370 and the corresponding critical value is t=-1.727 How is using the critical value of t=-1.796 “more conservative” than using the critical value of t=-1.727

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Textbook Question

Degrees of Freedom In Exercise 20 “Blanking Out on Tests,” using the “smaller of n1-1 and n2-1” for the number of degrees of freedom results in df=15 Find the number of degrees of freedom using Formula 9-1. In general, how are hypothesis tests and confidence intervals affected by using Formula 9-1 instead of the “smaller of n1-1 and n2-1 ”?

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