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Chapter 9: Problem 9

Listed below are heights (in.) of mothers and their first daughters. The data are from a journal kept by Francis Galton. (See Data Set 5 "Family Heights" in Appendix B.) Use a \(0.05\) significance level to test the claim that there is no difference in heights between mothers and their first daughters. $$ \begin{array}{|l|c|c|c|c|c|c|c|c|c|c|} \hline \text { Height of Mother } & 68.0 & 60.0 & 61.0 & 63.5 & 69.0 & 64.0 & 69.0 & 64.0 & 63.5 & 66.0 \\ \hline \text { Height of Daughter } & 68.5 & 60.0 & 63.5 & 67.5 & 68.0 & 65.5 & 69.0 & 68.0 & 64.5 & 63.0 \\ \hline \end{array} $$

Short Answer

Expert verified
Based on the t-test at a 0.05 significance level, if H_0 is not rejected, there is not enough evidence to suggest a difference in heights between mothers and daughters.

Step by step solution

01

State the Hypotheses

Formulate the null hypothesis (H_0) and the alternative hypothesis (H_a). H_0: mu_M = mu_D (There is no difference in heights between mothers and daughters).H_a: mu_M e mu_D (There is a difference in heights between mothers and daughters).
02

Calculate the Differences

Find the difference between each mother's height and her daughter's height. Record these differences.68.0 - 68.5 = -0.560.0 - 60.0 = 061.0 - 63.5 = -2.563.5 - 67.5 = -469.0 - 68.0 = 164.0 - 65.5 = -1.569.0 - 69.0 = 064.0 - 68.0 = -463.5 - 64.5 = -166.0 - 63.0 = 3
03

Calculate the Mean and Standard Deviation of Differences

Calculate the mean (D) and standard deviation (s) of the differences.Sum of differences = -0.5 + 0 - 2.5 - 4 + 1 - 1.5 + 0 - 4 - 1 + 3 = -9.5Mean of differences (D) = sum of differences / n = -9.5 / 10 = -0.95Standard deviation (s) is calculated using:s = sqrt((sum((difference - mean_difference)^2))/(n-1))
04

Perform the t-Test Calculation

Use the formula for the t-statistic:t = D / (s / sqrt(n)) = (-0.95) / (s / sqrt(10))Note: Here, you'd replace s with the previously calculated standard deviation.
05

Find the t-Critical Value

Determine the critical t-value from the t-distribution table at the 0.05 significance level for a two-tailed test with df = n-1 = 10 - 1 = 9 degrees of freedom.
06

Compare t-Statistic with t-Critical

Compare the calculated absolute value of the t-statistic with the critical t-value.If |t| > t_critical, reject H_0.If |t| 鈮� t_critical, do not reject H_0.
07

Conclude the Hypothesis Test

Make a conclusion based on the comparison:e.g., Based on the calculated t-statistic and the critical t-value, if H_0 is not rejected, the conclusion is that there is not enough evidence to suggest there is a difference in heights between mothers and daughters at the 0.05 significance level.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

t-test
The t-test is a statistical method used to determine if there is a significant difference between the means of two groups. It helps in understanding whether the differences are likely due to random chance or if they are statistically significant. In this exercise, we are using a paired t-test because we are comparing the heights of mothers and their first daughters. The paired t-test specifically deals with paired samples, where each pair is closely related or matched in some direct way. This test calculates the difference between each paired observation, computes the mean and standard deviation of these differences, and then uses these statistics to determine the t-statistic. The t-statistic is then compared to a critical value from the t-distribution table to decide whether the null hypothesis should be rejected or not.
significance level
The significance level, usually denoted by \( \alpha \) or alpha, represents the probability of making a Type I error, which is rejecting the null hypothesis when it is actually true. In most scientific studies, a significance level of 0.05 is commonly used. This means there is a 5% risk of concluding that a difference exists when there is no actual difference. For this exercise, we use a 0.05 significance level to test the claim that there is no difference in heights between mothers and their first daughters. If our p-value, derived from the t-test statistic, is less than 0.05, we reject the null hypothesis. If it is greater than 0.05, we do not have enough evidence to reject the null hypothesis, suggesting that any observed difference could be due to random variation.
paired samples
Paired samples are two sets of observations that are taken from the same group but at different times or under different conditions. In this problem, we are working with heights of mothers and their first daughters. Each mother's height is paired with her daughter's height, making them a matched pair. The nature of paired samples helps in reducing variability because the differences within each pair are used rather than the individual measurements themselves. This makes paired sample t-tests more powerful than unpaired tests in detecting differences that exist. For paired samples, we calculate the difference between each pair, find the mean and standard deviation of these differences, and then use these to perform a t-test. By focusing on the differences within pairs, this method accounts for any potential pairing-related effects, making the analysis more precise.

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Most popular questions from this chapter

When the author visited Dublin, Ireland (home of Guinness Brewery employee William Gosset, who first developed the \(t\) distribution), he recorded the ages of randomly selected passenger cars and randomly selected taxis. The ages can be found from the license plates. (There is no end to the fun of traveling with the author.) The ages (in years) are listed below. We might expect that taxis would be newer, so test the claim that the mean age of cars is greater than the mean age of taxis. $$ \begin{array}{|l|l|l|l|r|r|r|r|r|r|r|r|r|r|l|l|c|l|l|l|l|l|l|l|l|l|l|l|} \hline \text { Car Ages } & 4 & 0 & 8 & 11 & 14 & 3 & 4 & 4 & 3 & 5 & 8 & 3 & 3 & 7 & 4 & 6 & 6 & 1 & 8 & 2 & 15 & 11 & 4 & 1 & 6 & 1 & 8 \\ \hline \text { Taxi Ages } & 8 & 8 & 0 & 3 & 8 & 4 & 3 & 3 & 6 & 11 & 7 & 7 & 6 & 9 & 5 & 10 & 8 & 4 & 3 & 4 & & & & & & & \\ \hline \end{array} $$

Researchers from the University of British Columbia conducted a study to investigate the effects of color on cognitive tasks. Words were displayed on a computer screen with background colors of red and blue. Results from scores on a test of word recall are given below. Higher scores correspond to greater word recall. a. Use a \(0.05\) significance level to test the claim that the samples are from populations with the same mean. b. Construct a confidence interval appropriate for the hypothesis test in part (a). What is it about the confidence interval that causes us to reach the same conclusion from part (a)? c. Does the background color appear to have an effect on word recall scores? If so, which color appears to be associated with higher word memory recall scores? $$ \begin{array}{|l|l|} \hline \text { Red Background } & n=35, \bar{x}=15.89, s=5.90 \\ \hline \text { Blue Background } & n=36, \bar{x}=12.31, s=5.48 \\ \hline \end{array} $$

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, \(P\) -value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. A study investigated survival rates for in-hospital patients who suffered cardiac arrest. Among 58,593 patients who had cardiac arrest during the day, 11,604 survived and were discharged. Among 28,155 patients who suffered cardiac arrest at night, 4139 survived and were discharged (based on data from "Survival from In-Hospital Cardiac Arrest During Nights and Weekends," by Peberdy et al., Journal of the American Medical Association, Vol. 299, No. 7). We want to use a \(0.01\) significance level to test the claim that the survival rates are the same for day and night. a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval. c. Based on the results, does it appear that for in-hospital patients who suffer cardiac arrest, the survival rate is the same for day and night?

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, \(P\) -value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. A study was conducted to determine the proportion of people who dream in black and white instead of color. Among 306 people over the age of 55 , 68 dream in black and white, and among 298 people under the age of 25,13 dream in black and white (based on data from "Do We Dream in Color?" by Eva Murzyn, Consciousness and Cognition, Vol. 17, No. 4). We want to use a \(0.01\) significance level to test the claim that the proportion of people over 55 who dream in black and white is greater than the proportion of those under 25 . a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval. c. An explanation given for the results is that those over the age of 55 grew up exposed to media that was mostly displayed in black and white. Can the results from parts (a) and (b) be used to verify that explanation?

A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are given in the accompanying table (based on "An Analysis of Factors That Contribute to the Efficacy of Hypnotic Analgesia," by Price and Barber, Journal of Abnormal Psychology, Vol. 96, No. 1 ). The values are before and after hypnosis; the measurements are in centimeters on a pain scale. Higher values correspond to greater levels of pain. Construct a \(95 \%\) confidence interval for the mean of the "before/after" differences. Does hypnotism appear to be effective in reducing pain? $$ \begin{array}{l|c|c|c|c|c|c|c|c|} \hline \text { Subject } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } & \text { F } & \text { G } & \text { H } \\ \hline \text { Before } & 6.6 & 6.5 & 9.0 & 10.3 & 11.3 & 8.1 & 6.3 & 11.6 \\ \hline \text { After } & 6.8 & 2.4 & 7.4 & 8.5 & 8.1 & 6.1 & 3.4 & 2.0 \\ \hline \end{array} $$
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