/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 114 Did survival rate for passengers... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Did survival rate for passengers on the Titanic really depend on the type of ticket they had? Following are the data for the 2201 people on board listed by whether they survived and what type of ticket they had. Does survival appear to be independent of ticket class? (Test the hypothesis at \(\alpha=0.05 .)\) What is the \(P\) -value of the test statistic? $$\begin{array}{lccccc} & \text { Crew } & \text { First } & \text { Second } & \text { Third } & \text { Total } \\\\\hline \text { Alive } & 212 & 202 & 118 & 178 & 710 \\\\\text { Dead } & 673 & 123 & 167 & 528 & 1491 \\\\\hline \text { Total } & 885 & 325 & 285 & 706 & 2201\end{array}$$

Short Answer

Expert verified
Survival is not independent of ticket class; the p-value is likely less than 0.05.

Step by step solution

01

Define the Hypotheses

We are testing whether survival is independent of ticket class. The null hypothesis (\(H_0\)) states that survival is independent of ticket class, while the alternative hypothesis (\(H_1\)) states that survival depends on ticket class.
02

Calculate Expected Frequencies

To find the expected frequency for each cell, use the formula \(E = \frac{ ( ext{Row Total} imes ext{Column Total} )}{ ext{Grand Total} }\). For example, the expected frequency for "Alive, Crew" is \(E = \frac{(710)(885)}{2201} \approx 285.37\). Calculate similarly for all cells.
03

Compute the Chi-square Test Statistic

The Chi-square test statistic is calculated using \( \chi^2 = \sum \frac{(O - E)^2}{E} \), where \(O\) is the observed frequency, and \(E\) is the expected frequency. Add up the results of all cells.
04

Determine Degrees of Freedom

Degrees of freedom for this test is calculated as \(( ext{Number of Rows} - 1 ) \times ( ext{Number of Columns} - 1 )\). Here, \( df = (2-1)(4-1) = 3\).
05

Find the Critical Value and P-value

Using the Chi-square distribution table and \(\alpha = 0.05\), the critical value for 3 degrees of freedom is approximately 7.815. Compare the calculated \(\chi^2\) to this value. Also, find the p-value using a calculator or software.
06

Make a Decision

If the calculated \(\chi^2\) is greater than the critical value or if the p-value is less than \(\alpha = 0.05\), reject \(H_0\). Otherwise, do not reject \(H_0\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Hypothesis Testing
Hypothesis testing is a fundamental aspect of statistics that helps us make decisions based on data analysis. In the context of this exercise, we want to determine if a passenger's survival on the Titanic was influenced by the type of ticket they held. To do this, we set up two hypotheses:
  • The **null hypothesis** (\(H_0\)) suggests that survival is independent of the ticket class. In other words, the class of ticket a passenger had does not influence their likelihood of survival.
  • The **alternative hypothesis** (\(H_1\)) suggests that survival depends on the ticket class.
If evidence strongly contradicts the null hypothesis, we reject it in favor of the alternative hypothesis. This evidence is judged through statistical tests, which calculate probabilities (like the p-value) to guide decision-making. A common threshold used to determine statistical significance is \(\alpha = 0.05\), meaning there's only a 5% chance that the observed data could occur if the null hypothesis is true. Initiating hypothesis testing involves these logical steps and calculations to help us interpret complex data.
Degrees of Freedom
In statistics, degrees of freedom (\(df\)) refers to the number of values in a calculation that are free to vary. It's crucial in determining the validity of statistical tests like the Chi-square test. In the Titanic example, we calculate degrees of freedom to determine the appropriate Chi-square distribution against which the test statistic will be compared. To compute this, the formula for degrees of freedom in a contingency table is \((\text{Number of Rows} - 1) \times (\text{Number of Columns} - 1)\).Given:
  • Number of rows: 2 (representing "Alive" and "Dead" categories)
  • Number of columns: 4 (representing Crew, First, Second, and Third ticket classes)
This gives us degrees of freedom as \((2-1)(4-1) = 3\). This degree of freedom is the basis for determining the critical value in the Chi-square distribution, which in turn helps us decide if our observed results significantly differ from the expected ones.
P-value
The p-value is a metric used to determine the significance of statistical results in hypothesis testing. It helps us understand the strength of evidence against the null hypothesis. For the Titanic survival data, after calculating the Chi-square test statistic, we find the p-value to understand how likely it is to observe such extreme data if the null hypothesis were true. The smaller the p-value, the stronger the evidence against the null hypothesis.Here's how it is used:
  • If the p-value is less than or equal to \(\alpha = 0.05\), the evidence against the null hypothesis is strong enough to reject it. This suggests a relationship between ticket class and survival.
  • If the p-value is greater than \(\alpha = 0.05\), it indicates insufficient evidence to reject the null hypothesis, pointing to an assumption that ticket class and survival are independent.
A proper understanding and interpretation of the p-value are key in drawing meaningful conclusions from hypothesis tests.
Expected Frequency
Expected frequency is the hypothesized distribution of data across different categories if the null hypothesis were true. It aids in determining if observed data significantly deviates from what is "expected" when variables are independent. Here's how to compute it:The formula for expected frequency in a contingency table is \(E = \frac{(\text{Row Total} \times \text{Column Total})}{\text{Grand Total}}\).For each cell of the Titanic data table, calculate expected frequencies to compare against observed data:
  • For "Alive, Crew" class: \(E = \frac{(710)(885)}{2201} \approx 285.37\)
  • Repeat this for all other combinations in the table.
Comparing expected frequencies with observed frequencies using the Chi-square formula helps determine how divergent our data are from what might be expected if survival and ticket class were independent. This process is central to evaluating whether the data supports rejecting or retaining the null hypothesis.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Patients in a hospital are classified as surgical or medical. A record is kept of the number of times patients require nursing service during the night and whether or not these patients are on Medicare. The data are presented here: $$\begin{array}{ccc} &{\text { Patient Category }} \\\\\hline \text { Medicare } & \text { Surgical } & \text { Medical } \\\\\text { Yes } & 46 & 52 \\\\\text { No } & 36 & 43\end{array}$$ Test the hypothesis (using \(\alpha=0.01\) ) that calls by surgicalmedical patients are independent of whether the patients are receiving Medicare. Find the \(P\) -value for this test.

Consider the test of \(H_{0}: \sigma^{2}=5\) against \(H_{1}: \sigma^{2}<5 .\) Approximate the \(P\) -value for each of the following test statistics. (a) \(x_{0}^{2}=25.2\) and \(n=20\) (b) \(x_{0}^{2}=15.2\) and \(n=12\) (c) \(x_{0}^{2}=4.2\) and \(n=15\)

The life in hours of a battery is known to be approximately normally distributed with standard deviation \(\sigma=1.25\) hours. A random sample of 10 batteries has a mean life of \(\bar{x}=40.5\) hours. (a) Is there evidence to support the claim that battery life exceeds 40 hours? Use \(\alpha=0.05 .\) (b) What is the \(P\) -value for the test in part (a)? (c) What is the \(\beta\) -error for the test in part (a) if the true mean life is 42 hours? (d) What sample size would be required to ensure that \(\beta\) does not exceed 0.10 if the true mean life is 44 hours? (e) Explain how you could answer the question in part (a) by calculating an appropriate confidence bound on life.

An article in Food Chemistry ["A Study of Factors Affecting Extraction of Peanut (Arachis Hypgaea L.) Solids with Water" (1991, Vol. \(42(2),\) pp. \(153-165)\) ] reported that the percent protein extracted from peanut milk as follows: $$\begin{array}{llllllll}78.3 & 77.1 & 71.3 & 84.5 & 87.8 & 75.7 & 64.8 & 72.5 \\\78.2 & 91.2 & 86.2 & 80.9 & 82.1 & 89.3 & 89.4 & 81.6\end{array}$$ (a) Can you support a claim that the mean percent protein extracted exceeds 80 percent? Use \(\alpha=0.05\) (b) Is there evidence that the percent protein extracted is normally distributed? (c) What is the \(P\) -value of the test statistic computed in part (a)?

When \(X_{1}, X_{2}, \ldots, X_{n}\) are independent Poisson random variables, each with parameter \(\lambda,\) and \(n\) is large, the sample mean \(\bar{X}\) has an approximate normal distribution with mean \(\lambda\) and variance \(\lambda / n .\) Therefore, $$Z=\frac{\bar{X}-\lambda}{\sqrt{\lambda / n}}$$ has approximately a standard normal distribution. Thus, we can test \(H_{0}: \lambda=\lambda_{0}\) by replacing \(\lambda\) in \(Z\) by \(\lambda_{0} .\) When \(X_{i}\) are Poisson variables, this test is preferable to the large-sample test of Section \(9-2.3,\) which would use \(S / \sqrt{n}\) in the denominator because it is designed just for the Poisson distribution. Suppose that the number of open circuits on a semiconductor wafer has a Poisson distribution. Test data for 500 wafers indicate a total of 1038 opens. Using \(\alpha=0.05,\) does this suggest that the mean number of open circuits per wafer exceeds \(2.0 ?\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.