/*! 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 90 In an experiment to study the de... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In an experiment to study the dependence of hypertension on smoking habits, the following data were taken on 180 individuals Test the hypothesis that the presence or absence of hy pertension is independent of smoking habits. Use 0.05 level of significance.

Short Answer

Expert verified
Without specific data, only the steps for performing a chi-square independence test can be outlined. Given the data, these steps would help determine if hypertension and smoking habits are dependent or independent with a significance level of 0.05.

Step by step solution

01

State the hypotheses

The null hypothesis (H0) is that smoking habits and hypertension are independent. The alternative hypothesis (Ha) is that smoking habits and hypertension are not independent.
02

Set up the data into a contingency table

We have two categories (smoking habits and hypertension). We arrange these in a table to count the instances of each category and combination. However, as no specific data is given, we can only note down this step without actual execution.
03

Compute expected frequencies

For each cell in the table, calculate the expected frequency. The expected frequency is calculated as (row total * column total) / overall total.
04

Calculate chi-square test statistic

For each cell in the table, calculate (\(Observed - Expected)^2/Expected\). Sum these numbers to get the chi-square test statistic. This again needs concrete data to be executed.
05

Calculate degrees of freedom and lookup critical value

Degrees of freedom for a contingency table is given by (Number of rows - 1) * (Number of columns - 1). For a 0.05 significance level, and given degrees of freedom, the critical chi-square value can be looked up in a chi-square distribution table.
06

Compare and conclude

If the calculated chi-square test statistic is greater than the critical value, reject the null hypothesis. If it is less, do not reject the null hypothesis. This means that if the null is rejected, hypertension and smoking habits are not independent at the 0.05 significance level. Conversely, if the null is not rejected, there's not enough evidence to say that the two variables are dependent.

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.

Null Hypothesis
The null hypothesis in statistical analysis represents a statement of no effect or no difference - it's our starting assumption. When dealing with a chi-square test for independence, such as examining the relationship between hypertension and smoking habits, we posit the null hypothesis (\( H_0 \) ) to assert that there is no association between the two variables. We're essentially saying that if we observe any connection in our sample data, this could easily have happened by chance in a population where hypertension and smoking habits are truly unrelated.

In our case, the null hypothesis is that the presence or absence of hypertension is independent of smoking habits. To determine if we should reject this hypothesis, we conduct the chi-square test. If the test results suggest that our observed data is significantly different from what we would expect under the null hypothesis, then we would consider rejecting it in favor of the alternative hypothesis, which claims that there is an association between hypertension and smoking.
Contingency Table
A contingency table, also known as a cross-tabulation or crosstab, is a type of table used in statistics to summarize the relationship between several categorical variables. In the chi-square test for independence, we use a contingency table to display the frequency distribution of the variables.

For instance, to explore the relationship between smoking habits and hypertension, we would create a two-dimensional table. Each row of the table represents a category of one variable (e.g., presence or absence of hypertension), and each column represents a category of the other variable (e.g., smoker or non-smoker). The cells of the table are filled with the observed frequencies - the actual counts of individuals who fall into each combination of categories.
Expected Frequency
Expected frequency is a crucial concept in running a chi-square test for independence. It represents the number of observations that one would expect in each cell of the contingency table if the null hypothesis were true - meaning if the variables are indeed independent of each other.

To calculate the expected frequency for each cell, you multiply the total for the row by the total for the column and then divide by the overall total. The formula looks like this: \( Expected\text{ }Frequency = \frac{(Row\text{ }Total \times Column\text{ }Total)}{Grand\text{ }Total} \)

These expected frequencies are then compared with the observed frequencies to see if there are any significant deviations, which could indicate dependence between the variables.
Degrees of Freedom
Degrees of freedom (df) in statistics represent the number of values in a calculation that are free to vary. When conducting a chi-square test, the degrees of freedom depend on the number of categories in each variable. Specifically, for a contingency table, the degrees of freedom are calculated as: \( df = (Number\text{ }of\text{ }rows - 1) \times (Number\text{ }of\text{ }columns - 1) \)

This number is critical because it helps us determine the appropriate critical value from the chi-square distribution that our test statistic needs to exceed to reject the null hypothesis. The degrees of freedom reflect the sample size and the number of constraints imposed upon the sample data. In our exercise with smoking habits and hypertension, if we had two categories each for smoking and hypertension, then we'd have \( df = (2 - 1) \times (2 - 1) = 1 \) degree of freedom.
Significance Level
The significance level, denoted as \( alpha \), is a threshold that we set before conducting a hypothesis test to decide whether we will reject the null hypothesis. It's the probability of making the error of rejecting the null hypothesis when it is actually true (Type I error).

A common choice for the significance level is 0.05, meaning there is a 5% chance of concluding that there is an effect or difference when there is none. When we conduct a chi-square test, if our calculated test statistic exceeds the critical chi-square value that corresponds to our chosen significance level and degrees of freedom, we then reject the null hypothesis. This would indicate that our findings are statistically significant and not likely a result of random chance.

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

A random sample of 400 voters in a certain city are asked if they favor an additional \(4 \%\) gasoline sales tax to provide badly needed revenues for street repairs. If more than 220 but fewer than 260 favor the sales \(\operatorname{tax}\), we shall conclude that \(60 \%\) of the voters are for it. (a) Find the probability of committing a type I error if \(60 \%\) of the voters favor the increased tax. (b) What is the probability of committing a type II error using this test procedure if actually only \(48 \%\) of the voters are in favor of the additional gasoline \(\operatorname{tax}^{7}\)

\( \quad \wedge\) machine is supposed to mix peanuts, hazelnuts, cashews, and pecans in the ratio 5: 2: 2: 1 . A can containing 500 of these mixed nuts was found to have 269 peanuts, 112 hazelnuts, 74 cashews, and 45 pecans. At the 0.05 level of significance, test the hypothesis that the machine is mixing the nuts in the ratio \(5: 2 ; 2: 1\).

In a controlled laboratory experiment, scientists at the University of Minnesota discovered that \(25 \%\) of a certain strain of rats subjected to a \(20 \%\) coffee bean diet and then force-fed a powerful cancer-causing chemical later developed cancerous tumors. Would we have reason to believe that the proportion of rats developing tumors when subjected to this diet has increased if the experiment were repeated and 16 of 48 rats developed tumors? Use a 0.05 level of significance.

In Exercise 1.19 on page \(28,\) test the goodness of fit between the observed class frequencies and the corresponding expected frequencies of a normal distribution with \(p=1.8\) and \(a=0.4,\) using a 0.01 level of significance.

10.36 \text { A large automobile manufacturing company is } trying to decide whether to purchase brand \(A\) or brand \(B\) tires for its new models. To help arrive at a decision, an experiment is conducted using 12 of each brand. The tires are run until they wear out. The results are $$\begin{aligned}\text { Brand } A: & x_{1}=37,900 \text { kilometers, } \\ & s_{1}=5,100 \text { kilometers. } \\\\\text { Brand B: } \quad x_{1} &=39,800 \text { kilometers, } \\\& s_{2}=5,900 \text { kilometers. }\end{aligned}$$ Test the hypothesis that there is no difference in the average wear of 2 brands of tires. Assume the populations to be approximately normally distributed with equal variances. Use a P-value.

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.