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The paper referenced in the previous exercise also gave the accompanying data on the age at which smoking started for a sample of 1031 men who smoked low- tar cigarettes. $$ \begin{array}{cc} \text { Age } & \text { Frequency } \\ \hline<16 & 237 \\ 16-17 & 258 \\ 18-20 & 320 \\ \geq 21 & 216 \\ \hline \end{array} $$ a. Use a chi-square goodness-of-fit test to test the null hypothesis \(H_{0}: p_{1}=.25, p_{2}=.2, p_{3}=.3, p_{4}=.25\) where \(p_{1}=\) proportion of male low-tar cigarette smokers who started smoking before age 16, and \(p_{2}\), \(p_{3}\), and \(p_{4}\) are defined in a similar way for the other three age groups. b. The null hypothesis from Part (a) specifies that half of male smokers of low-tar cigarettes began smoking between the ages of 16 and \(20 .\) Explain why \(p_{2}=.2\) and \(p_{3}=.3\) is consistent with the ages between 16 and 20 being equally likely to be when smoking started.

Short Answer

Expert verified
The calculated Chi-Square value was 21.64, which is greater than the critical value of 7.81. Hence, we reject the null hypothesis and conclude that the actual distribution does not match the hypothesized distribution. Furthermore, the null hypothesis is consistent since the sum of the proportions for the age groups between 16 and 20 equals to 0.5, which corresponds to the claim that half of the total smokers started smoking between 16 and 20.

Step by step solution

01

Compute Expected Frequencies

Calculate the expected frequency for each age group under the null hypothesis. This is done by multiplying the total sample size by the hypothesized proportion for each group. Total sample size is 1031. Thus, Expected Frequency for group <16 is \(1031 * 0.25 = 257.75\), for group 16-17 is \(1031 * 0.2 = 206.2\), for group 18-20 is \(1031 * 0.3 = 309.3\), andfor group ≥21 is \(1031 * 0.25 = 257.75\)
02

Perform Chi-Square Calculations

The chi-square statistic is computed using the formula \(\sum (O - E)^2 / E\), where O is the observed frequency and E is the expected frequency. We will calculate it for each group then sum up all the values. Chi-square for group <16 is \( ((237 - 257.75)^2)/ 257.75 = 1.66 \),for group 16-17 is \( ((258 - 206.2)^2)/ 206.2 = 13.00 \),for group 18-20 is \( ((320 - 309.3)^2)/ 309.3 = 0.36 \), for group ≥21 is \( ((216 - 257.75)^2)/ 257.75 = 6.62\). Total Chi-square is \((1.66+13.00+0.36+6.62)=21.64\)
03

Compare Chi-Square Statistic to Critical Value

With 3 degrees of freedom (4 categories - 1), the critical value of Chi-square for a significance level of 0.05 is about 7.81. Our computed Chi-square of 21.6 is greater than this, so we reject the null hypothesis.
04

Explain the Hypothesized Proportions

It is given that half of the smokers started smoking between ages 16 and 20. That interval contains two groups, 16-17 and 18-20. The sum of the proportions for these two groups is \(0.2 + 0.3 = 0.5\), or half, which align with what was given. Hence, if the age to start smoking were equally likely between 16 and 20, then the proportions of people starting in those groups would add to 0.5.

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Key Concepts

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

Expected Frequency
In a Chi-Square Goodness-of-Fit Test, 'Expected Frequency' is a key term. It refers to the number that is predicted or expected for each category based on a specific probability distribution under the null hypothesis.
To find this, we multiply the total sample size by the proportion specified in the hypothesis for each category. In our smoking example, we have aged-based proportions like 0.25 for both the age groups \(<16\) and \( \geq21\). These proportions help us predict how many smokers we expect to fall into each age group if the null hypothesis is true.
So, if we have a total of 1031 smokers, the expected frequencies are calculated as follows:

  • For age group \(<16\): \(257.75\) males \( (1031 \times 0.25=257.75) \).
  • For age group \(16-17\): \(206.2\) males \( (1031 \times 0.2=206.2) \).
  • For age group \(18-20\): \(309.3\) males \( (1031 \times 0.3=309.3) \).
  • For age group \( \geq21\): \(257.75\) males, the same as the first group.
These expected frequencies provide a baseline to compare against the actually observed data, revealing how closely the observed data fits the expected distribution.
Null Hypothesis
The null hypothesis in a Chi-Square Goodness-of-Fit Test is an essential logical statement. It posits that there is no significant difference between the observed frequencies and those expected by the predicted distribution of categories.
In simpler terms, it claims that any difference we see between what we expect and what we observe is due to random chance and not because the distribution is wrong.
In our cigarette smoking age exercise, the null hypothesis \( (H_0) \) is defined by the proportions \( p_1, p_2, p_3, \) and \( p_4 \). These proportions represent the assumption that the age groups \(<16\), \(16-17\), \(18-20\), and \( \geq21\) make up 25%, 20%, 30%, and 25% of the sample, respectively.
By establishing these proportions, the null hypothesis makes a clear statement: the actual ages at which people start smoking follow this predicted pattern. By testing this hypothesis, we can understand if our assumptions about these proportions are correct or if there's a significant deviation suggesting some other pattern.
Observed Frequency
When conducting a Chi-Square Goodness-of-Fit Test, 'Observed Frequency' is what you actually see in the data you are analyzing. It's the count of occurrences within each category from your samples. Observed frequencies are critical because they help you determine whether there is a significant difference from the expected frequencies.
In our example, the observed frequencies are the number of males falling into different age groups:\(<16, 16-17, 18-20,\) and \( \geq21\). We have:
  • \(<16\): 237 males
  • \(16-17\): 258 males
  • \(18-20\): 320 males
  • \(\geq21\): 216 males
With these numbers, we can structure the Chi-Square calculation, looking for variations between what's observed and what's expected under the null hypothesis. Any significant deviation could mean our original assumptions aren't holding up well to scrutiny.
Degrees of Freedom
Understanding the 'Degrees of Freedom' (df) is crucial when performing statistical tests like the Chi-Square Goodness-of-Fit test. Simply put, degrees of freedom refer to the number of values that are free to vary in the final calculation of statistics after certain restrictions are applied. For this test, it's calculated as the number of categories minus one.
In our smoking example, there are four age categories (\(<16, 16-17, 18-20, \geq21\)). Thus, the degrees of freedom is \(4-1=3\).
Why does this matter? The degrees of freedom are used to determine the critical value from the Chi-Square distribution table. This critical value sets the threshold to decide whether to reject the null hypothesis. In our test case, with 3 degrees of freedom at a 0.05 significance level, the critical value is approximately 7.81. Our calculated Chi-Square statistic must be compared against this threshold to determine the hypothesis's validity. Hence, understanding degrees of freedom is integral to making informed decisions based on your statistical test outcomes.

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Most popular questions from this chapter

The paper "Sociochemosensory and Emotional Functions" (Psychological Science [2009]: \(1118-1124\) ) describes an interesting experiment to determine if college students can identify their roommates by smell. Forty-four female college students participated as subjects in the experiment. Each subject was presented with a set of three t-shirts that were identical in appearance. Each of the three t-shirts had been slept in for at least 7 hours by a person who had not used any scented products (like scented deodorant, soap, or shampoo) for at least 48 hours prior to sleeping in the shirt. One of the three shirts had been worn by the subject's roommate. The subject was asked to identify the shirt worn by her roommate. This process was then repeated with another three shirts, and the number of times out of the two trials that the subject correctly identified the shirt worn by her roommate was recorded. The resulting data is given in the accompanying table. $$ \begin{array}{l|ccc} \hline \text { Number of Correct Identifications } & 0 & 1 & 2 \\ \hline \text { Observed Count } & 21 & 10 & 13 \\ \hline \end{array} $$ a. Can a person identify her roommate by smell? If not, the data from the experiment should be consistent with what we would have expected to see if subjects were just guessing on each trial. That is, we would expect that the probability of selecting the correct shirt would be \(1 / 3\) on each of the two trials. It would then be reasonable to regard the number of correct identifications as a binomial variable with \(n=2\) and \(p=1 / 3\). Use this binomial distribution to compute the proportions of the time we would expect to see 0,1, and 2 correct identifications if subjects are just guessing. b. Use the three proportions computed in Part (a) to carry out a test to determine if the numbers of correct identifications by the students in this study are significantly different than what would have been expected by guessing. Use \(\alpha=.05 .\) (Note: One of the expected counts is just a bit less than \(5 .\) For purposes of this exercise, assume that it is \(\mathrm{OK}\) to proceed with a goodness-of-fit test.)

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