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Chapter 12: Problem 16

The polling organization Ipsos conducted telephone surveys in March of 2004,2005, and 2006 . In each year, 1001 people age 18 or older were asked about whether they planned to use a credit card to pay federal income taxes that year. The data given in the accompanying table are from the report "Fees Keeping Taxpayers from Using Credit Cards to Make Tax Payments" (IPSOS Insight, March 24, 2006 ). Is there evidence that the proportion falling in the three credit card response categories is not the same for all three years? Test the relevant hypotheses using a .05 significance level. $$ \begin{array}{l|rrr} & 2004 & 2005 & 2006 \\ \hline \text { Definitely/Probably Will } & 40 & 50 & 40 \\ \text { Might/Might Not/Probably Not } & 180 & 190 & 160 \\ \text { Definitely Will Not } & 781 & 761 & 801 \\ \hline \end{array} $$

Short Answer

Expert verified
To answer the question, carry out a chi-square test of independence to check the hypothesis about the distribution of responses over the years. If the p-value from the test is less than the 0.05 significance level, reject the null hypothesis, indicating a significant difference in the distribution of responses over the years. Conversely, if the p-value is greater than 0.05, there is insufficient evidence to reject the null hypothesis, meaning the distribution of responses is the same for each year.

Step by step solution

01

State the Hypotheses

Firstly, state the null and alternative hypotheses: \n\nThe null hypothesis \(H_0\): The responses are independent of the years. Meaning there is no difference in distribution for different years. \n\nThe alternative hypothesis \(H_a\): The responses are not independent of the years. Meaning there is a difference in distribution for different years.
02

Calculate the Expected Frequencies

Next, calculate the expected frequencies for each cell under the assumption of null hypothesis: \n\n The expected frequency is calculated by \(E = (row \ total * column \ total) / grand \ total\).
03

Calculate Test Statistic

The test statistic for a Chi-square test is calculated by \(\chi_{obs}^2 = \sum \frac{(O - E)^2}{E}\), where \(O\) is the observed frequency and \(E\) is the expected frequency. Calculate this for each cell and sum up all to get the final test statistic.
04

Determine the P-value

Determine the p-value associated with the observed value of the test statistic. If this calculated p-value is less than the significance level (0.05), reject the null hypothesis in favor of the alternative.
05

Draw the Conclusion

Based on the p-value result, draw conclusion about the hypotheses. If p-value < 0.05, reject the null hypothesis indicating a significant difference in the distribution of responses among years. Conversely, if p-value > 0.05, there is not enough evidence to reject the null hypothesis, so the distribution of responses is the same for each year.

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Key Concepts

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

Null and alternative hypotheses
When attempting to make sense of data using a Chi-square test, it's crucial to start with stating the null and alternative hypotheses. The null hypothesis (\(H_0\)) proposes that there is no effect or difference. In the context of the Ipsos survey, our null hypothesis is that the proportion of responses falls equally into the three credit card categories across the different years. Conversely, the alternative hypothesis (\(H_a\text{ or }H_1\text{ in some texts}\text{ which should be read as }H sub 1\)) suggests that the distribution of responses does differ between years, implying a potential trend or change in taxpayer credit card use over time.
Expected frequency calculation
The Chi-square test compares the expected frequencies 鈥� the frequencies we would expect if the null hypothesis were true 鈥� with the observed frequencies from the collected data. To calculate expected frequency (\(E\text{ (expected)}\)), we use the formula \(E = (row \text{ total }\times column \text{ total}) / grand \text{ total}\text{ (sum of all observations)}\). It's like sprinkling the overall frequencies evenly across the tabulated categories based on how large each row and column is. For the Ipsos survey data, we would calculate the expected frequency for each year and response category combination to see if the actual data (observed frequencies) align with what we would predict under no change scenario.
Test statistic calculation
Once we have our expected frequencies, the next step is to calculate the Chi-square test statistic (\(\chi^2\text{ (read as 'chi-square')}\)). This involves summing the squared difference between each observed frequency (\(O\text{ (observed)}\)) and its corresponding expected frequency, divided by the expected frequency: \[ \chi^2 = \sum \frac{(O - E)^2}{E} \.\] This formula gives us a single number that represents the aggregate deviation of observed data from what we'd expect if the null hypothesis were true. A larger value of the test statistic indicates a greater disparity between observed and expected data, which may lead to the rejection of the null hypothesis.
P-value determination
The p-value is a crucial concept in hypothesis testing. After calculating the test statistic, we determine the p-value, which tells us the probability of observing our data (or something more extreme) assuming the null hypothesis is true. If this p-value is less than the chosen significance level (usually 0.05), we have grounds to reject the null hypothesis. In simpler terms, a low p-value suggests that our observed data is quite unusual under the assumption of no effect or no difference (the null hypothesis) and may indicate that the alternative hypothesis is more likely to be true.
Statistical significance
Finally, we assess the statistical significance, which relates to whether the test results from our Chi-square analysis are due to chance or if they reflect a true effect in our studied population. If our p-value is below the threshold significance level (commonly 0.05), we say the results are statistically significant. This means there's a less than 5% probability that the observed distribution happened just by chance, under the null hypothesis. For the Ipsos survey data, concluding statistical significance would imply that credit card use for tax payments has indeed shifted significantly over the years examined.

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Most popular questions from this chapter

The article "Linkage Studies of the Tomato" (Transactions of the Royal Canadian Institute [1931]:1-19) reported the accompanying data on phenotypes resulting from crossing tall cut-leaf tomatoes with dwarf potato-leaf tomatoes. There are four possible phenotypes: (1) tall cut-leaf, (2) tall potato-leaf, (3) dwarf cutleaf, and (4) dwarf potato-leaf. $$ \begin{array}{c|cccc} & {\text { Phenotype }} \\ & 1 & 2 & 3 & 4 \\ \hline \text { Frequency } & 926 & 288 & 293 & 104 \\ \hline \end{array} $$ Mendel's laws of inheritance imply that \(p_{1}=9 / 16, p_{2}=\) \(3 / 16, p_{3}=3 / 16\), and \(p_{4}=1 / 16\). Are the data from this experiment consistent with Mendel's laws? Use a .01 significance level.

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.

A certain genetic characteristic of a particular plant can appear in one of three forms (phenotypes). A researcher has developed a theory, according to which the hypothesized proportions are \(p_{1}=.25, p_{2}=.50\), and \(p_{3}=.25 .\) A random sample of 200 plants yields \(X^{2}=\) \(4.63 .\) a. Carry out a test of the null hypothesis that the theory is correct, using level of significance \(\alpha=.05\). b. Suppose that a random sample of 300 plants had resulted in the same value of \(X^{2}\). How would your analysis and conclusion differ from those in Part (a)?

From the given information in each case below, state what you know about the \(P\) -value for a chi-square test and give the conclusion for a significance level of \(\alpha=.01\). a. \(X^{2}=7.5, \mathrm{df}=2\) b. \(X^{2}=13.0, \mathrm{df}=6\) c. \(\quad X^{2}=18.0, \mathrm{df}=9\) d. \(\quad X^{2}=21.3, \mathrm{df}=4\) e. \(\quad X^{2}=5.0, \mathrm{df}=3\)

The authors of the paper "Movie Character Smoking and Adolescent Smoking: Who Matters More, Good Guys or Bad Guys?" (Pediatrics [2009]: 135-141) classified characters who were depicted smoking in movies released between 2000 and \(2005 .\) The smoking characters were classified according to sex and whether the character type was positive, negative or neutral. The resulting data is given in the accompanying table. Assume that it is reasonable to consider this sample of smoking movie characters as representative of smoking movie characters. Do the data provide evidence of an association between sex and character type for movie characters who smoke? Use \(\alpha=.05\). $$ \begin{array}{lccc} & & \text { Character Type } \\ \hline \text { Sex } & \text { Positive } & \text { Negative } & \text { Neutral } \\ \hline \text { Male } & 255 & 106 & 130 \\ \text { Female } & 85 & 12 & 49 \\ \hline \end{array} $$
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