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Chapter 6: Problem 164

Quebec vs Texas Secession In Example 6.4 on page 408 we analyzed a poll of 800 Quebecers, in which \(28 \%\) thought that the province of Quebec should separate from Canada. Another poll of 500 Texans found that \(18 \%\) thought that the state of Texas should separate from the United States. \({ }^{38}\) (a) In the sample of 800 people, about how many Quebecers thought Quebec should separate from Canada? In the sample of 500 , how many Texans thought Texas should separate from the US? (b) In these two samples, what is the pooled proportion of Texans and Quebecers who want to separate? (c) Can we conclude that the two population proportions differ? Use a two- tailed test and interpret the result.

Short Answer

Expert verified
About 224 Quebecers and 90 Texans want their regions to secede. The pooled proportion is approximately 0.2415 or 24.15%. The z-score is approximately 3.82, which is greater than 3, suggesting that the population proportions do differ significantly.

Step by step solution

01

Calculation of Sample Proportions

First, calculate the number of people who want their respective regions to secede. In Quebec, \(28\%\) of 800 people thought Quebec should separate from Canada. This can be calculated as \((28/100)*800 \approx 224\). Similarly, in Texas, \(18\%\) of 500 people thought Texas should separate from the US. This can be calculated as \((18/100)*500 \approx 90\). The calculated numbers are the estimates of people in both samples who want their regions to secede.
02

Pooled Proportion Calculation

To calculate the pooled proportion, the total number of successes from both samples is computed and then divided by the total size of both samples. The total number of people who want their regions to secede is \(224+90 = 314\). The total sample size is \(800 + 500 = 1300\). Hence the pooled proportion is \(314/1300 \approx 0.2415\).
03

Two-tailed Test Execution and Interpretation

The two-tailed test checks if the proportions in both samples are significantly different. Find the standard error by using the formula \(\sqrt{p(1-p)(1/n1+1/n2)}\) where p is the pooled proportion and n1 and n2 are the sample sizes. Substituting the given values, the standard error is \(\sqrt{0.2415(1-0.2415)(1/800+1/500)} \approx 0.0262\). Find the z-score by subtracting the proportions and dividing by the standard error. This gives \(|0.28-0.18|/0.0262 \approx 3.82\). Generally, a z-score above 3 or below -3 indicates that the proportions are significantly different. Hence, it can be concluded that the population proportions do differ significantly.

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Key Concepts

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

Sample Proportions Calculation
Understanding the sample proportion calculation is crucial in statistics, especially when analyzing poll results or survey data. A sample proportion represents a percentage of participants in a study that have a particular characteristic. For instance, let's consider a scenario where we want to find out how many people support a particular policy within a group.

To calculate the sample proportion, you would follow a simple formula: \[ \text{Sample Proportion} = \frac{\text{Number of people with the characteristic}}{\text{Total number of people in the sample}} \]
This can also be expressed as a percentage. In the case of the Quebecers who support secession, with a sample size of 800 and 28% supporting secession, the number of supporters is calculated by multiplying 28% (or 0.28) by 800, resulting in approximately 224 individuals.
Pooled Proportion
When dealing with two or more samples, the pooled proportion is a tool that combines the information from these samples to give an overall proportion. This is particularly helpful when we wish to compare two groups and see if there is a significant difference regarding a specific characteristic.

The formula used to calculate the pooled proportion is:\[ \text{Pooled Proportion} = \frac{\text{Total number of successes in all samples}}{\text{Total sample size across all samples}} \]
In the context of the Quebec and Texas secession polls, we add the number of individuals supporting secession in Quebec (224) with those in Texas (90) to get a total number of 'successes.' Then, we divide this total by the combined sample size (1300). As a result, we obtain an overall proportion of the study population who favor secession, allowing us to conduct further comparative analyses.
Two-tailed Test
A two-tailed test in statistics is used when we're interested in determining whether there is a difference between two groups in either direction. This type of test is applied when the direction of the difference isn't specified beforehand. In the context of comparing proportions from two different populations, such as Quebecers and Texans on the issue of secession, a two-tailed test can assess whether the observed difference in sample proportions is statistically significant.

The key steps in this test involve calculating the standard error and then determining the z-score. The z-score measures the number of standard errors the observed difference is away from zero (no difference). If the z-score falls beyond the threshold set for significance (often at z-scores greater than 3 or less than -3), we can conclude that there is a significant difference between the two population proportions. The z-score in our example of secession preferences is approximately 3.82, indicating that the difference in secession support between the two populations is indeed significant.

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Most popular questions from this chapter

Systolic Blood Pressure and Survival Status Use technology and the ICUAdmissions dataset to find a \(95 \%\) confidence interval for the difference in systolic blood pressure (Systolic) upon admission to the Intensive Care Unit at the hospital based on survival of the patient (Status with 0 indicating the patient lived and 1 indicating the patient died.) Interpret the answer in context. Is "No difference" between those who lived and died a plausible option for the difference in mean systolic blood pressure? Which group had higher systolic blood pressures on arrival?

When we want \(95 \%\) confidence and use the conservative estimate of \(p=0.5,\) we can use the simple formula \(n=1 /(M E)^{2}\) to estimate the sample size needed for a given margin of error ME. In Exercises 6.40 to 6.43, use this formula to determine the sample size needed for the given margin of error. A margin of error of 0.04 .

Use a t-distribution and the given matched pair sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distribution of the differences is relatively normal. Assume that differences are computed using \(d=x_{1}-x_{2}\). Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1}<\mu_{2}\) using the paired data in the following table: $$ \begin{array}{lllllllll} \hline \text { Treatment } 1 & 16 & 12 & 18 & 21 & 15 & 11 & 14 & 22 \\ \text { Treatment } 2 & 18 & 20 & 25 & 21 & 19 & 8 & 15 & 20 \\ \hline \end{array} $$

Gender Bias In a study \(^{52}\) examining gender bias, a nationwide sample of 127 science professors evaluated the application materials of an undergraduate student who had ostensibly applied for a laboratory manager position. All participants received the same materials, which were randomly assigned either the name of a male \(\left(n_{m}=63\right)\) or the name of a female \(\left(n_{f}=64\right) .\) Participants believed that they were giving feedback to the applicant, including what salary could be expected. The average salary recommended for the male applicant was \(\$ 30,238\) with a standard deviation of \(\$ 5152\) while the average salary recommended for the (identical) female applicant was \(\$ 26,508\) with a standard deviation of \(\$ 7348\). Does this provide evidence of a gender bias, in which applicants with male names are given higher recommended salaries than applicants with female names? Show all details of the test.

Do Babies Prefer Speech? Psychologists in Montreal and Toronto conducted a study to determine if babies show any preference for speech over general noise. \(^{61}\) Fifty infants between the ages of \(4-13\) months were exposed to both happy-sounding infant speech and a hummed lullaby by the same woman. Interest in each sound was measured by the amount of time the baby looked at the woman while she made noise. The mean difference in looking time was 27.79 more seconds when she was speaking, with a standard deviation of 63.18 seconds. Perform the appropriate test to determine if this is sufficient evidence to conclude that babies prefer actual speaking to humming.
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