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3.29 Offshore drilling, Part 1. A 2010 survey asked 827 randomly sampled registered voters in California "Do you support? Or do you oppose? Drilling for oil and natural gas off the Coast of California? Or do you not know enough to say?" Below is the distribution of responses, separated based on whether or not the respondent graduated from college. (a) What percent of college graduates and what percent of the non-college graduates in this sample do not know enough to have an opinion on drilling for oil and natural gas off the Coast of California? \begin{tabular}{lcc} & \multicolumn{2}{c} { College Grad } \\ \cline { 2 - 3 } & Yes & No \\ \hline Support & 154 & 132 \\ Oppose & 180 & 126 \\ Do not know & 104 & 131 \\ \hline Total & 438 & 389 \end{tabular} the (b) Conduct a hypothesis test to determine if data provide strong evidence that the proportion of college graduates who do not have an opinion on this issue is different than that of non-college graduates.

Short Answer

Expert verified
23.74% of college grads and 33.67% of non-college grads do not know. Test suggests proportion difference is significant.

Step by step solution

01

Identify the necessary data

From the table, we find that 104 college graduates and 131 non-college graduates do not know enough to have an opinion. The total number of college graduates surveyed is 438, and the total number of non-college graduates surveyed is 389.
02

Calculate the percentage for each group

For college graduates: \( \frac{104}{438} \times 100 \approx 23.74\% \). For non-college graduates: \( \frac{131}{389} \times 100 \approx 33.67\% \). Thus, approximately 23.74% of college graduates and 33.67% of non-college graduates do not know enough to have an opinion.
03

Formulate Hypotheses

To find if there is a difference in proportions, set your null hypothesis as \( H_0: p_1 = p_2 \) (the proportions are the same) and the alternative hypothesis as \( H_a: p_1 eq p_2 \) (the proportions are different), where \( p_1 \) is the proportion for college grads, and \( p_2 \) is for non-college grads.
04

Calculate the test statistic

Use the formula for the test statistic for the difference in proportions: \[ z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{ \hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \]Where \( \hat{p_1} = \frac{104}{438} \), \( \hat{p_2} = \frac{131}{389} \), and \( \hat{p} = \frac{104 + 131}{438 + 389} \). Calculate these values to find the test statistic.
05

Determine Critical Value and Conclusion

Find the critical value for a typical significance level (say \( \alpha = 0.05 \)) from the standard normal distribution, which is 1.96 for a two-tailed test. Compare the calculated z-value to the critical values. If the z-value is beyond the range of -1.96 to 1.96, reject the null hypothesis, indicating there is a statistically significant difference in proportions.
06

Interpretation

Based on the comparison in Step 5, decide whether the result provides strong evidence to support the hypothesis that the proportions are different. If the z-value falls outside the critical range, then conclude that there is a significant difference in the proportions of opinions between college graduates and non-college graduates.

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Key Concepts

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

Proportions
In hypothesis testing, proportions are a way to express quantities that belong to a certain category out of a total. Here, we calculate proportions to understand what fraction of the surveyed college graduates and non-college graduates fall into each opinion category about offshore drilling. To find these proportions, divide the count of individuals in each category by the total number surveyed in their respective groups.

For instance, to determine how many college graduates don't feel informed enough to have an opinion on drilling, the calculation will be: \( \frac{104}{438} \). This results in approximately 23.74%. This means about 23.74% of the interviewed college graduates didn't know enough to form an opinion.
Similarly, for non-college graduates, the calculation \( \frac{131}{389} \) gives approximately 33.67%. Understanding these proportions helps us see the distribution of opinions in different demographic groups, which is a crucial step in any survey analysis. It tells us about the differences and similarities in perceptions or opinions of these two groups.
Survey Analysis
Survey analysis involves collecting and interpreting data to gain insights. In this scenario, a survey was conducted to gather opinions from two distinct groups about offshore drilling. The primary goal in survey analysis is to identify trends or differences in the collected data, helping draw meaningful conclusions.

Surveys play a crucial role in understanding public opinion or behaviors, especially when concerning issues like environmental policy or education. The data set included responses from a random sample of participants, ensuring that the results are representative of the broader population.
In the given survey, categories like 'Support', 'Oppose', and 'Do not know' allow for a detailed look at public sentiment. Survey analysis not only helps in knowing what percentage of people hold a particular view but also aids in comparing attitudes of different segments, such as college and non-college graduates. This kind of analysis is frequently utilized by policymakers and researchers in formulating strategies or further studies.
Statistical Significance
Statistical significance is a crucial aspect of hypothesis testing. It tells us whether the observed difference in survey outcomes is real or due to random chance. In hypothesis testing, we usually set a significance level, such as \( \alpha = 0.05 \), against which we measure our results.

To determine statistical significance here, we conduct a hypothesis test comparing proportions of opinions between college graduates and non-college graduates who are unsure about drilling. We start with a null hypothesis \( H_0 \): there is no difference in proportions, implying any observed difference is by chance.
Using a test statistic, calculated from the sample data, we compare this against critical values from a standard normal distribution. If our calculated statistic falls outside the predefined critical range (for a two-tailed test, typically between -1.96 and +1.96), it suggests the difference in proportions is statistically significant.
In simpler terms, statistical significance in this context helps us confidently say whether the education level influences how informed respondents feel about offshore drilling.
Data Interpretation
Data interpretation is seeing beyond raw numbers, extracting meaningful patterns, and drawing conclusions. After gathering survey data and performing necessary calculations, interpreting these results is what transforms data into actionable insights.

From the calculated proportions and hypothesis tests, we found a noticeable difference between college graduates and non-college graduates regarding their knowledge about offshore drilling. By determining statistical significance, we can assert whether the observed differences are meaningful or coincidental.
For data interpretation, look at both the survey questions and demographic details, along with calculated results. This helps in building a narrative around why certain groups reacted differently. Understanding context is key, as it aids in diagnosing if external factors (like misinformation or lack of access to education) may affect the perceptions.
The ability to interpret data impacts how results might shape public policies or drive further investigation. Clear interpretation also ensures that data presentation is understandable to stakeholders or decision-makers who might not be experts in statistics.

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Most popular questions from this chapter

Vegetarian college students. Suppose that \(8 \%\) of college students are vegetarians. Determine if the following statements are true or false, and explain your reasoning. (a) The distribution of the sample proportions of vegetarians in random samples of size 60 is approximately normal since \(n \geq 30\). (b) The distribution of the sample proportions of vegetarian college students in random samples of size 50 is right skewed. (c) A random sample of 125 college students where \(12 \%\) are vegetarians would be considered unusual. (d) A random sample of 250 college students where \(12 \%\) are vegetarians would be considered unusual. (e) The standard error would be reduced by one-half if we increased the sample size from 125 to 250 .

Offshore drilling, Part II. Results of a poll evaluating support for drilling for oil and natural gas off the coast of California were introduced in Exercise \(3.29 .\) \begin{tabular}{lcc} & \multicolumn{2}{c} { College Grad } \\ \cline { 2 - 3 } & Yes & No \\ \hline Support & 154 & 132 \\ Oppose & 180 & 126 \\ Do not know & 104 & 131 \\ \hline Total & 438 & 389 \end{tabular} (a) What percent of college graduates and what percent of the non-college graduates in this sample support drilling for oil and natural gas off the Coast of California? (b) Conduet a hypothesis test to determine if the data provide strong evidence that the proportion of college graduates who support off-shore drilling in California is different than that of noncollege graduates.

Browsing on the mobile device. A 2012 survey of 2,254 American adults indicates that \(17 \%\) of cell phone owners do their browsing on their phone rather than a computer or other device. (a) According to an online article, a report from a mobile research company indicates that 38 percent of Chinese mobile web users only access the internet through their cell phones. " Conduct a hypothesis test to determine if these data provide strong evidence that the proportion of Americans who only use their cell phones to access the internet is different than the Chinese proportion of \(38 \%\) (b) Interpret the p-value in this context. (c) Calculate a \(95 \%\) confidence interval for the proportion of Americans who access the internet on their cell phones, and interpret the interval in this context.

True or false, Part. II. Determine if the statements below are true or false. For each false statement, suggest an alternative wording to make it a true statement. (a) As the degrees of freedom increases, the mean of the chi-square distribution increases. (b) If you found \(X^{2}=10\) with \(d f=5\) you would fail to reject \(H_{0}\) at the \(5 \%\) significance level. (c) When finding the p-value of a chi-square test, we always shade the tail areas in both tails. (d) As the degrees of freedom increases, the variability of the chi-square distribution decreases.

Prop 19 in California. In a 2010 Survey USA poll, \(70 \%\) of the 119 respondents between the ages of 18 and 34 said they would vote in the 2010 general election for Prop \(19,\) which would change California law to legalize marijuana and allow it to be regulated and taxed. At a \(95 \%\) confidence level, this sample has an \(8 \%\) margin of error. Based on this information, determine if the following statements are true or false, and explain your reasoning." (a) We are \(95 \%\) confident that between \(62 \%\) and \(78 \%\) of the California voters in this sample support Prop \(19 .\) (b) We are \(95 \%\) confident that between \(62 \%\) and \(78 \%\) of all California voters between the ages of 18 and 34 support Prop \(19 .\) (c) If we considered many random samples of 119 California voters between the ages of 18 and 34 , and we calculated \(95 \%\) confidence intervals for each, \(95 \%\) of them will include the true population proportion of Californians who support Prop \(19 .\) (d) In order to decrease the margin of error to \(4 \%\), we would need to quadruple (multiply by 4) the sample size. (e) Based on this confidence interval, there is sufficient evidence to conclude that a majority of California voters between the ages of 18 and 34 support Prop \(19 .\)

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