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Conspiracy Some Americans believe that the entire \(9 / 11\) catastrophe was planned and executed by federal officials in order to provide the United States with a pretext for going to war in the Middle East and as a means of consolidating and extending the power of the then-current administration. This group of Americans is larger than you think. A Scripps-Howard poll of \(n=1010\) adults in August of 2006 found that \(36 \%\) of American consider such a scenario very or somewhat likely! \({ }^{23}\) In a follow-up poll, a random sample of \(n=100\) adult Americans found that 26 of those sampled agreed that the conspiracy theory was either likely or somewhat likely. Does this sample contradict the reported \(36 \%\) figure? Test at the \(\alpha=.05\) level of significance.

Short Answer

Expert verified
Answer: Yes, based on the results of the hypothesis test at the 0.05 level of significance, the sample of 100 adult Americans does contradict the reported 36% figure. It appears that the proportion of Americans who believe in the conspiracy theory is significantly different from 36%.

Step by step solution

01

State the Hypotheses

Null hypothesis (H0): The proportion of Americans who believe in the conspiracy theory is equal to 36% (P = 0.36). Alternative hypothesis (H1): The proportion of Americans who believe in the conspiracy theory is not equal to 36% (P ≠ 0.36).
02

Choose the Test Statistic and Significance Level

Since we are dealing with proportions, we will use the z-test for proportions. We will use a significance level of α = 0.05.
03

Calculate the Test Statistic

First, calculate the sample proportion (p̂): p̂ = x / n = 26 / 100 = 0.26 Next, compute the standard error (SE): SE = sqrt[(P * (1 - P)) / n] = sqrt[(0.36 * (1 - 0.36)) / 100] = 0.0483 Now, calculate the z-score: z = (p̂ - P) / SE = (0.26 - 0.36) / 0.0483 = -2.069
04

Determine the Critical Region

Since this is a two-tailed test, we need to find the critical values of z for α = 0.05. The critical values are z = -1.96 and z = 1.96. (You can find these values in a z-table or using a calculator)
05

Make a Decision

Since the calculated z-score (-2.069) is less than the critical value (-1.96), we reject the null hypothesis (H0).
06

Conclusion

Based on the results of the hypothesis test at the 0.05 level of significance, we can conclude that the sample of 100 adult Americans does contradict the reported 36% figure. It appears that the proportion of Americans who believe in the conspiracy theory is significantly different from 36%.

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Key Concepts

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

Z-Test for Proportions
When we want to compare the proportion of a sample to a known population proportion, we use what's known as a z-test for proportions. It’s a statistical method that helps us determine whether the observed difference between sample and population proportions is significant or just due to random chance. In simple terms, it's saying, 'Let's see if this sample looks like what we'd expect based on what we know about the bigger group.'

In our exercise, we're comparing the proportion of Americans who believe in a specific conspiracy theory, with an assumption about this proportion in the general population. Calculating the z-score gives us a picture of where our sample proportion lies in comparison to the assumed population proportion. A high z-score (far from zero) might indicate that the sample proportion is not typical of the population proportion.
Null and Alternative Hypotheses
The null hypothesis (\(H_0\)) and alternative hypothesis (\(H_1\)) are vital components in hypothesis testing. They are like contradictory statements we're trying to judge between based on our sample data.

The null hypothesis is our starting point; it's the idea that there's no effect or no difference, and in this case, it would be the belief that the proportion of the population that subscribes to the conspiracy theory is the initially reported 36%. The alternative hypothesis is what we suspect might be true instead — that the proportion who believe in the conspiracy differs from 36%. By conducting the test, we’re essentially trying to see if we have enough evidence to challenge the status quo (null hypothesis) in favor of our suspicion (alternative hypothesis).
Level of Significance
The level of significance, denoted by \(\alpha\), is a threshold we set to decide how much evidence is required to reject the null hypothesis. Think of it as the 'strictness' level of our test. The most common levels are 0.05, 0.01, and 0.10. The lower the significance level, the stronger evidence we need before we can reject the null hypothesis.

In our exercise, the significance level is set at 0.05. This means we would consider results that have less than a 5% probability of occurring if the null hypothesis were true as strong enough evidence against the null hypothesis. It’s like saying, 'We're 95% sure the difference we're seeing isn't just random chance.'
P-Value Interpretation
The p-value tells us how likely it is to get a result as extreme as ours if the null hypothesis were true. It’s a pithy way of seeing how surprising our data is. A small p-value (typically less than our \(\alpha\) level) suggests that our sample is unusual and that maybe the null hypothesis doesn't hold water.

For instance, a p-value lower than 0.05 would lead us to reject the null hypothesis, indicating that our findings are statistically significant and not likely due to random fluctuation. If, however, our p-value were higher than 0.05, we wouldn't have enough evidence to reject the null hypothesis, and we'd have to conclude that our sample doesn't provide strong enough evidence to support the claim that the true proportion differs from 36%.
Standard Error Calculation
The standard error (SE) measures the variability in the sample proportions if we took many samples. It’s crucial to calculate it because it helps us understand the precision of our sample proportion. Smaller SE means more confidence in our sample as a reflection of the population. You can think of it as the margin of error in a poll.

To calculate SE for proportions, as seen in the exercise, we use the formula \( SE = \sqrt{\frac{P(1 - P)}{n}} \) where \(P\) is the population proportion and \(n\) is the sample size. The SE is then used to normalize the difference between the sample proportion and the population proportion. The result is the z-score, which ultimately tells us how many standard errors away our sample proportion is from the population proportion.

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Most popular questions from this chapter

Independent random samples of 36 and 45 observations are drawn from two quantitative populations, 1 and \(2,\) respectively. The sample data summary is shown here: $$\begin{array}{lcc} & \text { Sample 1 } & \text { Sample 2 } \\\\\hline \text { Sample Size } & 36 & 45 \\\\\text { Sample Mean } & 1.24 & 1.31 \\\\\text { SampleVariance } & .0560 & .0540\end{array}$$ Do the data present sufficient evidence to indicate that the mean for population 1 is smaller than the mean for population 2 ? Use one of the two methods of testing presented in this section, and explain your conclusions

Colored Contacts Refer to Exercise \(9.37 .\) Contact lenses, worn by about 26 million Americans, come in many styles and colors. Most Americans wear soft lenses, with the most popular colors being the blue varieties \((25 \%),\) followed by greens \((24 \%),\) and then hazel or brown. A random sample of 80 tinted contact lens wearers was checked for the color of their lenses. Of these people, 22 wore blue lenses and only 15 wore green lenses. a. Do the sample data provide sufficient evidence to indicate that the proportion of tinted contact lens wearers who wear blue lenses is different from \(25 \% ?\) Use \(\alpha=.05 .\) b. Do the sample data provide sufficient evidence to indicate that the proportion of tinted contact lens wearers who wear green lenses is different from \(24 \% ?\) Use \(\alpha=.05\) c. Is there any reason to conduct a one-tailed test for either part a or b? Explain.

Adolescents and Social Stress In a study to compare ethnic differences in adolescents' social stress, researchers recruited subjects from three middle schools in Houston, Texas. \({ }^{21}\) Social stress among four ethnic groups was measured using the Social Attitudinal Familial and Environment Scale for Children (SAFE-C). In addition, demographic information abou the 316 students was collected using self-administered questionnaires. A tabulation of student responses to a question regarding their socioeconomic status (SES) compared with other families in which the students chose one of five responses (much worse off, somewhat worse off, about the same, better off, or much better off) resulted in the tabulation that follows. a. Do these data support the hypothesis that the proportion of adolescent African Americans who state that their SES is "about the same" exceeds that for adolescent Hispanic Americans? b. Find the \(p\) -value for the test. c. If you plan to test using \(\alpha=.05,\) what is your conclusion?

Bass Fishing The pH factor is a measure of the acidity or alkalinity of water. A reading of 7.0 is neutral; values in excess of 7.0 indicate alkalinity; those below 7.0 imply acidity. Loren Hill states that the best chance of catching bass occurs when the \(\mathrm{pH}\) of the water is in the range 7.5 to \(7.9 .{ }^{17}\) Suppose you suspect that acid rain is lowering the \(\mathrm{pH}\) of your favorite fishing spot and you wish to determine whether the \(\mathrm{pH}\) is less than 7.5 a. State the alternative and null hypotheses that you would choose for a statistical test. b. Does the alternative hypothesis in part a imply a one- or a two-tailed test? Explain. c. Suppose that a random sample of 30 water specimens gave \(\mathrm{pH}\) readings with \(\bar{x}=7.3\) and \(s=.2\) Just glancing at the data, do you think that the difference \(\bar{x}-7.5=-.2\) is large enough to indicate that the mean \(\mathrm{pH}\) of the water samples is less than \(7.5 ?\) (Do not conduct the test.) d. Now conduct a statistical test of the hypotheses in part a and state your conclusions. Test using \(\alpha=.05 .\) Compare your statistically based decision with your intuitive decision in part c.

Generation Next Born between 1980 and 1990, Generation Next was the topic of Exercise \(8.60 .^{16}\) In a survey of 500 female and 500 male students in Generation Next, 345 of the females and 365 of the males reported that they decided to attend college in order to make more money. a. Is there a significant difference in the population proportions of female and male students who decided to attend college in order to make more money? Use \(\alpha=.01\). b. Can you think of any reason why a statistically significant difference in these population proportions might be of practical importance? To whom might this difference be important?

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