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Are most student government leaders extroverts? According to Myers-Briggs estimates, about \(82 \%\) of college student government leaders are extroverts. (Source: Myers-Briggs Type Indicator Atlas of Type Tables.) Suppose that a Myers-Briggs personality preference test was given to a random sample of 73 student government leaders attending a large national leadership conference and that 56 were found to be extroverts. Does this indicate that the population proportion of extroverts among college student government leaders is different (either way) from \(82 \%\) ? Use \(\alpha=0.01\).

Short Answer

Expert verified
No, there is not enough evidence to say the proportion differs from 82%.

Step by step solution

01

State the Hypotheses

We need to determine if the proportion of extrovert student government leaders differs from 82%. Our null hypothesis is: \( H_0: p = 0.82 \), meaning there is no difference. The alternative hypothesis is: \( H_a: p eq 0.82 \), meaning the proportion is different from 82%.
02

Collect the Sample Information

We have a sample size \( n = 73 \), and the number of extroverts found is \( x = 56 \). Thus, the sample proportion is \( \hat{p} = \frac{56}{73} \approx 0.767 \).
03

Check Normality Assumption

We need to check if \( np_0 \) and \( n(1-p_0) \) are both greater than 5 to use the normal approximation. Here, \( np_0 = 73 \times 0.82 = 59.86 \) and \( n(1-p_0) = 73 \times 0.18 = 13.14 \). Both values are greater than 5, so the normal approximation is valid.
04

Calculate the Test Statistic

Use the formula for the test statistic: \( Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \). Substituting in the known values: \( Z = \frac{0.767 - 0.82}{\sqrt{\frac{0.82 \times 0.18}{73}}} \approx -0.98 \).
05

Determine the Critical Value and Decision

We are conducting a two-tailed test with \( \alpha = 0.01 \). Thus, the critical values for \( Z \) are \( \pm Z_{\alpha/2} = \pm 2.576 \). Since \( -0.98 \) is within the range \(-2.576\) to \(2.576\), we do not reject the null hypothesis.
06

Conclusion

Based on the test, we do not have enough evidence to reject the null hypothesis. Thus, we conclude that there is not enough evidence to say that the proportion of extroverts among student government leaders is different from 82%.

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Key Concepts

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

Proportion Test
A proportion test is a statistical method used to determine if a sample proportion is significantly different from a known or hypothesized population proportion. This is especially useful when you're trying to see if a certain characteristic in a sample group holds true for the broader population.

For this exercise, we're testing whether the proportion of extroverts among student government leaders differs from the known proportion of 82%. We start by defining our null hypothesis as the assumption that there is no difference between the sample proportion and the population proportion (i.e., the null hypothesis is that the proportion is still 0.82).

The test involves calculating a test statistic from the data, which will tell us how far our sample proportion is from the hypothesized population proportion in standard deviation units. This helps us decide whether to reject or fail to reject the null hypothesis depending on the significance level chosen, which in this case is 0.01.
Normal Approximation
Normal approximation is a technique used in hypothesis testing when dealing with proportions. It allows us to approximate the sample distribution to a normal distribution, which simplifies the calculation of the test statistic. However, for this approximation to be valid, certain conditions need to be met.

First, we check if both the expected number of successes (in our case, the number of extroverts) and failures (non-extroverts) are greater than 5, calculated as \( np_0 \) and \( n(1-p_0) \) respectively. If these values are greater than 5, we can safely use the normal approximation.

In our example problem, we calculated \( np_0 \) as 59.86 and \( n(1-p_0) \) as 13.14, both comfortably above 5, thus allowing us to use the normal distribution to approximate the distribution of the sample proportion.
Sample Proportion
The sample proportion is a simple calculation representing the fraction of the sample with the characteristic of interest. In this problem, it is the proportion of extroverts in the sample of student government leaders.

We calculate the sample proportion \( \hat{p} \) by dividing the number of extroverts by the total number of individuals in the sample. Here, \( \hat{p} = \frac{56}{73} \approx 0.767 \).

This sample proportion is a central part of our hypothesis test, as it forms the basis for comparing against the known population proportion (0.82 in our scenario). It helps in establishing whether there is a significant difference from the hypothesized population proportion.
Two-Tailed Test
A two-tailed test is a method used in hypothesis testing when we are interested in deviations on both sides of the hypothesized value, in either direction.

For our study, we want to test whether the actual proportion of extroverts differs from the hypothesized 82%, regardless of whether it's higher or lower. This is a typical situation for a two-tailed test.

In practice, this means we are looking for evidence that the sample could fall into the extremes (either high or low) of the normal distribution range we derive from our calculations. With a significance level \( \alpha = 0.01 \), the critical values are set as \( \pm 2.576 \) for the standard normal distribution.

If the calculated test statistic falls outside this interval, we would reject the null hypothesis. In our feature exercise, the calculated \( Z \)-statistic is \(-0.98\), which falls within this interval, therefore leading us to retain the null hypothesis and conclude that there is no statistically significant difference.

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Most popular questions from this chapter

Based on information from Harper's Index, \(r_{1}=\) 37 people out of a random sample of \(n_{1}=100\) adult Americans who did not attend college believe in extraterrestrials. However, out of a random sample of \(n_{2}=100\) adult Americans who did attend college, \(r_{2}=47\) claim that they believe in extraterrestrials. Does this indicate that the proportion of people who attended college and who believe in extraterrestrials is higher than the proportion who did not attend college? Use \(\alpha=0.01\).

Is the national crime rate really going down? Some sociologists say yes! They say that the reason for the decline in crime rates in the \(1980 \mathrm{~s}\) and \(1990 \mathrm{~s}\) is demographics. It seems that the population is aging, and older people commit fewer crimes. According to the FBI and the Justice Department, \(70 \%\) of all arrests are of males aged 15 to 34 years. (Source: True Odds, by J. Walsh, Merritt Publishing.) Suppose you are a sociologist in Rock Springs, Wyoming, and a random sample of police files showed that of 32 arrests last month, 24 were of males aged 15 to 34 years. Use a \(1 \%\) level of significance to test the claim that the population proportion of such arrests in Rock Springs is different from \(70 \%\).

Compare statistical testing with legal methods used in a U.S. court setting. Then discuss the following topics in class or consider the topics on your own. Please write a brief but complete essay in which you answer the following questions. (a) In a court setting, the person charged with a crime is initially considered to be innocent. The claim of innocence is maintained until the jury returns with a decision. Explain how the claim of innocence could be taken to be the null hypothesis. Do we assume that the null hypothesis is true throughout the testing procedure? What would the alternate hypothesis be in a court setting? (b) The court claims that a person is innocent if the evidence against the person is not adequate to find him or her guilty. This does not mean, however, that the court has necessarily proved the person to be innocent. It simply means that the evidence against the person was not adequate for the jury to find him or her guilty. How does this situation compare with a statistical test for which the conclusion is "do not reject" (i.e., accept) the null hypothesis? What would be a type II error in this context? (c) If the evidence against a person is adequate for the jury to find him or her guilty, then the court claims that the person is guilty. Remember, this does not mean that the court has necessarily proved the person to be guilty. It simply means that the evidence against the person was strong enough to find him or her guilty. How does this situation compare with a statistical test for which the conclusion is to "reject" the null hypothesis? What would be a type I error in this context? (d) In a court setting, the final decision as to whether the person charged is innocent or guilty is made at the end of the trial, usually by a jury of impartial people. In hypothesis testing, the final decision to reject or not reject the null hypothesis is made at the end of the test by using information or data from an (impartial) random sample. Discuss these similarities between statistical hypothesis testing and a court setting. (e) We hope that you are able to use this discussion to increase your understanding of statistical testing by comparing it with something that is a well. known part of our American way of life. However, all analogies have weak points. It is important not to take the analogy between statistical hypothesis testing and legal court methods too far. For instance, the judge does not set a level of significance and tell the jury to determine a verdict that is wrong only \(5 \%\) or \(1 \%\) of the time. Discuss some of these weak points in the analogy between the court setting and hypothesis testing.

Let \(x\) be a random variable that represents assembly times for the Ford Taurus. The Wall Street Journal reported that the average assembly time is \(\mu=38\) hours. A modification to the assembly procedure has been made. Experience with this new method indicates that \(\sigma=1.2\) hours. It is thought that the average assembly time may be reduced by this modification. A random sample of 47 new Ford Taurus automobiles coming off the assembly line showed the average assembly time of the new method to be \(\bar{x}=37.5\) hours. Does this indicate that the average assembly time has been reduced? Use \(\alpha=0.01\).

In the following data pairs, \(A\) represents birth rate and \(B\) represents death rate per 1000 resident population. The data are paired by counties in the Midwest. A random sample of 16 counties gave the following information. (Reference: County and City Data Book, U.S. Department of Commerce.) \(\begin{array}{l|cccccccc} \hline \text { A: } & 12.7 & 13.4 & 12.8 & 12.1 & 11.6 & 11.1 & 14.2 & 15.1 \\\ \hline B: & 9.8 & 14.5 & 10.7 & 14.2 & 13.0 & 12.9 & 10.9 & 10.0 \\ \hline \\ \hline A: & 12.5 & 12.3 & 13.1 & 15.8 & 10.3 & 12.7 & 11.1 & 15.7 \\ \hline B: & 14.1 & 13.6 & 9.1 & 10.2 & 17.9 & 11.8 & 7.0 & 9.2 \\ \hline \end{array}\) Do the data indicate a difference (either way) between population average birth rate and death rate in this region? Use \(\alpha=0.01\).

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