/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 Suppose you are interested in wh... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose you are interested in whether more than \(50 \%\) of voters in California support a proposition (Prop X). After the vote, you find the total number that support it and the total number that oppose it.

Short Answer

Expert verified
The solution will depend on the specific data provided in the question but the steps outlined provide a general approach to the problem of testing a hypothesis about a proportion.

Step by step solution

01

- Define Null and Alternative Hypotheses

The null hypothesis is that the proportion of voters who support is equal to 0.5. This can be represented as: \( H_0 : p = 0.5 \). The alternative hypothesis is that the proportion of voters who support the proposition is more than 0.5. This can be represented as: \( H_1 : p > 0.5 \)
02

- Compute Test Statistic

Next, the test statistic must be computed. This will involve using the observed proportions and the assumed proportion under the null hypothesis (0.5 in this case). The formula for the test statistic in this case is \( Z = \frac{(p - 0.5)}{\sqrt{\frac{0.5 * (1 - 0.5)}{n}}} \).
03

- Make Decision

Compare the observed test statistic to the critical value for the chosen significance level. If the observed test statistic is greater than the critical value, reject the null hypothesis, indicating strong evidence that more than 50% of voters support the proposition. If not, fail to reject the null hypothesis, indicating insufficient evidence that more than 50% support the proposition.
04

- Interpret the Result

Finally, interpret the decision in the context of the problem. For example, if the null hypothesis was rejected, this would conclude that the data provides strong evidence that more than 50% of voters in California support proposition X.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Null and Alternative Hypotheses
Understanding the null and alternative hypotheses is crucial in hypothesis testing. The null hypothesis (\( H_0 \) is a general statement or default position that there is no difference or effect. In the context of our example, the null hypothesis asserts that exactly half (\( 50\text{%} \) of the voters in California support the proposition (Prop X), which is stated mathematically as \( H_0 : p = 0.5 \).

On the other hand, the alternative hypothesis (\( H_1 \) represents what the researcher is trying to prove — that the proportion in question is different than the null hypothesis claims. In our exercise, the alternative hypothesis claims that more than half of voters support Prop X: \( H_1 : p > 0.5 \). This hypothesis testing is directional, meaning the interest is specifically in finding whether support exceeds 50\text{%}, not just whether it differs.\
Test Statistic Calculation
The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It's used to assess the degree of compatibility between the null hypothesis and the sample data. To calculate the test statistic, one would use the formula: \[ Z = \frac{(p - 0.5)}{\sqrt{\frac{0.5 * (1 - 0.5)}{n}}} \]

In this formula, \( p \) represents the sample proportion of voters who support the proposition, while \( n \) is the total number of voters. The denominator is the standard error of the sampling distribution. If \( p \) is significantly greater than 0.5 according to this formula, the test statistic \( Z \) will be a large positive number, indicating that the null hypothesis may not adequately describe the data.
Significance Level and Decision Making
In hypothesis testing, the significance level, often denoted by \( \alpha \) (alpha), is the probability of rejecting the null hypothesis when it is actually true. Common levels of significance are 0.05, 0.01, and 0.10. This level determines the critical value which the test statistic must exceed for us to reject the null hypothesis.\

Decision Rule\

\The decision rule is a pre-determined plan that specifies the conditions under which the null hypothesis will be rejected or not rejected based on the comparison between the test statistic and critical value. If the test statistic is greater than the critical value corresponding to our chosen significance level, we reject the null hypothesis, concluding there's significant evidence to support the alternative.
Interpreting Statistical Results
After calculating the test statistic and comparing it to the critical value, it's vital to interpret the results in the context of the original question. Rejection of the null hypothesis implies there is statistical evidence to support the claim of the alternative hypothesis. For example, if we rejected \( H_0 \) in our voter proposition case, we would conclude there's evidence that more than 50\text{%} of the sampled voters support Prop X.\
\
It's important to understand that rejecting the null hypothesis doesn't prove the alternative hypothesis is true; it merely suggests there is enough evidence to consider it plausible. Additionally, failing to reject the null hypothesis does not prove it is true either. It simply implies that there's not enough evidence to support the alternative hypothesis given the data at the designated level of significance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In the study referenced in exercise \(10.33\), researchers also collected data on use of apps to monitor diet and calorie intake. The data are reported in the table. Test the hypothesis that diet app use and gender are associated. Use a \(0.05\) significance level. $$ \begin{array}{ccc} \text { Use } & \text { Male } & \text { Female } \\ \hline \text { Yes } & 43 & 241 \\ \hline \text { No } & 50 & 84 \\ \hline \end{array} $$

In a 2016 article published in the Journal of American College Health, Heller et al. surveyed a sample of students at an urban community college. Students' ages and frequency of alcohol use per month are recorded in the following table. Because some of the expected counts are less than 5, we should combine some groups. For this question, combine the frequencies \(10-29\) days and Every day into one group. Label this group \(10+\) days and show your new table. Then test the new table to see whether there is an association between age group and alcohol use using a significance level of \(0.05\). Assume this is a random sample of students from this college. $$ \begin{array}{lcccc} \hline & & \text { Alcohol Use } \\ \text { Age } & \text { None } & \text { 1-9 days } & \text { 10-29 days } & \text { Every day } \\ \hline 18-20 & 182 & 100 & 27 & 4 \\ \hline 21-24 & 142 & 109 & 35 & 4 \\ \hline 25-29 & 49 & 41 & 5 & 2 \\ \hline 30+ & 76 & 32 & 8 & 2 \end{array} $$

A vaccine is available to prevent the contraction of human papillomavirus (HPV). The Centers for Disease Control and Prevention recommends this vaccination for all young girls in two doses. In a 2015 study reported in the Journal of American College Health, Lee et al. studied vaccination rates among Asian American and Pacific Islander (AAPI) women and non-Latina white women. Data are shown in the table. Test the hypothesis that vaccination rates and race are associated. Use a \(0.05\) significance level. $$ \begin{array}{|lcc|} \hline \text { Completed HPV vaccinations } & \text { AAPI } & \text { White } \\\ \hline \text { Yes } & 136 & 1170 \\ \hline \text { No } & 216 & 759 \\ \hline \end{array} $$

In July 2013 , Jeff Witmer obtained a data set from the Tampa Bay Times after the Zimmerman case was decided. Zimmerman shot and killed Trayvon Martin (an unarmed black teenager) and was acquitted. The data set concerns "stand your ground" cases with male defendants. Some of these were fatal attacks, and some were not fatal. Many of those charged used guns, but some used various kinds of knives or other methods. $$ \begin{array}{|lccc|} \hline & & \text { Accused } \\ \hline & \text { Nonwhite } & \text { White } & \text { All } \\ \hline \text { Not Convicted } & 48 & 80 & 128 \\ \hline \text { Convicted } & 19 & 38 & 57 \\ \hline \text { All } & 67 & 118 & 185 \\ \hline \end{array} $$ a. What percentage of the nonwhite defendants were convicted? b. What percentage of the white defendants were convicted? c. Test the hypothesis that conviction is independent of race at the \(0.05\) level. Assume you have a random sample.

Professional musicians listened to five violins being played, without seeing the instruments. One violin was a Stradivarius, and the other four were modern-day violins. When asked to pick the Stradivarius (after listening to all five), 39 got it right and 113 got it wrong. a. Use the chi-square goodness-of-fit test to test the hypothesis that the experts are not simply guessing. Use a significance level of \(0.05\). b. Perform a one-proportion \(z\) -test with the same data, using a one-tailed alternative that the experts should get more than \(20 \%\) correct. Use a significance level of \(0.05\). c. Compare your p-values and conclusions.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.