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Chapter 8: Problem 51

A proponent of a new proposition on a ballot wants to know whether the proposition is likely to pass. Suppose a poll is taken, and 580 out of 1000 randomly selected people support the proposition. Should the proponent use a hypothesis test or a confidence interval to answer this question? Explain. If it is a hypothesis test, state the hypotheses and find the test statistic, p-value, and conclusion. If a confidence interval is appropriate, find the approximate \(95 \%\) confidence interval. In both cases, assume that the necessary conditions have been met.

Short Answer

Expert verified
Based on the calculation, the approximate \(95 \%\) confidence interval for the proportion of the population that supports the proposition is between \(54.9 \%\) and \(61.1 \%\). This means that the proponent can be \(95 \%\) confident that the true proportion of voters that support the proposition lies within this interval.

Step by step solution

01

Calculate Sample Proportion

First, calculate the sample proportion (\(p\)) of individuals who support the proposition. This is done by dividing the number of individuals who support the proposition (580) by the total number of individuals polled (1000). So, \(p = \dfrac{580}{1000} = 0.58.\)
02

Calculating Standard Error

The standard error (SE) for the sample proportion is computed using the formula \(SE = \sqrt{\dfrac{p(1-p)}{n}}\), where \(n\) is the number of individuals polled. Rocking this formula, \(SE = \sqrt{\dfrac{0.58(1-0.58)}{1000}} = 0.0157.\)
03

Calculate the Confidence Interval

The \(95\%\) confidence interval is calculated using the formula \((p-Z_SE, p+Z_SE)\), where \(Z is the critical value for a \(95 \%\) confidence interval. Based on the standard normal distribution, \(Z is approximately 1.96. Plugging in the values, the confidence interval is \((0.58-1.96*0.0157, 0.58+1.96*0.0157) = (0.549, 0.611)\).\)

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Most popular questions from this chapter

A researcher studying extrasensory perception (ESP) tests 300 students. Each student is asked to predict the outcome of a large number of coin flips. For each student, a hypothesis test using a \(5 \%\) significance level is performed. If the \(\mathrm{p}\) -value is less than or equal to \(0.05\), the researcher concludes that the student has ESP. Assuming that none of the 300 students actually have ESP, about how many would you expect the researcher to conclude do have ESP? Explain.

In problem \(8.16\), a college chemistry instructor thinks the use of embedded tutors will improve the succes: rate in introductory chemistry courses. The instructor carried out a hypothesis test and found that the observed value of the test statistic was \(2.33 .\) The \(\mathrm{p}\) -value associated with this test statistic is \(0.0099 .\) Explain the meaning of the p-value in this context. Based on this result, should the instructor believe the success rate has improved?

Refer to Exercise \(8.97 .\) Suppose 14 out of 20 voters in Pennsylvania report having voted for an independent candidate. The null hypothesis is that the population proportion is \(0.50 .\) What value of the test statistic should you report?

Suppose a poll is taken that shows 220 out of 400 randomly selected Twitter users feel that Twitter should do more to decrease hateful and abusive content on the site. Test the hypothesis that the majority (more than \(50 \%\) ) of Twitter users feel the site should do more to decrease hateful and abusive content on the site. Use a significance level of \(0.01\).

An economist is testing the hypothesis that the employment rate for law school graduates is different from \(86.7 \%\). The economist is using a \(5 \%\) significance level and these hypotheses: \(\mathrm{H}_{0}: p=0.867\) and \(\mathrm{H}_{\mathrm{a}}: p \neq 0.867 .\) Explain what the \(5 \%\) significance level means in context.
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