Editors preparing a report on the economy are trying to estimate the percentage of businesses that plan to hire additional employees in the next 60 days. They are willing to accept a margin of error of 4% but want 99% confidence. How many randomly selected employers will they need to contact?
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Step 1
The formula is: n = (Z^2 * P * (1-P)) / E^2 where: - n is the sample size - Z is the Z-score (which corresponds to the desired confidence level) - P is the estimated proportion of the population - E is the desired margin of error For a 99% confidence level, the Show more…
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Editors preparing a report on the economy are trying to estimate the percentage of businesses that plan to hire additional employees in the next 60 days. They are willing to accept a margin of error of 5% but want 95% confidence. How many randomly selected employers will they need to contact?
Hubert A.
Editors preparing a report on the economy are trying to estimate the percentage of businesses that plan to hire additional employees in the next 60 days. They are willing to accept a margin of error of 3% but want 99% confidence. How many randomly selected employers will they need to contact? Question: n= __________ (round up to the nearest integer.)
Robin C.
Please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. (b) Check Requirements What sampling distribution will you use? Do you think the sample size is sufficiently large? Explain. Compute the value of the sample test statistic and corresponding $z$ value. (c) Find the $P$ -value of the test statistic. Sketch the sampling distribution and show the area corresponding to the $P$ -value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level $\alpha$ ? (e) Interpret your conclusion in the context of the application. Myers-Briggs: Extroverts 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 $\alpha=0.01$.
Hypothesis Testing
Testing a Proportion $p$
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