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A researcher has estimated that at most 49% of American adults skip breakfast on weekdays. In...

A researcher has estimated that at most 49% of American adults skip breakfast on weekdays. In a random sample of 280 American adults, 154 said they skip breakfast on weekdays. Is there sufficient evidence to show that more than 49% of American adults skip breakfast on weekdays? Use a 0.05 level of significance.

I. Set up Hypotheses

II. Calculate the test statistic

III. Calculate the p-value

IV. Make a Statistical Decision (Reject or Do not reject H0)

V. Make a Conclusion

math   Statistics-And-Probability   
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Answer #1

0-55 = 54 280 Hypothesss : Ho: p=o.ug Ha p > 0.49 11) Test statistics IP(1-P) 0.55-0.49 o.agkais) 280 - 2.01 11D pualue = P(Z

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