Copy of 5Math 243 Hwk sheet on Ch

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University of Oregon *

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243

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Mathematics

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Feb 20, 2024

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docx

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Uploaded by GrandWildcatMaster786

Math 243 Hwk sheet on ch.5 Name: M Maximum Points: 5 pts Show all your work and interpret the results for full credit. ● There are 11 questions ● Please make sure to round the decimals to 4 places. Percentages to 2 decimal places. ● EACH QUESTION WORTH 1 point. This will be converted to 5 pts in grade book ● Use Proper symbols and formulas where needed. Interpret the results. ● Explain the answer in the context of the given question. This means to write a complete sentence after you find the answer. Failure to do so will result in partial credit. 1. Pg.277 pb.5.7 Refer to Pg.270-272 for formulas. In 2013, the Pew Research Foundation reported that “45% of U.S. adults report that they live with one or more chronic conditions”. However, this value was based on a sample, so it may not be a perfect estimate for the population parameter of interest on its own. The study reported a standard error of about 1.2%, and a normal model may reasonably be used in this setting. Create a 95% confidence interval for the proportion of U.S. adults who live with one or more chronic conditions. Also, interpret the confidence interval in the context of the study. sample proportion=0.45 standard error=0.012 margin of error=0.02352 we are 95% confident that the proportion of U.S adults who live with one or more chronic conditions is between 0.42648=42.65% and 0.47352=47.35% Lakshmi Balaji Math 243 Hwk sheet on ch.5 -confidence interval Page 1
2. Refer to Pg.269-274; also pg.302-306 The Pew research poll asked whether respondents expected to see a female president in their lifetime. 78% of the 1,835 respondents said “yes”. a) Construct a 90% confidence interval for the proportion of Americans who expect to see a female president in their lifetime and interpret this interval in the context of the data. **First Identify the unknown population parameter: P +/- z* * sqrt(P * (1 - P)/n) = 0.78 +/- 1.645 * sqrt(0.78 * 0.22/1835) = 0.78 +/- 0.0159 = 0.7641, 0.7959 we are 90% confident that the true proportion of americans who expect to see a female president is between 76.41% and 79.58% b) How would you expect the width of a 98% confidence interval to compare to the interval you calculated in part (a)? Explain your reasoning. The width of the 98% confidence interval will be wider than the width of 90% confidence interval. c) Now construct the 98% confidence interval. P +/- z* * sqrt(P * (1 - P)/n) = 0.78 +/- 2.33 * sqrt(0.78 * 0.22/1835) = 0.78 +/- 0.0225 = 0.7575, 0.8025 we are 98% confident that the true proportion of americans who expect to see a female president is between 75.75% and 80.25% d) In general, what does 98% confidence interval mean to you? Lakshmi Balaji Math 243 Hwk sheet on ch.5 -confidence interval Page 2
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