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meClock Plus Home Due Sunday by 11:59pm Points 6 Submitting an external tool Available Jul 7 at 12am-Aug 2 at 11:59pm 27 days Announcements Grades Chapter 8: Confidence intervals (Proportions) Score: 0/6 0/6 answered Done Syllabus Modules Question 1 > B0/1 pt 9399 Assignments Discussions Files You are interested in constructing a 99% confidence interval for the proportion of all caterpillars that eventually become butterflies. Of the 376 randomly selected caterpillars observed, 55 lived to become butterflies. Round answers to 4 decimal places where possible. a. With 99% confidence the proportion of all caterpillars that lived to become a butterfly is between and Student Support Tutoring Services Transfer Services b. If many groups of 376 randomly selected caterpillars were observed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population proportion of caterpillars that become butterflies and about percent will not contain the true population proportion. e Tutoring Hint: Hints Video [1 Submit Question

Answer 1

1. a)

b) 99%, 1%

Let p denotes the population proportion. Let p denotes the sample proportion and n denotes the sample size. Here, 55 0.1463, n = 376, standard error = 0.146 * (1 – 0.146) 376 = 0.018224 0.0182 376 At 99 percent confidence level, Margin of error is © * (1-P) ME= 0.146 * (1 – 0.146) 376 * 21-0.99 * 20.005 = 0.146 * (1 – 0.146) 376 * 2.575829 = 0.046943 -0.047 n a 99 percent confidence interval for the population proportion p is © * (1-P) = P + * 2 1-0.99 2 n = 0.1463 + 0.146 * (1 – 0.146) 376 * 20.005 = 0.1463 + 0.146 * (1 – 0.146) 376 * 2.575829 = 0.1463 + 0.046943 + (0.099334, 0.193219) ~(0.0993,0.1932)

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3.

Let p denotes the population proportion. Let p denotes the sample proportion and n denotes the sample size. Here, 280 P 0.5600, n = 500, standard error 0.560 * (1 – 0.560) 500 0.022199 ~ 0.0222 500 At 95 percent confidence level, Margin of error is *(1-P) ME= * 2 1-0.95 0.560 * (1 – 0.560) 500 * 20.025 0.560 * (1 – 0.560) 500 * 1.959964 = 0.043509 0.044 n a 95 percent confidence interval for the population proportion p is * (1 – Ô) = = P + * 2 1-0.95 n = 0.5600 + 0.560 * (1 – 0.560) 500 * 20.025 = 0.5600+ 0.560 * (1 – 0.560) 500 * 1.959964 = 0.5600 + 0.043509 (0.516491, 0.603509) ~ (0.516,0.604)

4.

Let p denotes the population proportion. Let o denotes the sample proportion and n denotes the sample size. Here, P = 0.840, n = 331, standard error = 0.840 * (1 – 0.840) 331 = 0.020150 0.0202 At 95 percent confidence level, Margin of error is p = (1-0) ME= * 21-0.95 0.840 * (1 – 0.840) 331 * 20.025 = 0.840 * (1 – 0.840) * 1.959964 = 0.039494 0.039 331 n a 95 percent confidence interval for the population proportion p is © * (1-) = P + * 2 1-0.95 2 n = 0.8400 + 0.840 * (1 – 0.840) 331 * 20.025 = 0.8400 + 0.840 * (1 – 0.840) 331 * 1.959964 = 0.8400 + 0.039494 + (0.800506,0.879494) ~(0.801,0.879)

Let p denotes the population proportion. Let p denotes the sample proportion and n denotes the sample size. Here, 42 P 0.1400, n = 300, standard error = 0.140 * (1 – 0.140) 300 = 0.020033 0.0200 300 At 95 percent confidence level, Margin of error is p = (1-0) ME= 0.140 * (1 – 0.140) 300 * 21-0.95 * 20.025 = 0.140 * (1 – 0.140) * 1.959964 = 0.039265 0.039 300 n a 95 percent confidence interval for the population proportion p is p = (1-p) = P + * 2 1-0.95 n = 0.1400 + 0.140 * (1 – 0.140) 300 * 20.025 = 0.1400 + 0.140 * (1 – 0.140) 300 * 1.959964 = 0.1400 = 0.039265 = (0.100735, 0.179265) ~ (0.1007,0.1793)