A determine the value of the sample proportion of

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a. Determine the value of the sample proportion of Californians who have visited Yosemite. Is 𝑝 ^ this value higher or lower than the reported population value? = 240 /1000 = 0.24 𝑝 ^ This value is lower than the reported population value. b. Determine the expected value of the sample proportion . µ 𝑝 ^ = = 0.28 µ 𝑝 ^ 𝑝 c. Determine the standard deviation of the sample proportion . σ 𝑝 ^ = = 0.0142 σ 𝑝 ^ 0.28(1−0.28) 1000 d. Determine that the condition for normality is satisfied. 1000(0.28) = 280 & 1000(1-.28) = 720 Both values are greater than 10, therefore the conditions for normality are satisfied. e. Determine the probability the sample proportion exceeds 0.24. P( > 0.24) = P(Z > ) = P(Z > -2.82) = 1-0.0024 = 0.9976 𝑝 ^ 0.24−0.28 0.0142 P(x > 0.24) = 0.9976 f. Determine the probability the sample proportion is between 0.2 and 0.3.
P(0.2<x<0.3) = 0.9205 Formulas for Confidence Intervals Confidence interval for μ when σ is known Use Inverse Normal Table - between 𝑋 ± 𝑍 · σ 𝑛 Confidence interval for μ when σ is unknown Use Inverse t table df=n-1 𝑋 ± ? · ? 𝑛 4. 20 students were asked how many units they were taking at a community college. 5 5 5 8 9 10 10 13 15 15 15 15 15 16 18 19 20 21 22 24 a. Calculate the sample mean. This is a point estimator for the population mean, μ . Explain what this means. = 14. This means that the average population taking courses at the community 𝑋 college is taking 14 units. b. Find a 95% confidence interval for the population mean, assuming σ = 5 . Calculate and explain the margin of error Confidence interval = 2.19 Margin of error= 16.19 and -11.81 14(+-) 1.96 X 5 20 2.19 + 14= 16.19

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