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a.Determine the value of the sample proportionof Californians who have visited Yosemite. Is𝑝^this value higher or lower than the reported population value?= 240 /1000 = 0.24𝑝^This value is lower than the reported population value.b.Determine the expected value of the sample proportion.µ𝑝^== 0.28µ𝑝^𝑝c.Determine the standard deviation of the sample proportion.σ𝑝^== 0.0142σ𝑝^0.28(1−0.28)1000d.Determine that the condition for normality is satisfied.1000(0.28) = 280 & 1000(1-.28) = 720Both 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 = 0.9976𝑝^0.24−0.280.0142P(x > 0.24) = 0.9976f.Determine the probability the sample proportion is between 0.2 and 0.3.
P(0.2<x<0.3) = 0.9205Formulas for Confidence IntervalsConfidence interval forμwhenσis knownUseInverse Normal Table- between𝑋‾± 𝑍 ·σ𝑛Confidence interval forμwhenσis unknownUseInverse t tabledf=n-1𝑋‾± ? ·?𝑛4.20 students were asked how many units they were taking at a community college.55589101013151515151516181920212224a.Calculate the sample mean. This is apoint estimatorfor the population mean,μ. Explainwhat this means.=14. This means that the average population taking courses at the community𝑋‾college is taking 14 units.b.Find a 95% confidence interval for the population mean,assuming σ = 5. Calculate andexplain the margin of errorConfidence interval = 2.19 Margin of error= 16.19 and -11.8114(+-) 1.96 X5202.19 + 14= 16.19