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Hello! I need help with this:Calculation of the confidence interval Statistics.The confidence interval should be calculated for the percentage of people who chose the answer spruce:Sample: 313Answers:Spruce - 272Pine - 41Confidence level - 0.9

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We have to calculate a 90% confidence interval for the proportion that chose the answer "Spruce".

The sample proportion is p = 0.869:


p=(X)/(n)=(272)/(313)\approx0.869

The standard error of the proportion is:


\begin{gathered} \sigma_p=\sqrt{(p(1-p))/(n)} \\ \\ \sigma_p=\sqrt{(0.869\cdot0.131)/(313)} \\ \\ \sigma_p\approx√(0.0003637) \\ \sigma_p\approx0.019 \end{gathered}

The critical z-value for a 90% confidence interval is z = 1.645.

The margin of error (MOE) can be calculated as:


MOE=z\cdot\sigma_p=1.645\cdot0.019\approx0.031

Then, the lower and upper bounds of the confidence interval are:


\begin{gathered} LL=p-z\sigma_p=0.869-0.031=0.838 \\ UL=p+z\sigma_p=0.869+0.031=0.900 \end{gathered}

Answer: The 90% confidence interval for the population proportion is (0.838, 0.900).

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