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Question
a board - game club meets each week at the local library. people bring different board games to play, like checkers and chess. out of the 50 people that meet each week, 67% prefer to play checkers over chess. construct an 83% confidence interval for the population mean of people that prefer to play checkers over chess.
ci=(56.06%, 77.94%)
ci=(53.97%, 80.03%)
ci=(49.88%, 84.12%)
ci=(57.89%, 76.14%)
Step1: Find $\alpha/2$ and $z_{\alpha/2}$
Determine $\alpha$ from confidence level.
Step2: Calculate fraction value
Compute $\frac{\hat{p}(1 - \hat{p})}{n}$.
Step3: Compute square - root
Find $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$.
Step4: Calculate margin of error
Multiply $z_{\alpha/2}$ and square - root value.
Step5: Find confidence interval bounds
Subtract and add margin of error from/to $\hat{p}$.
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We are given a sample of $n = 50$ people, and the sample - proportion $\hat{p}=0.67$.
For an $83\%$ confidence interval, the significance level $\alpha=1 - 0.83 = 0.17$. Then $\alpha/2=0.085$.
The $z$ - value $z_{\alpha/2}=z_{0.085}$. Looking up in the standard normal table, $z_{0.085}\approx1.37$.
The formula for the confidence interval for a proportion is $\hat{p}\pm z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$.
First, calculate $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$:
Step1: Calculate the value inside the square - root
$\hat{p}(1 - \hat{p})=0.67\times(1 - 0.67)=0.67\times0.33 = 0.2211$
$\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.2211}{50}=0.004422$
Step2: Calculate the square - root value
$\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.004422}\approx0.0665$
Step3: Calculate the margin of error
The margin of error $E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=1.37\times0.0665\approx0.0911$
Step4: Calculate the lower and upper bounds of the confidence interval
The lower bound $= \hat{p}-E=0.67 - 0.0911 = 0.5789$
The upper bound $=\hat{p}+E=0.67 + 0.0911 = 0.7611$
So the $83\%$ confidence interval is $(57.89\%,76.11\%)$
The answer is $\text{CI}=(57.89\%,76.11\%)$