An airline transports over 10,000 passengers daily, and the airline is curious what proportion of their passengers use mobile boarding passes instead of paper passes. They take an SRS of 80 passengers and find that 60 of them use mobile boarding passes.

Based on this sample, which of the following is a $95 \%$ confidence interval for the proportion of passengers who use mobile boarding passes?

Choose 1 answer:
(A) $(0.625,0.875)$
(B) $(0.637,0.863)$
(c) $(0.655,0.845)$
(D) $(0.670,0.830)$

Question

An airline transports over 10,000 passengers daily, and the airline is curious what proportion of their passengers use mobile boarding passes instead of paper passes. They take an SRS of 80 passengers and find that 60 of them use mobile boarding passes.

Based on this sample, which of the following is a $95 \%$ confidence interval for the proportion of passengers who use mobile boarding passes?

Choose 1 answer:
(A) $(0.625,0.875)$
(B) $(0.637,0.863)$
(c) $(0.655,0.845)$
(D) $(0.670,0.830)$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Calculate the sample proportion (p̂) by dividing the number of passengers using mobile boarding passes by the total number of passengers sampled: $p̂ = \frac{60}{80} = 0.75$. ‖ ‖ step 2 ⋮ Find the standard error (SE) of the sample proportion using the formula $SE = \sqrt{\frac{p̂(1 - p̂)}{n}}$, where $n$ is the sample size: $SE = \sqrt{\frac{0.75(1 - 0.75)}{80}}$. ‖ ‖ step 3 ⋮ Calculate the value of the standard error: $SE = \sqrt{\frac{0.75 \cdot 0.25}{80}} = \sqrt{\frac{0.1875}{80}} = \sqrt{0.00234375} \approx 0.0484$. ‖ ‖ step 4 ⋮ Determine the z-score for a 95% confidence interval, which is approximately 1.96 for a two-tailed test. ‖ ‖ step 5 ⋮ Calculate the margin of error (ME) using the z-score and the standard error: $ME = z \cdot SE = 1.96 \cdot 0.0484 \approx 0.0949$. ‖ ‖ step 6 ⋮ Find the lower and upper bounds of the 95% confidence interval by subtracting and adding the margin of error to the sample proportion: lower bound $= p̂ - ME = 0.75 - 0.0949 \approx 0.6551$ and upper bound $= p̂ + ME = 0.75 + 0.0949 \approx 0.8449$. ‖ ‖ step 7 ⋮ Round the lower and upper bounds to three decimal places: lower bound $\approx 0.655$ and upper bound $\approx 0.845$. ‖ ***C*** ∻Key Concept∻ ⚹Confidence Interval for a Proportion⚹ ∻Explanation∻ ⚹The confidence interval for a proportion is calculated using the sample proportion, the z-score corresponding to the desired confidence level, and the standard error of the sample proportion. It provides a range of values within which the true population proportion is likely to fall.⚹◊What formula can be used to calculate the confidence interval for the proportion of passengers using mobile boarding passes in this scenario?⍭ Generate me a similar question◊

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