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John Stud · Answer 1 · 2022-09-10T03:48:27+0000

Answer:

The variable of interest is the proportion of flips that land the correct way when flipped randomly.

The necessary conditions
$n\pi \geq 10$ and
$n(1-\pi) \geq 10$ are present.

The 98% confidence interval for the overall proportion of bottles that land correctly when flipped randomly is (0.131, 0.167).

Explanation:

Variable of Interest:

Proportion of flips that land the correct way when flipped randomly.

In a sample with a number n of people surveyed with a probability of a success of
$\pi$ , and a confidence level of
$1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{(\pi(1-\pi))/(n)}$

In which

z is the zscore that has a pvalue of
$1 - (\alpha)/(2)$ .

Necessary conditions:

The necessary conditions are:

$n\pi \geq 10$

$n(1-\pi) \geq 10$

You observe 2180 random, independent flips, and 325 land the correct way.

This means that
$n = 2180, \pi = (325)/(2180) = 0.149$

Necessary conditions

$n\pi = 2180*0.149 = 325 \geq 10$

$n(1-\pi) = 2180*0.851 = 1855 \geq 10$

The necessary conditions
$n\pi \geq 10$ and
$n(1-\pi) \geq 10$ are present.

98% confidence level

So
$\alpha = 0.02$ , z is the value of Z that has a pvalue of
1 - (0.02)/(2) = 0.99 , so
Z = 2.33 .

The lower limit of this interval is:

$\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.149 - 2.33\sqrt{(0.149*0.851)/(2180)} = 0.131$

The upper limit of this interval is:

$\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.149 + 2.33\sqrt{(0.149*0.851)/(2180)} = 0.167$

The 98% confidence interval for the overall proportion of bottles that land correctly when flipped randomly is (0.131, 0.167).

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