The phenomenon of bottle flipping has hit the nation. You design a machine that can randomly flip a bottle. You observe 2180 random, independent flips, and 325 land the correct way. Estimate the overall proportion of bottles that land correctly when flipped randomly. Use a 98% confidence level. Round everything to 3 decimal places. a) State the parameter of interest. Verify that the necessary conditions are present in order to carry out the procedure.

Answer:

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

The necessary conditions [tex]n\pi \geq 10[/tex] and [tex]n(1-\pi) \geq 10[/tex] are present.

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

Step-by-step 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 [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

Necessary conditions:

The necessary conditions are:

[tex]n\pi \geq 10[/tex]

[tex]n(1-\pi) \geq 10[/tex]

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

This means that [tex]n = 2180, \pi = \frac{325}{2180} = 0.149[/tex]

Necessary conditions

[tex]n\pi = 2180*0.149 = 325 \geq 10[/tex]

[tex]n(1-\pi) = 2180*0.851 = 1855 \geq 10[/tex]

The necessary conditions [tex]n\pi \geq 10[/tex] and [tex]n(1-\pi) \geq 10[/tex] are present.

98% confidence level

So [tex]\alpha = 0.02[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.02}{2} = 0.99[/tex], so [tex]Z = 2.33[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.149 - 2.33\sqrt{\frac{0.149*0.851}{2180}} = 0.131[/tex]

The upper limit of this interval is:

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

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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