A math teacher was frustrated at the number of students leaving their graphing calculator behind in her classroom at the end of class without a way to locate the student. A random sample of 50 students is selected, and of the students questioned, 32 had their names written on their graphing calculators.

Required:
What would be a 99% confidence interval for the proportion of all students at this school who have their names written on their graphing calculators?

Respuesta :

Answer:

The 99% confidence interval for the proportion of all students at this school who have their names written on their graphing calculators is (0.4652, 0.8148).

Step-by-step explanation:

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].

A random sample of 50 students is selected, and of the students questioned, 32 had their names written on their graphing calculators.

This means that [tex]n = 50, \pi = \frac{32}{50} = 0.64[/tex]

99% confidence level

So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.64 - 2.575\sqrt{\frac{0.64*0.36}{50}} = 0.4652[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.64 + 2.575\sqrt{\frac{0.64*0.36}{50}} = 0.8148[/tex]

The 99% confidence interval for the proportion of all students at this school who have their names written on their graphing calculators is (0.4652, 0.8148).