Question

A country's education department reported that in 2015, 70.8% of students enrolled in college or a trade school within 12 months of graduating high school. In 2017, a random sample of 154 individuals who graduated from high school 12 months prior was selected. From this sample, 94 students were found to be enrolled in college or a trade school. Complete parts a through c. a. Construct a 90% confidence interval to estimate the actual proportion of students enrolled in college or a trade school within 12 months of graduating from high school in 2017. and an upper limit of The confidence interval has a lower limit of (Round to three decimal places as needed.) ما
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Answer 1

Answer:

The confidence interval has a lower limit of 0.546 and an upper limit of 0.675.

Step-by-step explanation:

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.

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

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

In 2017, a random sample of 154 individuals who graduated from high school 12 months prior was selected. From this sample, 94 students were found to be enrolled in college or a trade school.

This means that $n = 154, \pi = \frac{94}{154} = 0.6104$

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6104 - 1.645\sqrt{\frac{0.6104*0.3896}{154}} = 0.546$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6104 + 1.645\sqrt{\frac{0.6104*0.3896}{154}} = 0.675$

The confidence interval has a lower limit of 0.546 and an upper limit of 0.675.