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4 years ago

|15.87 – 48.09| + |-79.67 + 42.31| A. 55.13 B. 60.58 C. 61.70 D. 69.58

Mathematics

1 answer:

gogolik [260]4 years ago

5 0

It would be d

i hope i helped

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3 years ago

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3 years ago

A home improvement store has at least $18,000 worth of lawnmowers to sell during the summer. If the cost of each lawnmower is $4

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Hi!

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3 years ago

A recent survey showed that among 100 randomly selected college seniors, 20 plan to attend graduate school and 80 do not. Determ

Hunter-Best [27]

Answer:

The 68% confidence interval for the population proportion of college seniors who plan to attend graduate school is (0.16, 0.24).

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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

A recent survey showed that among 100 randomly selected college seniors, 20 plan to attend graduate school and 80 do not.

This means that $n = 100, \pi = \frac{20}{100} = 0.2$

68% confidence level

So $\alpha = 0.32$ , z is the value of Z that has a pvalue of $1 - \frac{0.32}{2} = 0.84$ , so $Z = 0.995$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2 - 0.995\sqrt{\frac{0.2*0.8}{100}} = 0.16$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2 + 0.995\sqrt{\frac{0.2*0.8}{100}} = 0.24$

The 68% confidence interval for the population proportion of college seniors who plan to attend graduate school is (0.16, 0.24).

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3 years ago