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3 years ago

100p for an answer please on a test

Mathematics

1 answer:

Pavlova-9 [17]3 years ago

4 0

Where’s the rest of the question? Cant answer unless I can see if she’s correct or not.

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The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

Korolek [52]

Answer:

A sample of 499 is needed.

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 $1 - \frac{\alpha}{2}$ .

The margin of error is given by:

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

In this question, we have that:

$\pi = 0.21$

90% confidence level

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

How large a sample would be required in order to estimate the fraction of tenth graders reading at or below the eighth grade level at the 90% confidence level with an error of at most 0.03

We need a sample of n, which is found when $M = 0.03$ . So

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

$0.03 = 1.645\sqrt{\frac{0.21*0.79}{n}}$

$0.03\sqrt{n} = 1.645\sqrt{0.21*0.79}$

$\sqrt{n} = \frac{1.645\sqrt{0.21*0.79}}{0.03}$

$(\sqrt{n})^2 = (\frac{1.645\sqrt{0.21*0.79}}{0.03})^2$

$n = 498.81$

Rounding up

A sample of 499 is needed.

8 0

2 years ago

on Wednesday jean's nursery received a shipment of 60 flowering crabapple trees. jean has ordered 80 trees. what percent of her

WITCHER [35]

75% when you have 60/80

7 0

3 years ago

There are 9,481 eligible voters in a precinct. 500 were selected at random and asked to indicate whether they planned to vote fo

maria [59]

Answer:

The confidence limits for the proportion that plan to vote for the Democratic incumbent are 0.725 and 0.775.

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 $1 - \frac{\alpha}{2}$ .

Of the 500 surveyed, 350 said they were going to vote for the Democratic incumbent.

This means that $n = 500, \pi = \frac{350}{500} = 0.75$