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3 years ago

11

determine the sample sieze erequired to estimate teh mean score on a standardizes test within 4 [pints of the trrue mean with 98

% confidence

1 answer:

Andreas93 [3]3 years ago

5 0

Answer: Minimum sample size = 110

Step-by-step explanation:

Formula to find minimum sample size :

$n=(\dfrac{z\times s}{E})^2$ , where z = critical z-value for confidence level, s= sample standard deviation , E = margin of error

Given: E= 4 , s= 18

Critical z-value for 98% confidence = 2.327

So, $n=(\dfrac{2.327\times18}{4})^2$

$=(\dfrac{41.886}{4})^2\\\\=(10.4715)^2=109.65231225\approx110$

Minimum sample size = 110

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4 years ago

A physical therapist wants to determine the difference in the proportion of men and women who participate in regular sustained p

Studentka2010 [4]

Using the z-distribution, it is found that the needed sample sizes are:

a) 242

b) 1842

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , 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 $\frac{1+\alpha}{2}$ .

The margin of error is:

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

99% confidence level, hence $\alpha = 0.99$ , z is the value of Z that has a p-value of $\frac{1+0.99}{2} = 0.995$ , so $z = 2.575$ .

Item a:

The estimate is:

$\pi = 0.223 - 0.189 = 0.034$

The sample size is <u>n for which M = 0.03</u>, then:

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

$0.03 = 2.575\sqrt{\frac{0.034(0.966)}{n}}$

$0.03\sqrt{n} = 2.575\sqrt{0.034(0.966)}$

$\sqrt{n} = \frac{2.575\sqrt{0.034(0.966)}}{0.03}$

$(\sqrt{n})^2 = \left(\frac{2.575\sqrt{0.034(0.966)}}{0.03}\right)^2$

$n = 241.97$

Rounding up, a sample of 242 is needed.

Item b:

No prior estimates, hence $\pi = 0.5$ is used.

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

$0.03 = 2.575\sqrt{\frac{0.5(0.5)}{n}}$

$0.03\sqrt{n} = 2.575\sqrt{0.5(0.5)}$

$\sqrt{n} = \frac{2.575\sqrt{0.5(0.5)}}{0.03}$

$(\sqrt{n})^2 = \left(\frac{2.575\sqrt{0.5(0.5)}}{0.03}\right)^2$

$n = 1841.8$

Rounding up, a sample of 1842 is needed.

For more on the z-distribution, you can check brainly.com/question/25404151

5 0

3 years ago

PLEASE HELP!!!!!!!!!!!!

hammer [34]

Answer:

The answer to your question is: Option B is correct

Step-by-step explanation:

Which expression is equivalent to the square root of 80?

Find the prime factors of 80

80 = 2⁴5 then $\sqrt{80} = 4\sqrt{5}$ = $\sqrt{2^{2}2^{2} 5 }$

Option A which is 4 square root of 2 times the square root of 5 This option is incorrect

Option B which is square root of 2^4 times the square root of 5 This option is correct

Option C which is the square root of 2 times the square root of 2 times the square root of 5 This option is incorrect

Option D which is the square root of 2^2 times the square root of 2^2 times the square root of 5^2 This option is incorrect

6 0

3 years ago