Respuesta :
Answer:
d. 1,600.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1 - 0.99}{2} = 0.005[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1 - \alpha[/tex].
That is z with a pvalue of [tex]1 - 0.005 = 0.995[/tex], so Z = 2.575.
Now, find the margin of error M as such
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
Based on previous research, the standard deviation of the distribution of the age at which children begin to walk is estimated to be 1.5 months.
This means that [tex]\sigma = 1.5[/tex]
Of the following, which is the smallest sample size that will result in a margin of error of 0.1 month or less for the confidence interval?
The sample size has to be n or larger. n is found when [tex]M = 0.1[/tex]. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.1 = 2.575\frac{1.5}{\sqrt{n}}[/tex]
[tex]0.1\sqrt{n} = 2.575*1.5[/tex]
Multiplying both sides by 10
[tex]\sqrt{n} = 2.575*15[/tex]
[tex](\sqrt{n})^2 = (2.575*15)^2[/tex]
[tex]n = 1492[/tex]
So the sample size has to be at least 1492, which means that of the possible options, the smallest sample size is 1600, given by option d.
The sample size should be at least 1492, So the possible options, the smallest sample size is 1600, option D is the correct answer
Based on previous research, the standard deviation of the distribution of the age at which children begin to walk is estimated to be 1.5 months. A random sample of children will be selected, and the age at which each child begins to walk will be recorded. A 99% confidence interval for the average age at which children begin to walk will be constructed.
What is the margin of error?
The margin of error tells you how many percentages points your results will differ from the real population value.
[tex]M=z\frac{\sigma}{\sqrt{n} }[/tex]
We need to find our α level, that is the subtraction from 1 by the confidence interval for the average age divided by 2.
[tex]\alpha = \frac{1-0.99}{2}\\ =0.005[/tex]
Now, we need to find z which is 1-α
[tex]1-\alpha \\=1-0.005\\\rm z=2.575[/tex]
The margin of error M
[tex]M=z\frac{\sigma}{\sqrt{n} }[/tex]
Here, [tex]\sigma[/tex] is the standard deviation of the population.
n is the size of the sample.
So,
[tex]\rm M=z\frac{\sigma}{\sqrt{n} } \\\rm0.1=2.575\frac{1.5}{\sqrt{n} } \\\rm0.1\times\sqrt{n} =2.575\times{1.5}\\\rm\sqrt{n} =2.575\times{1.5}\\\rm(\sqrt{n} )^{2} =(2.575\times{1.5})^{2} \\\rm n=1492[/tex]
Hence, the sample size should be at least 1492, So the possible options, the smallest sample size is 1600, option D is the correct answer.
Learn more about the margin of error here:
https://brainly.com/question/6979326
