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Sagot :
Answer:
She needs a sample size of 25.
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 - 0.68}{2} = 0.16[/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 - 0.16 = 0.84[/tex], so Z = 0.995.
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.
The population SD is 2 grams.
This means that [tex]\sigma = 2[/tex]
What is the minimum sample size she needs to create a confidence interval that has a width of 0.4 grams?
She needs a sample size of n.
n is found when M = 0.4. So
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.4 = 0.995\frac{2}{\sqrt{n}}[/tex]
[tex]\sqrt{n} = \frac{0.995*2}{0.4}[/tex]
[tex](\sqrt{n})^2 = (\frac{0.995*2}{0.4})^2[/tex]
[tex]n = 24.8[/tex]
Rounding up:
She needs a sample size of 25.
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