Answer:
We need a sample size of at least 719
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{1-0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]
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.
How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?
This is at least n, in which n is found when [tex]M = 0.3, \sigma = 4.103[/tex]. So
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
[tex]0.3 = 1.96*\frac{4.103}{\sqrt{n}}[/tex]
[tex]0.3\sqrt{n} = 1.96*4.103[/tex]
[tex]\sqrt{n} = \frac{1.96*4.103}{0.3}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}[/tex]
[tex]n = 718.57[/tex]
Rouding up
We need a sample size of at least 719
Using the sample size formula, the number of samples require to vary the population is 719.
Using the sample size relation :
Substituting the values into the relation :
n = [tex] (\frac{1.96 \times 4.103}{0.3}) ^{2}[/tex]
n = [tex] (\frac{8.04188}{0.3}) ^{2}[/tex]
n = [tex] (26.806266)^{2}[/tex]
n = 718.57
Hence, the minimum sample size required is 719.
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