Respuesta :
Answer:
The sample size should be at least 79.
Step-by-step explanation:
Let Y be be the resting glocose level of a member of the population chosen randomly, then Y is a random variable with unkown mean λ and Standard deviation σ = 27.
Let X be the sample mean of a sample of lenght n. X has the same mean as Y and the standard deviation is [tex] \sigma = \frac{27}{\sqrt{n}} . [/tex]
If n is reasonable high, the Central Limit Theorem states that X has distribution approximately Normal, with mean λ and Standard deviation [tex] sigma = \frac{27}{\sqrt{n}} . [/tex]
If we standarize X, we get a random variable W
[tex] W = \frac{X-\lambda}{\frac{27}{\sqrt{n}}} \simeq N(0,1) [/tex]
The values of W are tabulated and can be found on the attached file. We want a 95% confidence interval, so we want Z such that
P(-Z < w < Z) = 0.95
Using the symmetry of the normal density function, we get that
P(W < Z) = 0.975
If we look at the table, we will find that Z = 1.96, therefore we have
[tex] P(-1.64 < \frac{X-\lambda}{\frac{27}{\sqrt{n}}} < 1.64) = 0.95 [/tex]
Equivalently,
[tex] 0.95 = P(-1.64*\frac{27}{\sqrt{n}} < X- \lambda < 1.64*\frac{27}{\sqrt{n}}) [/tex]
Taking out the X and the sign, after reverting the inequalities, we obtain
[tex] P(X -1.64*\frac{27}{\sqrt{n}} < \lambda < X +1.64*\frac{27}{\sqrt{n}}) = 0.95 [/tex]
Thus, a confidence interval with 95% confidence is
[tex] CI = [ X-1.64*\frac{27}{\sqrt{n}}, X+1.64*\frac{27}{\sqrt{n}}] [/tex]
The (absolute) margin of error of this interval is [tex] 1.64*\frac{27}{\sqrt{n}} , [/tex] we want that number to be at most 5, so we take n such that
[tex] 1.64*\frac{27}{\sqrt{n}} = 5 \, \rightarrow \frac{1}{\sqrt{n}} = \frac{5}{1.64*27} \, \rightarrow \sqrt{n} = 8.85 \, \rightarrow n = 78.42 [/tex]
We take n = 79. I hope that works for you!
