Respuesta :
Answer:
The minimum number is 1456.
Step-by-step explanation:
The sheer Stength for spot welds is a random variable with unknown mean μ and standard deviation σ = 10.5. Let X be the sample mean of a sample size of n spot welds. The Central Limit Theorem states that X has approrximately Normal Distribution with mean μ and standard deviation 10.5/(√n). Lets denote with W the standarization of X. W has distribution approximately Normal with mean 0 and standard deviation 1. W is given by the following formula
[tex] W = \frac{X-\mu}{\frac{10.5}{\sqrt{n}}} [/tex]
The cummulative distribution function of W, noted by Ф, has its values tabulated, and they can be found on the attached file. We want a confidence interval of 99%, so we should find Z such that P(-Z < W < Z) = 0.99. Since the Normal density function is symmetric, this is equivalent to find Z such that P(W < Z) = 0.995, in other words, Ф(Z) = 0.995
If we watch the table, we will find that the value of Z corresponding to the value 0.995 is 3.27. This means that
[tex] 0.99 = P(-3.27 < \frac{X-\mu}{\frac{10.5}{\sqrt{n}}} < 3.27) [/tex]
This is equivalent to
[tex] P(-3.27* \frac{10.5}{\sqrt{n}} < X - \mu < 3.27* \frac{10.5}{\sqrt{n}}) =0.99 [/tex]
Now, we take out the X and the sign, reverting the inequalities, and we obtain
[tex] P(X - 3.27* \frac{10.5}{\sqrt{n}} < \mu < X + 3.27* \frac{10.5}{\sqrt{n}}) = 0.99 [/tex]
Thus, a 99% confidence interval for μ is [tex] CI = [X - 3.27* \frac{10.5}{\sqrt{n}} , X + 3.27* \frac{10.5}{\sqrt{n}}] [/tex].
In order for the sample mean to be within 0.9 psi from the true mean, we need n such that
[tex] 3.27* \frac{10.5}{\sqrt{n}} < 0.9 \, \rightarrow \frac{1}{\sqrt{n}} < \frac{0.9}{3.27*10.5} \, \rightarrow \sqrt{n} > 38.15 \, \rightarrow n > 1455.4[/tex]
We take n = 1456 in order to assure that the sample mean is within 0.9 psi of its true mean.
