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
![W = \frac{X-\mu}{\frac{10.5}{\sqrt{n}}}](https://tex.z-dn.net/?f=%20W%20%3D%20%5Cfrac%7BX-%5Cmu%7D%7B%5Cfrac%7B10.5%7D%7B%5Csqrt%7Bn%7D%7D%7D%20)
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
![0.99 = P(-3.27 < \frac{X-\mu}{\frac{10.5}{\sqrt{n}}} < 3.27)](https://tex.z-dn.net/?f=%200.99%20%3D%20P%28-3.27%20%3C%20%5Cfrac%7BX-%5Cmu%7D%7B%5Cfrac%7B10.5%7D%7B%5Csqrt%7Bn%7D%7D%7D%20%3C%203.27%29%20)
This is equivalent to
![P(-3.27* \frac{10.5}{\sqrt{n}} < X - \mu < 3.27* \frac{10.5}{\sqrt{n}}) =0.99](https://tex.z-dn.net/?f=%20P%28-3.27%2A%20%5Cfrac%7B10.5%7D%7B%5Csqrt%7Bn%7D%7D%20%3C%20X%20-%20%5Cmu%20%3C%203.27%2A%20%5Cfrac%7B10.5%7D%7B%5Csqrt%7Bn%7D%7D%29%20%3D0.99%20)
Now, we take out the X and the sign, reverting the inequalities, and we obtain
![P(X - 3.27* \frac{10.5}{\sqrt{n}} < \mu < X + 3.27* \frac{10.5}{\sqrt{n}}) = 0.99](https://tex.z-dn.net/?f=%20P%28X%20-%203.27%2A%20%5Cfrac%7B10.5%7D%7B%5Csqrt%7Bn%7D%7D%20%3C%20%5Cmu%20%3C%20X%20%2B%203.27%2A%20%5Cfrac%7B10.5%7D%7B%5Csqrt%7Bn%7D%7D%29%20%3D%200.99%20)
Thus, a 99% confidence interval for μ is
.
In order for the sample mean to be within 0.9 psi from the true mean, we need n such that
![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](https://tex.z-dn.net/?f=%203.27%2A%20%5Cfrac%7B10.5%7D%7B%5Csqrt%7Bn%7D%7D%20%3C%200.9%20%5C%2C%20%5Crightarrow%20%5Cfrac%7B1%7D%7B%5Csqrt%7Bn%7D%7D%20%3C%20%5Cfrac%7B0.9%7D%7B3.27%2A10.5%7D%20%5C%2C%20%5Crightarrow%20%5Csqrt%7Bn%7D%20%3E%2038.15%20%5C%2C%20%5Crightarrow%20n%20%3E%201455.4)
We take n = 1456 in order to assure that the sample mean is within 0.9 psi of its true mean.