Answer:
The probability content for the confidence interval (3.49, 3.69) is 96%.
Step-by-step explanation:
The sample selected to determine the mean number of times the pet owners visited their veterinarian each year is of size, <em>n</em> = 475.
The sample selected is quite large, i.e. <em>n</em> > 30.
According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.
Then, the mean of the distribution of sample mean is given by,
; for <em>n</em> → ∞.
And the standard deviation of the distribution of sample means is given by,
; for <em>n</em> → ∞.
So, the random variable <em>X</em>, defined as the number of visits to the veterinarian each year, follows a Normal distribution.
The (1 - <em>α</em>)% confidence interval for the population mean (<em>μ</em>) is:
![CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=CI%3D%5Cbar%20x%5Cpm%20z_%7B%5Calpha%2F2%7D%5Ctimes%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
The information provided is:
![\bar x=3.59\\s=1.045\\](https://tex.z-dn.net/?f=%5Cbar%20x%3D3.59%5C%5Cs%3D1.045%5C%5C)
The confidence interval is (3.49, 3.69).
The margin of error of the confidence interval for the population mean is:
![MOE=\frac{UL-LL}{2}= z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=MOE%3D%5Cfrac%7BUL-LL%7D%7B2%7D%3D%20z_%7B%5Calpha%2F2%7D%5Ctimes%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
Compute the MOE as follows:
![MOE=\frac{UL-LL}{2}=\frac{3.69-3.49}{2}=0.10](https://tex.z-dn.net/?f=MOE%3D%5Cfrac%7BUL-LL%7D%7B2%7D%3D%5Cfrac%7B3.69-3.49%7D%7B2%7D%3D0.10)
Compute the critical value of <em>z</em> as follows:
![MOE=z_{\alpha/2}\times \frac{s}{\sqrt{n}}\\0.10=z_{\alpha/2}\times \frac{1.045}{\sqrt{475}}\\z_{\alpha/2}=2.0856\\z_{\alpha/2}\approx2.09](https://tex.z-dn.net/?f=MOE%3Dz_%7B%5Calpha%2F2%7D%5Ctimes%20%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%5C%5C0.10%3Dz_%7B%5Calpha%2F2%7D%5Ctimes%20%5Cfrac%7B1.045%7D%7B%5Csqrt%7B475%7D%7D%5C%5Cz_%7B%5Calpha%2F2%7D%3D2.0856%5C%5Cz_%7B%5Calpha%2F2%7D%5Capprox2.09)
Compute the value of
as follows:
![P(-z_{\alpha/2}](https://tex.z-dn.net/?f=P%28-z_%7B%5Calpha%2F2%7D%3CZ%3Cz_%7B%5Calpha%2F2%7D%29%3DP%28-2.09%3CZ%3C2.09%29)
![=P(Z](https://tex.z-dn.net/?f=%3DP%28Z%3C2.09%29-P%28Z%3C-2.09%29%5C%5C%3DP%28Z%3C2.09%29-%5B1-P%28Z%3C2.09%29%5D%5C%5C%3D2P%28Z%3C2.09%29-1%5C%5C%3D%282%5Ctimes%200.9817%29-1%5C%5C%3D0.9634%5C%5C%5Capprox0.96)
Thus, the probability content for the confidence interval (3.49, 3.69) is 96%.