To construct a confidence interval using the given confidence level, do whichever of the following is appropriate. (a) Find the critical value z Subscript alpha divided by 2, (b) find the critical value t Subscript alpha divided by 2, or (c) state that neither the normal nor the t distribution applies. 95%; nequals200; sigma equals 19.0; population appears to be skewe

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and take appropriately huge random-samples (n > 30) from the population with replacement, then the distribution of the sample- means will be approximately normally-distributed.

Then, the mean of the sample means is given by,

[tex]\mu_{\bar x}=\mu[/tex]

And the standard deviation of the sample means is given by,

[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}[/tex]

The information provided is:

n = 200

σ = 19.0

Population is skewed.

As the sample selected is quite large, i.e. n = 200 > 30 the central limit theorem can be used to approximate the distribution of the sample mean by the normal distribution.

So, [tex]\bar X\sim N(\mu, \frac{\sigma^{2}}{n})[/tex].

Then to construct a confidence interval for mean we will use a z-interval.

And for 95% confidence level we will compute the critical value of z, i.e. [tex]z_{\alpha/2}[/tex].

Thus, the correct option is (a).