Question

You intend to estimate a population mean with a confidence interval. You believe the population to have a normal distribution. Your sample size is 63. While it is an uncommon confidence interval, find the critical valuebthat corresponds tona confidence interval of 81.2%.

Answer 1

Answer:

The critical value that corresponds to a confidence interval of 81.2% is 1.32.

Step-by-step explanation:

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

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

$\mu_{\bar x}=\mu$

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

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

In this case the sample selected is of size, n = 63.

As the sample size n = 63 > 30, the sampling distribution of sample mean will be approximately normal.

So, a z-interval will be used to estimate the population mean.

The confidence level is, 81.2%.

The value of α is:

$\alpha=1-\text{Confidence Level}\\\\\alpha=1-0.812\\\\\alpha=0.188$

The critical value is:

$z_{\alpha/2}=z_{0.188/2}=z_{0.094}=- 1.32\\\\z_{1-\alpha/2}=z_{1-0.188/2}=z_{0.906}= 1.32$

*Use a z-table.

Thus, the critical value that corresponds to a confidence interval of 81.2% is 1.32.