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,
[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]
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:
[tex]\alpha=1-\text{Confidence Level}\\\\\alpha=1-0.812\\\\\alpha=0.188[/tex]
The critical value is:
[tex]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[/tex]
*Use a z-table.
Thus, the critical value that corresponds to a confidence interval of 81.2% is 1.32.
