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Question 1209947: You intend to estimate a population mean with a confidence interval. You believe the population to have a normal distribution. Your sample size is 4.
While it is an uncommon confidence level, find the critical value that corresponds to a confidence level of 80.5%.
(Report answer accurate to three decimal places with appropriate rounding.)
ta/2 +/- =
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let $n$ be the sample size, which is $n=4$.
The confidence level is 80.5%, or 0.805.
The significance level $\alpha$ is $1 - 0.805 = 0.195$.
The area in each tail is $\alpha/2 = 0.195/2 = 0.0975$.
Since the population is normally distributed and the sample size is small ($n=4$), we need to use a t-distribution with $n-1 = 4-1 = 3$ degrees of freedom.
We need to find the t-value $t_{\alpha/2}$ such that the area to the right of $t_{\alpha/2}$ is 0.0975.
We can use a t-table or a calculator to find this value.
Using a t-table or calculator, we find that the t-value corresponding to an area of 0.0975 in the right tail with 3 degrees of freedom is approximately 1.589.
Therefore, $t_{\alpha/2} \approx 1.589$.
The critical value that corresponds to a confidence level of 80.5% is approximately 1.589.
Final Answer: The final answer is $\boxed{1.589}$
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