Assume that a sample is used to estimate a population mean mu. Find the 99% confidence interval for a sample of size 45 with a mean of 78.2 and a standard deviation of 10.7. Enter your answer as an open-interval (i.e., parentheses) accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

Question

99% C.I. =

Answer 1

To calculate the 99% confidence interval for the population mean $\mu$, we can use the formula for the confidence interval based on the sample mean, the critical value from the standard normal distribution (Z-distribution), and the standard error of the mean.

1. Calculate the standard error (SE): $$SE=\frac{s}{\sqrt{n}}$$ where $s$ is the sample standard deviation, and $n$ is the sample size.

   Given:

   • $s=10.7$
   • $n=45$

   $$SE=\frac{10.7}{\sqrt{45}}\approx \frac{10.7}{6.708}\approx 1.592$$

2. Find the critical value for a 99% confidence level: For a two-tailed test at a 99% confidence level, we look up the Z-score that corresponds to an area of $0.005$ in each tail (since $0.01/2=0.005$). This critical value is approximately $Z\approx 2.576$.

3. Calculate the margin of error (ME): $$ME=Z\times SE$$ $$ME=2.576\times 1.592\approx 4.102$$

4. Construct the confidence interval: The confidence interval can be calculated using the sample mean and the margin of error. $$\text{Confidence Interval}=(\overline{x}-ME,\overline{x}+ME)$$ Where $\overline{x}=78.2$.

   $$\text{Lower limit}=78.2-4.102\approx 74.1$$ $$\text{Upper limit}=78.2+4.102\approx 82.3$$

Thus, the 99% confidence interval for the population mean $\mu$ is:

$$\text{99\% C.I.}=(74.1,82.3)$$