A student wanted to construct a 99% confidence interval for the mean age of students in her statistics class. She randomly selected nine students. Their mean age was 19.1 years with a sample standard deviation of 1.5 years. What is the 99% confidence interval for the population mean?
A)[17.42,20.78]
B)[17.48,20.72]
C)[14.23,23.98]
D)[0.44,3.80]

Question

Quizplus · Accepted Answer

Given the sample size (n=9), sample mean ($\bar{x}=19.1$), and sample standard deviation (s=1.5), along with the confidence level (99%), we can calculate the confidence interval for the population mean using the t-distribution (since the population standard deviation is unknown and the sample size is small). The degrees of freedom (df) is $n-1 = 8$. For a 99% confidence level, the t-value from the t-distribution table corresponding to df=8 is approximately 3.355 (this value might slightly vary depending on the source, but it's around this value for a two-tailed test). The formula for the confidence interval is:$$\bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}}$$Plugging in the values:$$19.1 \pm 3.355 \times \frac{1.5}{\sqrt{9}}$$$$19.1 \pm 3.355 \times 0.5$$$$19.1 \pm 1.6775$$This calculation results in the interval $[17.4225, 20.7775]$, which rounds to $[17.42, 20.78]$, matching option A.

A Student Wanted to Construct a 99% Confidence Interval for the Mean