No matter how I approach it..i don't get it!
A recent study of 25 students reported that they spend an average of $18.53 per week on gas. The standard deviation of the sample was $3.00. Find the 95% confidence interval of the true mean.
No matter how I approach it..i don't get it!
A recent study of 25 students reported that they spend an average of $18.53 per week on gas. The standard deviation of the sample was $3.00. Find the 95% confidence interval of the true mean.
The formula you need can be found here: Confidence Intervals
The formula will also be in your class notes and textbook. Please show what you've tried and say where you get stuck.
By the Central Limit Theorem
$\displaystyle Z=\displaystyle\huge\frac{\bar{X}-\mu}{\left(\frac{\sigma}{\sqrt{n}}\right)}$
where $\displaystyle \sigma$ is the population SD, $\displaystyle \mu$ is the population mean, $\displaystyle n$ is the sample size,
$\displaystyle \bar{X}$ is the sample mean and $\displaystyle \frac{\sigma}{\sqrt{n}}$ is the sample SD.
The 95% interval ranges from 2.5% to 97.5%.
Locate the Z-score for 97.5% and use the values given to rearrange the equation for $\displaystyle \mu$