# Interval estimate to infer the population mean with a 95% confidence level

An industrial designer wants to determine the average time it takes for an adult to assemble a toy. 24 people were randomly chosen to assemble the toy and the time taken (in minutes) were as follows:

17, 13, 18, 19, 17, 21, 29, 22, 16, 28, 21, 15 26, 23, 24, 20, 8, 17, 17, 21, 32, 18, 25, 22

Question = Using interval estimate to infer the population mean with a 95% confidence level

So n = 24

We can also find that

Sample mean, $\bar x$ = 20.375

Sample variance, $s^2$ = 5.36342282

Standard deviation $s$ = 2.315903479

To find 95% confidence level, we have

Upper limit : 20.375 + 1.96 $\times$ $\frac\sigma{\sqrt{24}}$ = 22.52081603 Lower limit : 20.375 - 1.96 $\times$ $\frac\sigma{\sqrt{24}}$ = 18.22918397

But the answer is 18.11 ≤ µ ≤ 22.64 instead. May I know what is wrong here?

Note that the sample size is quite small and hence you need to use the $t$ statistic with $n-1$ degrees of freedom, instead of the $z$ score.
Actually, what you denoted as the sample variance is in fact the sample standard deviation $s=5.363$. The 95% confidence interval is given by $$\left[ \bar{x} - t_{0.025,23} \cdot \frac{s}{\sqrt{n}} ,\;\;\;\; \bar{x}+t_{0.025,23}\cdot \frac{s}{\sqrt{n}}\right].$$ We need to use $0.025=\frac{0.05}{2}$ because this is a two-tailed test. Looking the value of $t_{0.025,23}$ up in a table, we find $t_{0.025,23} = 2.069$. Hence the LCL is equal to $20.375 - 2.069\cdot\frac{5.363}{\sqrt{24}}=18.11$, and the UCL is equal to $20.375 + 2.069\cdot\frac{5.363}{\sqrt{24}}=22.64$.
• may I know did you used my Standard deviation value, $s$ = 2.315903479? It would be good if you can edit your answer to show the full solution to it. Apr 16, 2015 at 13:12