In a survey, 10 people were asked how much they spent on their child’s last birthday gift. The results were roughly bell-shaped with a mean of 48 and standard deviation of 13. Construct a confidence interval at a 99% confidence level.

Answer :

Given :

The sample mean $(x\overline{)}=48$

The sample standard deviation $(s)=13$

The sample size $(n)=10$

Confidence level = 99%

Solution :

For a 99% confidence level,

$$\alpha =1–0.99=0.01$$

The degrees of freedom :

$$\text{df}=n–1=10–1=9$$

Critical value :

$$t_c=t_{\frac{\alpha }{2},df}=t_{0.005,9}=3.25$$

The 99% confidence interval :

$$\text{The 99\% confidence interval}=\overline{x}\pm t_c\times \frac{s}{\sqrt{n}}$$ $$=48\pm 3.25\times \frac{13}{\sqrt{10}}$$ $$=48\pm 13.361$$