Question

A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normally distributed. $3.60 $4.50 $2.80 $6.30 $2.60 $5.20 $6.75 $4.25 $8.00 $3.00 Find the 95% confidence interval for the true mean.

Answer 1

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : $\mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7$

Standard deviation : $\sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}$

$=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}$

$=\dfrac{30.265}{10}=3.0265$

The confidence interval for the population mean (for sample size <30) is given by :-

$\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given significance level : $\alpha=1-0.95=0.05$

Critical value : $t_{n-1,\alpha/2}=t_{9,0.025}=2.262$

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

$4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)$

Hence, 95% confidence interval for the true mean= (2.54,6.86)