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 : [tex]\mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7[/tex]
Standard deviation : [tex]\sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}[/tex]
[tex]=\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}}[/tex]
[tex]=\dfrac{30.265}{10}=3.0265[/tex]
The confidence interval for the population mean (for sample size <30) is given by :-
[tex]\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}[/tex]
Given significance level : [tex]\alpha=1-0.95=0.05[/tex]
Critical value : [tex]t_{n-1,\alpha/2}=t_{9,0.025}=2.262[/tex]
We assume that the population is normally distributed.
Now, the 95% confidence interval for the true mean will be :-
[tex]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)[/tex]
Hence, 95% confidence interval for the true mean= (2.54,6.86)
