Question

A survey of several 9 to 11 year olds recorded the following amounts spent on a trip to the mall:

$24.73,$24.85,$22.86,$23.81,$21.31,$17.40

Construct the 80% confidence interval for the average amount spent by 9 to 11 year olds on a trip to the mall. Assume the population is approximately normal.

Step 1 of 4:

Calculate the sample mean for the given sample data. Round your answer to two decimal places.

Step 2 of 4:

Calculate the sample standard deviation for the given sample data. Round your answer to two decimal places.

Critical value?

:

Construct the 80%confidence interval. Round your answer to two decimal places.

Answer 1

Sr.	X	$x_{i}-\bar{x}$	$(x_{i}-\bar{x})^{2}$
1	24.73	2.2367	5.0027
2	24.85	2.3567	5.5539
3	22.86	0.3667	0.1344
4	23.81	1.3167	1.7336
5	21.31	-1.1833	1.4003
6	17.4	-5.0933	25.9420
Total	134.96	0	39.7669

n = 6

Sample Mean = $\frac{\sum x_{i}}{n}$ = 134.96 / 6 = 22.493 = 22.49

Sample SD = $\sqrt{\frac{\sum (x_{i}-\bar{x})^{2}}{n-1}}$ = $\sqrt{\frac{39.7669}{5}}$ = 2.8202 = 2.82

Try avoiding to round off the intermediate calculations to get more accuracy

Since we don't have the known population standard deviation we are going to use t-dist

(1 - $\alpha$) confidence interval for population mean

$(\bar{x}\bar{+}t_{\alpha/2,n-1}\frac{S_{x}}{\sqrt{n}})$

Where critical value $t_{\alpha/2,n-1}=t_{0.20/2,5}$ ((1 - $\alpha$) = 80% therefore $\alpha$ = 0.20)

= $t_{0.1,5}$ (using t-dist tables for df = 5 and prob = 0.1 or excel function 'tinv(0.20,5)' )

= 1.4759

Substituting the values in the interval

$(22.4933\bar{+}1.4759\frac{2.8202}{\sqrt{6}})$

$\mathbf{(20.79,24.19)}$