Question

The amount that a sample of eight customers spent for lunch ($) at a fast-food restaurant is shown below:

67, 52, 86, 52, 88, 32, 28, 43

a. Find the sample mean ???

b. Find the sample standard deviation S

c. Construct a 95% confidence interval estimate for the population mean ?

Answer 1

Solution:

Given that

a )The mean of sample is   $\bar x$

$\sum$x/n =  (67+ 52+ 86+ 52+ 88+ 32+ 28+ 43 / 8 )

= 448 / 8

= 56

The sample mean is 56

b ) Sample standard deviation is s

s = $\sqrt$1/(n-1)$\sum$(x - $\bar x$ )²

⁼  $\sqrt$1/7 (67 - 56 )²+ (52 - 56 )²+ (86 - 56 )²+ (52- 56 )² +( 88 - 56 )²+ (32 - 56 )² + (28 - 56 )²+ (43- 56 )²)

= $\sqrt$1/7 (121+16+900+16+1024+576+784+169)

= $\sqrt$3606/7

= $\sqrt$515.14

= 22.70

Sample standard deviation = 22.70

c ) $\bar x$ = 56

s = 22.70

n = 8

Degrees of freedom = df = n - 1 = 7

At 95% confidence level the t is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

t_{$\alpha$ /2,df} = t_0.025,7 = 2.365

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 2.365 * (22.70 / $\sqrt$8) = 18.9

The 95% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

56 - 18.98 < $\mu$ < 56 + 18.98

37.02 < $\mu$ < 74.98

(37.02, 74.98 )