Question

In an examination of holiday spending (known to be normally distributed) of a sample of 16 holiday shoppers at a local mall, an average of $54 was spent per hour of shopping. Based on the current sample, the standard deviation is equal to $21. Find a 90% confidence interval for the population mean level of spending per hour.

Answer 1

solution:

)Given that,

= 54

s =21

n = 16

Degrees of freedom = df = n - 1 =16 - 1 = 15

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t /2,df = t0.05,15 = 1.753    ( using student t table)

Margin of error = E = t/2,df * (s /n)

= 1.753 * (21 / 16) = 9.20

The 90% confidence interval the population mean is,

- E < < + E

54-9.2 < < 54+ 9.2

44.8 < < 63.2

(44.8 , 63.2)