Question

In a survey, 24 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $31 and standard deviation of $13. Construct a confidence interval at a 80% confidence level.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 31

sample standard deviation = s = 13

sample size = n = 24

Degrees of freedom = df = n - 1 = 23

At 80% confidence level the t is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

t _/2,df = t_0.10,23 = 1.319

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

= 1.319 * ( 13/ 24)

= 3.500

The 80% confidence interval estimate of the population mean is,

- E < < + E

31 - 3.500 < < 31 + 3.500

108.500 < < 34.500

(108.500 , 34.500)