Question

In a survey, 29 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $43 and standard deviation of $20. Construct a confidence interval at a 95% confidence level.

Give your answers to one decimal place.

±

Answer 1

Solution :

Given that,

Point estimate = sample mean = = $43

sample standard deviation = s = $20

sample size = n = 29

Degrees of freedom = df = n - 1 =29-1=28

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,28= 2.048

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

= 2.048 * (20 / 29)

Margin of error = E = 7.6

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

- E < < + E

43 - 7.6 < < 43 + 7.6

35.4  < < 50.6

(35.4,50.6)