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 $33 and standard deviation of $15. Use the theory-based inference applet to find the confidence interval at a 99% confidence level. Give your answers to one decimal place.

• We are 99% confident that the average (mean) amount that people spent on their child's last birthday gift, μμ, was between  dollars and  dollars.
  
• The sample statistic is  dollars.
  
• The margin of error is  dollars.
• The confidence interval can be written as
  dollars ±±  dollars.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = $ 33 dollars

sample standard deviation = s = $ 15 dollars

sample size = n = 29

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

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = t0.005,28 = 2.763

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

= 2.763 * ( 15 / 29)

Margin of error = E = $ 7.7 dollars

The 99% confidence interval estimate of the population mean is,

  ± E  

= $ 33   ± $ 7.7

= ( $ 25.3, $ 40.7 )

We are 99% confident that the average (mean) amount that people spent on their child's last birthday gift, μ, was between 25.3 dollars and 40.7 dollars.