In: Statistics and Probability
In a survey, 18 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 $5. Construct a confidence interval at a 95% confidence level.
Give your answers to one decimal place.
±±
Solution :
Given that,
= $33
s = $5
n = 18
Degrees of freedom = df = n - 1 = 18 - 1 = 17
At 95% confidence level the t is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
t /2,df = t0.025,17 = 2.110
Margin of error = E = t/2,df * (s /n)
= 2.110 * (5 / 18)
= 2.487
The 95% confidence interval estimate of the population mean is,
- E < < + E
33 - 2.487 < < 33 + 2.487
30.5 < < 35.5
(30.5,35.5)
The 95% confidence interval is 30.5 to 35.5