Question

In a survey, 13 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $30 and sample standard deviation of $20. Construct a confidence interval at a 99% confidence level. Give your answers to one decimal place.

Answer #1

Solution :

Given that,

= $30

s = $20

n = 13

Degrees of freedom = df = n - 1 = 13 - 1 = 12

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005,12 = 3.055

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

= 3.055 * ( 20/ 13)

= 16.9

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

- E < < + E

30 - 16.9 < < 30 + 16.9

13.1 < < 46.9

(13.1,46.9)

**The 99% confidence interval 13.1 to 46.9**

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