In a survey, 18 people were asked how much they spent on their child's last birthday...

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.

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Expert Solution

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 / $\sqrt$ n)

= 2.110 * (5 / $\sqrt$ 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

orchestra answered 1 year ago

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