Question

In a survey, 24 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $39 and standard deviation of $3. Construct a confidence interval at a 99% confidence level.

1. Enter your answer as an open-interval (i.e., parentheses) accurate to one decimal place.

   Confidence interval =

2. 
   
3. Express the same answer as a tri-linear inequality one decimal place.

   < ¯xx¯ <

4. 
   
5. Express the same answer using the point estimate and margin of error. Give your answers to one decimal place.

   ¯xx¯ =  ±±

Answer 1

A) Given data

Sample size (n)=24

Mean (X')= 39

Standard deviation (s)=3

We need to determine the 99% confidence interval

Degrees of freedom=n-1=24-1=23

t value for 23 degrees of freedom and Singnificance level (0.01) is

t =2.80

Margin of error (E) =(t×s)/(√n)

=(2.80×3)/(√24)

=8.5/4.898 =1.714

a) Confidence interval is

CI =X' ± E

=39 ± 1.714

Lower limit=39-1.714 =37.286

Upper limit=39+1.714=40.714

b) Finally CI is

37.286<(xx)'<40.714

c) Point estimate is=39

d) margin of error (E)=1.714