Question

1.)

In a survey, 49 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $31 and standard deviation of $5.

Construct a confidence interval at a 95% confidence level.

Give your answers to one decimal place.

±±

2.)

Assume that a sample is used to estimate a population mean μμ. Find the margin of error M.E. that corresponds to a sample of size 603 with a mean of 63.5 and a standard deviation of 9.4 at a confidence level of 99%.

Report M.E. accurate to one decimal place.

M.E. =

Answer 1

Solution:
1 ) Given:
n=49, Xbar=$31, σ=$5
(1-alpha)%= 95%
1-alpha= 0.95
alpha= 0.05
alpha/2=0.025
Z(alpha/2)=1.96 ..... From standard normal table.
Margin of error=Z(alpha/2) ×( σ /✓n)
= (1.96)×(5/✓49)
=1.4
95% confidence interval for the population mean is given as,
Xbar ± Margin of error=(31-1.4 ,31+1.4)
=(29.6,32.4)
95% confidence interval for the population mean is (29.6,32.4)
2)Solution:
Given:
n=603, Xbar=63.5, σ=9.4
(1-alpha)%= 99%
1-alpha= 0.99
alpha= 0.01
alpha/2=0.005
Z(alpha/2)=2.576 ..... From standard normal table.
Margin of error=Z(alpha/2) ×( σ /✓n)
= (2.576)×(9.4/✓603)
=0.9861
=1.0 ........ up to one decimal place.
Margin of error = 1.0