Question

1) In a survey, 11 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $50 and standard deviation of $14. Construct a confidence interval at a 90% confidence level. Give your answers to one decimal place.

2) Out of 200 people sampled, 132 had kids. Based on this, construct a 95% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to three places

Answer 1

Solution :

1) Given that,

Point estimate = sample mean = = 50

Population standard deviation =    = 14

Sample size = n = 11

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

Margin of error = E = Z/2 * ( /n)

= 1.645 * ( 14 /  11 )

= 6.9

At 90% confidence interval estimate of the population mean is,

  ± E

50 ± 6.9   

( 43.1, 56.9 )

2) Given that,

n = 200

x = 132

Point estimate = sample proportion = = x / n = 132 / 200 = 0.66

1 - = 1 - 0.66 = 0.34

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.960}

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 (((0.66 * 0.34) / 200 )

= 0.066

A 95% confidence interval for population proportion p is ,

± E   

= 0.66 ± 0.066

= ( 0.594, 0.726 )