Question

The following data is the results of a random sample of the ages (in years) of 12 houses in a community: 18, 22, 23, 19, 18, 24, 26, 20, 17, 12, 27, 28. Assume the population is normal. Construct a 90% confidence interval for the mean age of a house in this community. a) (18.8, 23.5) b) (18.7, 23.6) c) (18.9, 23.4) d) (19.0, 23.3) 7. If you want to estimate the mean age of a house in the community discussed in question 5 within 1.5 years with 98% confidence, how many houses should be included in the sample? Assume = 4.9 years

Answer 1

We have given data in increasing order will be,
12,17,18,18,19,20,22,23,24,26,27,28

Sample size:12

Mean (x̄): 21.1667

Standard deviation (s): 4.7065

t critical value with 90% confidence level and degree of freedom n-1 = 11 is 1.796

t critical value by using Excel command TINV(0.10,11)

We can use here,

=>(18.7,23.6)

Option b) is correct.

Question 2) Solution: E= margin of error = 1.5 years.

Population standard deviation =4.9 years

Z critical value for 98% confidence level is 2.33

We can use here sample size formula

  

=57.93

Therefore there will be 58 houses should be included in the sample