Question

A furniture mover calculates the actual weight as a proportion of estimated weight for a sample of 31 recent jobs. The sample mean is 1.13 and the sample standard deviation is 0.16.

a) Calculate a 95% confidence interval for the population mean using t tables.
b) Assume that the population standard deviation is known to be 0.16. Calculate a 95% confidence interval for the population mean using the z table.
c) Are the 95% confidence intervals for the population mean, using t tables and using the z table assuming that the population standard deviation is known to be 0.16, of roughly similar size? Explain.

Answer 1

solution Here n=31, x = 113 s=0:16 d-f=n-1=30, At 95%, staja 2.0023 The 95% confidence interval is a + tayo ) = 1134 2.ou23 035) =1137 0.0587 =(1.0713, 1.1887) b) Given o=0.16, At 95%, 20/2=1.96 the 95% c- I es T + Zoya Cosme) = 1.137196 039) =1:13+0.0563 =(1.0737, 1.1863) c) yes from both the intervals from part (a) (b) having similar size. because the sample size n is more than 3o. so this to and 2/2 is approximatly closer