Question

We take a random sample of four children, and measure their heights when they are five years old and also when they are six years old. The following are our observations. Height at 5 years old (cm) Height at 6 years old (cm) old (cm) Child 1 98 104 104 Child 2 1 02 108 108 Child 3 100 104 Child 4 103 111 Let up be the true mean of the difference in height for each child (calculated as "Height at 6 years old" minus "Height at 5 years old"). Find the upper limit of a 95% confidence interval for up Answer:

Answer 1

The difference between the height of each child at the age of 6 years and 5 years is computed in the table below: 5-year old Difference 98 102 100 103 6-year old 104 108 104 111 The average and standard deviation of difference is computed below: 6+6+4+8 UD = 0 1 .1=1 Š(x– Hp) D=1 4-1 |(6-6)²+(6-6)* +(4 - 6)? +(8-6)2 4-1 = 1.633 Degrees of freedom for t-test statistic = n - 1 = 4 - 1 = 3

Critical value of t-test statistic at 3 degrees of freedom and 5% level of significance is 3.1824. The screenshot of the MS-Excel function used to compute critical value is attached below: fx =T.INV.2T(0.05,3) The 95% confidence interval is computed below: = 6+(3.1824x 1.623) = 6+2.5984 = (6-2.5984,6+2.5984) =(3.4016,8.5984) Thus, the required confidence interval is (3.40,8.60)