Question

4. An experiment was performed to compare the blood flow pressure in the legs of a group of ten cyclists to the same in a group of ten runners. The data collected was assumed to be Normally distributed and had Yr = 14.5, S2 = 15.37 for runners and Y = 11.1, S2 = 15.84 for cyclists. (a) Find a 90% confidence interval for the difference in mean blood flow pressure between runners and cyclists fly – Mic. (b) Does the interval contain zero? Explain what the answer means for the confidence in the statement that the blood flow pressure is the same in the two groups.

Answer 1

Solution:- from the giver data - a) confidence interval for difference between two population means is given as below: > confidence interval - (11-22) + tx Jepa (w a) where Spe in pooled variance. Sp2 = ((n 1-1) x S 1 ² + (2-1) + S2²] (1 + n2-2) - Given that, XI= 14.5 x2 111 Si= 15.37 Su= 15.84 n =10 n2=10 df = n1+02-2 = 10+10 -2 = 18 Critical t value = 1.7341 { by using t-table}

Spr - ((61-1x S + (r2_1) x S) (ol+n2 - 2) Sp² (10-1) 15.37 +(10-1)× 15.84) (10+10-2) Sp2 - 15.6050 confidence interwas = (x - 324 tx confidence internal = (14.5- 11.1) + 1.7341 x 15.6050 confidence internal = 3.4 + 1.7341 x 1.7666 Confidence internal = 3.4 & 3.0635 lower limit - 3.4- 3.0635= 0.3365 upper limit & 3.4+ 3.0635= 6.4635 confidence interval - (0.3365, 6.4635). Above interwal does not contain zero. The blood flow pressue is not same in the tero groups as the confidence internas does not contain Tero.
Thank you.