Question

Strategic placement of lobster traps is one of the keys for a successful lobster fisherman. A study was conducted of the average distance separating traps long dash—called trap spacinglong dash—deployed by lobster fishermen. The​ trap-spacing measurements​ (in meters) for a sample of seven teams from fishing cooperative​ A, and a sample of eight teams from fishing cooperative B are repeated in the accompanying table. For this​ problem, we are interested in comparing the mean​ trap-spacing measurements of the two fishing cooperatives.​ (Since the sample from fishing cooperative A is listed​ first, treat it as the first​ sample.)

Find a 90% confidence interval for the target parameter.

The confidence interval is ( _ , _ )

(Round to one decimal as needed.)

Please try and show work. THANK YOU!

Table of Trap Spacing Measurements: The left side of the table is A cooperative (total # = 7) and the right side is B cooperative (total # = 8).

80	152
90	103
85	95
73	105
80	69
73	102
83	66
	116

Answer 1

Population parameter is the difference between mean trap spacing measurements of two fishing cooperatives that is $\mu 1-\mu 1$

90% confidence interval for $\mu 1-\mu 1$ is

(21 – 22) Ete** 1/n1+1/n2)

cl = 80.57, r2 = 101, s12 = 38.2857, s2 = 733.1429, n1 = 7, n2 = 8

s = V(nl - 1)s12 + (n 2 - 1)s22)/(n1 + n2-2)

=20.3086

degrees of freedom = n1+n2-2 = 7+8-2=13

For 90% confidence with df =13 , two tailed critical value of t is

(a = 0.10)

tc = 1.77 (from t table)

Thus,90% confidence interval for $\mu 1-\mu 1$ is

(80.57 – 101) + 1.77 * 20.3086 * (1/7 +1/8)

=(-20.435+18.60

=(- 39.0 , -1.8)

Calculation :

	x1	(x1-$\bar{x1}$)^2			x2	(x2-$\bar{x2}$)^2
	80	0.3249			152	2601
	90	88.9249			103	4
	85	19.6249			95	36
	73	57.3049			105	16
	80	0.3249			69	1024
	73	57.3049			102	1
	83	5.9049			66	1225
sum/ss	564	229.7143			116	225
$\bar{x1}$=sum/7	80.57143			sum/ss	808	5132
$s1^2$=ss/6		38.28571667		$\bar{x2}$=sum/8	101
				$s2^2$=ss/7		733.1428571