Question

Random samples of of outh bass and smallmouth bass were taken from a lake, and their lengths in millimeters) were deter mined. We wish to know if the mean standard length differs between the two species in this lake. The results were as follows:
Largemouth Bass Smallmouth Bass
x 164.8 272.8
s 96.4 40.0
n 97 125

Answer 1

Answer:

Hence , the means of Largemouth Bass and Smallmouth Bass are significantly different.

Step-by-step expl;anation:

From the question we are told that:

Largemouth Bass:

$\=x_1 =164.8$

$s_1=96.4$

$n_1=125$

Smallmouth Bass:

$\=x_2 =272.8$

$s_2=40$

$n_2=97$

Assume

$\alpha =0.05$

Generally The hypothesis is given as

H_0: The Largemouth Bass and Smallmouth Bass are equal  

H_1: The Largemouth Bass and Smallmouth Bass are  not equal

Generally the equation for Test statistics is mathematically given by

$T=frac{( \=x_2 - \=x_1 )}{\sqrt{\frac{s^{1}}{n_1} + \frac{s_1^{2}}{n_2}}}$

$T =\frac{(272.8 - 164.8)}{\sqrt{\frac{96.4^{2}}{125} + \frac{40^{2}}{97}}}$

$T=\frac{108}{9.530925}$

$T=11.33$

Therefore

From table

Critical Value

$T_{\alpha,n_2-1}$

$T_{0.05,96}=1.661$

Conclude

Since 11.33 is greater that 1.661 we eject the null hypothesis that the means are the same.

Hence , the means of Largemouth Bass and Smallmouth Bass are significantly different.